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 Quadratic_Equations/113899: Hi I am having great difficulty trying to solve the following:- I have tried puting the equation for the first part of the problem into my TI-83 graphing calculator..but I cannot find a proper solution .. in fact I have tried all I can think of, this problem is causing me to fall behind with my work. I have seen many examples of how to solve the problem for a rocket ..the fact that this is a two stage rocket..is causing me to not be able to solve it I have spent ages trying to work out the way to program my TI-83 calculator to solve the following... I would appreciate guidance or direction to a site that can show me how to solve the following;- the motion of two stage rocket is stated by it height y in metres above the ground since t seconds since launch. Flies vertically Ist stage burns for 5 seconds and cuts out. 2nd stage rocket is under gravity alone. modelled by the equations Stage 1 height y = 10t2 time 0< t <5 stage 2 height y = - 5t2 + 150t - 375 time t >5 Question asks you plot function on ti 83, using an appropriate window for time(s) hence sketch a position time graph for each stage of flight adding labels etc.. also ask you to plot use the n deriv function to plot rate of change of function. If I can find the basic means of plotting graph I should be able to answer the question but cannot find anything in manual that is similar to solving a question of this type. I appreciate yout time and effort and thanks !!!!!!!!1 solutions Answer 82894 by jim_thompson5910(28598)   on 2007-12-02 12:50:18 (Show Source): You can put this solution on YOUR website!In order to graph pieces of a graph simply type y1=(10x^2)(0< x <5) and y2=(5x^2 + 150x - 375)(x>5) Notice how the interval is right next to the function. This is how you enter piecewise functions into a TI calculator
Graphs/113911: Find an equation for the line with y-intercept 3 that is perpendicular to the line: y=2/3x-4
1 solutions

Answer 82892 by jim_thompson5910(28598)   on 2007-12-02 12:44:16 (Show Source):
You can put this solution on YOUR website!
Remember if a line has the y-intercept 3, then it goes through the point (0,3)

 Solved by pluggable solver: Finding the Equation of a Line Parallel or Perpendicular to a Given Line Remember, any two perpendicular lines are negative reciprocals of each other. So if you're given the slope of , you can find the perpendicular slope by this formula: where is the perpendicular slope So plug in the given slope to find the perpendicular slope When you divide fractions, you multiply the first fraction (which is really ) by the reciprocal of the second Multiply the fractions. So the perpendicular slope is So now we know the slope of the unknown line is (its the negative reciprocal of from the line ). Also since the unknown line goes through (0,3), we can find the equation by plugging in this info into the point-slope formula Point-Slope Formula: where m is the slope and (,) is the given point Plug in , , and Distribute Multiply Add to both sides to isolate y Make into equivalent fractions with equal denominators Combine the fractions Reduce any fractions So the equation of the line that is perpendicular to and goes through (,) is So here are the graphs of the equations and graph of the given equation (red) and graph of the line (green) that is perpendicular to the given graph and goes through (,)

 Inequalities/113912: x>-21 solutions Answer 82891 by jim_thompson5910(28598)   on 2007-12-02 12:41:56 (Show Source): You can put this solution on YOUR website!Do you want to graph? Start with the given inequality: Set up a number line: Now plot the point on the number line Now pick any test point you want, I'm going to choose x=0, and test the inequality Plug in Since this inequality is true, we simply shade the entire portion in which contains the point x=0 using the point as the boundary.This means we shade everything to the right of the point like this: Graph of with the shaded region in blue note: at the point , there is an open circle. This means the point is excluded from the solution set.
 Equations/113909: 4x-40=81 solutions Answer 82888 by jim_thompson5910(28598)   on 2007-12-02 12:25:04 (Show Source): You can put this solution on YOUR website! Start with the given equation Add 40 to both sides Combine like terms on the right side Divide both sides by 4 to isolate x Divide -------------------------------------------------------------- Answer: So our answer is
 Equations/113908: 7x=841 solutions Answer 82887 by jim_thompson5910(28598)   on 2007-12-02 12:21:52 (Show Source): You can put this solution on YOUR website! Start with the given equation Divide both sides by 7 to isolate x Divide -------------------------------------------------------------- Answer: So our answer is
 Equations/113907: x+58=671 solutions Answer 82886 by jim_thompson5910(28598)   on 2007-12-02 12:17:17 (Show Source): You can put this solution on YOUR website! Start with the given equation Subtract 58 from both sides Combine like terms on the right side -------------------------------------------------------------- Answer: So our answer is
 Complex_Numbers/113855: Combine the complex numbers: (7-i) + (-3 + 4i).1 solutions Answer 82885 by jim_thompson5910(28598)   on 2007-12-02 12:11:25 (Show Source): You can put this solution on YOUR website! Start with the given expression Group like terms (i.e. pair the real components with each other and pair the imaginary components with each other) Combine like terms. So it is now in form where and
Complex_Numbers/113856: Factor: 12a^2 + 5a - 2.
1 solutions

Answer 82884 by jim_thompson5910(28598)   on 2007-12-02 12:10:22 (Show Source):
You can put this solution on YOUR website!
Solved by pluggable solver: Factoring using the AC method (Factor by Grouping)
In order to factor , first multiply the leading coefficient 12 and the last term -2 to get -24. Now we need to ask ourselves: What two numbers multiply to -24 and add to 5? Lets find out by listing all of the possible factors of -24

Factors:

1,2,3,4,6,8,12,24,

-1,-2,-3,-4,-6,-8,-12,-24, List the negative factors as well. This will allow us to find all possible combinations

These factors pair up to multiply to -24.

(-1)*(24)=-24

(-2)*(12)=-24

(-3)*(8)=-24

(-4)*(6)=-24

Now which of these pairs add to 5? Lets make a table of all of the pairs of factors we multiplied and see which two numbers add to 5

||||||||
 First Number | Second Number | Sum 1 | -24 | 1+(-24)=-23 2 | -12 | 2+(-12)=-10 3 | -8 | 3+(-8)=-5 4 | -6 | 4+(-6)=-2 -1 | 24 | (-1)+24=23 -2 | 12 | (-2)+12=10 -3 | 8 | (-3)+8=5 -4 | 6 | (-4)+6=2

We can see from the table that -3 and 8 add to 5. So the two numbers that multiply to -24 and add to 5 are: -3 and 8

breaks down to this (just replace with the two numbers that multiply to -24 and add to 5, which are: -3 and 8)

Replace with

Group the first two terms together and the last two terms together like this:

Factor a 3a out of the first group and factor a 2 out of the second group.

Now since we have a common term we can combine the two terms.

Combine like terms.
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Notice how foils back to our original problem . This verifies our answer.

Complex_Numbers/113857: Factor completely: x^2 - 8x + 16.
1 solutions

Answer 82883 by jim_thompson5910(28598)   on 2007-12-02 12:08:23 (Show Source):
You can put this solution on YOUR website!
Solved by pluggable solver: Factoring using the AC method (Factor by Grouping)
In order to factor , first multiply the leading coefficient 1 and the last term 16 to get 16. Now we need to ask ourselves: What two numbers multiply to 16 and add to -8? Lets find out by listing all of the possible factors of 16

Factors:

1,2,4,8,16,

-1,-2,-4,-8,-16, List the negative factors as well. This will allow us to find all possible combinations

These factors pair up to multiply to 16.

1*16=16

2*8=16

4*4=16

(-1)*(-16)=16

(-2)*(-8)=16

(-4)*(-4)=16

note: remember two negative numbers multiplied together make a positive number

Now which of these pairs add to -8? Lets make a table of all of the pairs of factors we multiplied and see which two numbers add to -8

||||||
 First Number | Second Number | Sum 1 | 16 | 1+16=17 2 | 8 | 2+8=10 4 | 4 | 4+4=8 -1 | -16 | -1+(-16)=-17 -2 | -8 | -2+(-8)=-10 -4 | -4 | -4+(-4)=-8

We can see from the table that -4 and -4 add to -8. So the two numbers that multiply to 16 and add to -8 are: -4 and -4

breaks down to this (just replace with the two numbers that multiply to 16 and add to -8, which are: -4 and -4)

Replace with

Group the first two terms together and the last two terms together like this:

Factor a 1x out of the first group and factor a -4 out of the second group.

Now since we have a common term we can combine the two terms.

Combine like terms.
==============================================================================

which can also be written as since the factors repeat themselves

Notice how foils back to our original problem . This verifies our answer.

Rational-functions/113863: My question pertains to solving linear equations using subsitution method.
Solve:
5x+2y = 0
x-3y = 0
Please show me detailed steps on doing a problem of this type - Thank you kindly!
1 solutions

Answer 82882 by jim_thompson5910(28598)   on 2007-12-02 12:07:20 (Show Source):
You can put this solution on YOUR website!
 Solved by pluggable solver: Solving a linear system of equations by subsitution Lets start with the given system of linear equations Now in order to solve this system by using substitution, we need to solve (or isolate) one variable. I'm going to choose y. Solve for y for the first equation Subtract from both sides Divide both sides by 2. Which breaks down and reduces to Now we've fully isolated y Since y equals we can substitute the expression into y of the 2nd equation. This will eliminate y so we can solve for x. Replace y with . Since this eliminates y, we can now solve for x. Distribute -3 to Multiply Reduce any fractions Add to both sides Combine the terms on the right side Make 1 into a fraction with a denominator of 2 Now combine the terms on the left side. Multiply both sides by . This will cancel out and isolate x So when we multiply and (and simplify) we get <---------------------------------One answer Now that we know that , lets substitute that in for x to solve for y Plug in into the 2nd equation Multiply Add to both sides Combine the terms on the right side Multiply both sides by . This will cancel out -3 on the left side. Multiply the terms on the right side Reduce So this is the other answer <---------------------------------Other answer So our solution is and which can also look like (,) Notice if we graph the equations (if you need help with graphing, check out this solver) we get graph of (red) and (green) (hint: you may have to solve for y to graph these) intersecting at the blue circle. and we can see that the two equations intersect at (,). This verifies our answer. ----------------------------------------------------------------------------------------------- Check: Plug in (,) into the system of equations Let and . Now plug those values into the equation Plug in and Multiply Add Reduce. Since this equation is true the solution works. So the solution (,) satisfies Let and . Now plug those values into the equation Plug in and Multiply Add Reduce. Since this equation is true the solution works. So the solution (,) satisfies Since the solution (,) satisfies the system of equations this verifies our answer.

 Circles/113898: Give the equation for the circle with center C(3,-2) and radius 4.1 solutions Answer 82881 by jim_thompson5910(28598)   on 2007-12-02 12:05:10 (Show Source): You can put this solution on YOUR website!The general equation of a circle is where (h,k) is the center and r is the radius. So using this equation, we get Plug in h=3,k=-2, and r=4. These are given Rewrite as Square 4 to get 16 So the equation for the circle with center C(3,-2) and radius 4 is
 Coordinate-system/113889: Write the equation of the line that passes through point (–7, –15) with a slope of 3. 1 solutions Answer 82880 by jim_thompson5910(28598)   on 2007-12-02 12:01:35 (Show Source): You can put this solution on YOUR website! If you want to find the equation of line with a given a slope of which goes through the point (,), you can simply use the point-slope formula to find the equation: ---Point-Slope Formula--- where is the slope, and is the given point So lets use the Point-Slope Formula to find the equation of the line Plug in , , and (these values are given) Rewrite as Rewrite as Distribute Multiply and to get Subtract 15 from both sides to isolate y Combine like terms and to get ------------------------------------------------------------------------------------------------------------ Answer: So the equation of the line with a slope of which goes through the point (,) is: which is now in form where the slope is and the y-intercept is Notice if we graph the equation and plot the point (,), we get (note: if you need help with graphing, check out this solver) Graph of through the point (,) and we can see that the point lies on the line. Since we know the equation has a slope of and goes through the point (,), this verifies our answer.
 Coordinate-system/113888: Write the equation of the line with slope 4 and y-intercept (0, –5). Then graph the line.1 solutions Answer 82879 by jim_thompson5910(28598)   on 2007-12-02 12:00:34 (Show Source): You can put this solution on YOUR website!If we have the y-intercept (0, –5), then the line goes through the point (0, –5) If you want to find the equation of line with a given a slope of which goes through the point (,), you can simply use the point-slope formula to find the equation: ---Point-Slope Formula--- where is the slope, and is the given point So lets use the Point-Slope Formula to find the equation of the line Plug in , , and (these values are given) Rewrite as Distribute Multiply and to get Subtract 5 from both sides to isolate y Combine like terms and to get ------------------------------------------------------------------------------------------------------------ Answer: So the equation of the line with a slope of which goes through the point (,) is: which is now in form where the slope is and the y-intercept is Notice if we graph the equation and plot the point (,), we get (note: if you need help with graphing, check out this solver) Graph of through the point (,) and we can see that the point lies on the line. Since we know the equation has a slope of and goes through the point (,), this verifies our answer.
 Coordinate-system/113890: Write the equation of the line passing through (6, –5) and (–3, 4)1 solutions Answer 82878 by jim_thompson5910(28598)   on 2007-12-02 11:59:08 (Show Source): You can put this solution on YOUR website!First lets find the slope through the points (,) and (,) Start with the slope formula (note: is the first point (,) and is the second point (,)) Plug in ,,, (these are the coordinates of given points) Subtract the terms in the numerator to get . Subtract the terms in the denominator to get Reduce So the slope is ------------------------------------------------ Now let's use the point-slope formula to find the equation of the line: ------Point-Slope Formula------ where is the slope, and is one of the given points So lets use the Point-Slope Formula to find the equation of the line Plug in , , and (these values are given) Rewrite as Distribute Multiply and to get Subtract from both sides to isolate y Combine like terms and to get ------------------------------------------------------------------------------------------------------------ Answer: So the equation of the line which goes through the points (,) and (,) is: The equation is now in form (which is slope-intercept form) where the slope is and the y-intercept is Notice if we graph the equation and plot the points (,) and (,), we get this: (note: if you need help with graphing, check out this solver) Graph of through the points (,) and (,) Notice how the two points lie on the line. This graphically verifies our answer.
 Complex_Numbers/113906: Can I take average of a complex number and a real number? i.e. [(a+bi)+c]/2=?1 solutions Answer 82877 by jim_thompson5910(28598)   on 2007-12-02 11:57:06 (Show Source): You can put this solution on YOUR website!If you had something like then you will get a complex number as a result. So you'll get where since you can only add the real parts.
 Human-and-algebraic-language/113900: 1. solve these equations y=10-x 2x-y=-41 solutions Answer 82876 by jim_thompson5910(28598)   on 2007-12-02 11:54:49 (Show Source): You can put this solution on YOUR website!Start with the given system Plug in into the first equation. In other words, replace each with . Notice we've eliminated the variables. So we now have a simple equation with one unknown. Distribute the negative Combine like terms on the left side Add 10 to both sides Combine like terms on the right side Divide both sides by 3 to isolate x Divide Now that we know that , we can plug this into to find Substitute for each Simplify So our answer is and
Linear-equations/113894: x+y=4
-x+y=2
1 solutions

Answer 82875 by jim_thompson5910(28598)   on 2007-12-02 11:49:52 (Show Source):
You can put this solution on YOUR website!
 Solved by pluggable solver: Solving a System of Linear Equations by Elimination/Addition Lets start with the given system of linear equations In order to solve for one variable, we must eliminate the other variable. So if we wanted to solve for y, we would have to eliminate x (or vice versa). So lets eliminate x. In order to do that, we need to have both x coefficients that are equal but have opposite signs (for instance 2 and -2 are equal but have opposite signs). This way they will add to zero. So to make the x coefficients equal but opposite, we need to multiply both x coefficients by some number to get them to an equal number. So if we wanted to get 1 and -1 to some equal number, we could try to get them to the LCM. Since the LCM of 1 and -1 is -1, we need to multiply both sides of the top equation by -1 and multiply both sides of the bottom equation by -1 like this: Multiply the top equation (both sides) by -1 Multiply the bottom equation (both sides) by -1 So after multiplying we get this: Notice how -1 and 1 add to zero (ie ) Now add the equations together. In order to add 2 equations, group like terms and combine them Notice the x coefficients add to zero and cancel out. This means we've eliminated x altogether. So after adding and canceling out the x terms we're left with: Divide both sides by to solve for y Reduce Now plug this answer into the top equation to solve for x Plug in Multiply Subtract from both sides Combine the terms on the right side Multiply both sides by . This will cancel out on the left side. Multiply the terms on the right side So our answer is , which also looks like (, ) Notice if we graph the equations (if you need help with graphing, check out this solver) we get graph of (red) (green) (hint: you may have to solve for y to graph these) and the intersection of the lines (blue circle). and we can see that the two equations intersect at (,). This verifies our answer.

Human-and-algebraic-language/113901: 5. solve the following systems of equations
3x+5y=25
x+2y=10
1 solutions

Answer 82874 by jim_thompson5910(28598)   on 2007-12-02 11:48:29 (Show Source):
You can put this solution on YOUR website!
 Solved by pluggable solver: Solving a linear system of equations by subsitution Lets start with the given system of linear equations Now in order to solve this system by using substitution, we need to solve (or isolate) one variable. I'm going to choose y. Solve for y for the first equation Subtract from both sides Divide both sides by 5. Which breaks down and reduces to Now we've fully isolated y Since y equals we can substitute the expression into y of the 2nd equation. This will eliminate y so we can solve for x. Replace y with . Since this eliminates y, we can now solve for x. Distribute 2 to Multiply Reduce any fractions Subtract from both sides Combine the terms on the right side Make 1 into a fraction with a denominator of 5 Now combine the terms on the left side. Multiply both sides by . This will cancel out and isolate x So when we multiply and (and simplify) we get <---------------------------------One answer Now that we know that , lets substitute that in for x to solve for y Plug in into the 2nd equation Multiply Add to both sides Combine the terms on the right side Multiply both sides by . This will cancel out 2 on the left side. Multiply the terms on the right side Reduce So this is the other answer <---------------------------------Other answer So our solution is and which can also look like (,) Notice if we graph the equations (if you need help with graphing, check out this solver) we get graph of (red) and (green) (hint: you may have to solve for y to graph these) intersecting at the blue circle. and we can see that the two equations intersect at (,). This verifies our answer. ----------------------------------------------------------------------------------------------- Check: Plug in (,) into the system of equations Let and . Now plug those values into the equation Plug in and Multiply Add Reduce. Since this equation is true the solution works. So the solution (,) satisfies Let and . Now plug those values into the equation Plug in and Multiply Add Reduce. Since this equation is true the solution works. So the solution (,) satisfies Since the solution (,) satisfies the system of equations this verifies our answer.

 Human-and-algebraic-language/113902: 5. solve the following systems of equations x=3y+7 x+y=-51 solutions Answer 82873 by jim_thompson5910(28598)   on 2007-12-02 11:47:09 (Show Source): You can put this solution on YOUR website!Start with the given system Plug in into the first equation. In other words, replace each with . Notice we've eliminated the variables. So we now have a simple equation with one unknown. Combine like terms on the left side Subtract 7 from both sides Combine like terms on the right side Divide both sides by 4 to isolate y Divide Now that we know that , we can plug this into to find Substitute for each Simplify So our answer is and
 Human-and-algebraic-language/113903: 8. solve the following systems of equations y=3x-5 6x-3y=31 solutions Answer 82872 by jim_thompson5910(28598)   on 2007-12-02 11:45:18 (Show Source): You can put this solution on YOUR website! Start with the given system Plug in into the first equation. In other words, replace each with . Notice we've eliminated the variables. So we now have a simple equation with one unknown. Distribute Combine like terms on the left side Subtract 15 from both sides Combine like terms on the right side Divide both sides by -3 to isolate x Divide Now that we know that , we can plug this into to find Substitute for each Simplify So our answer is and
Linear-equations/113838: I am having trouble with linear equations by elimination.
x + y = 9
x + 4y = 15
I need to find x and y
Thank you very much
1 solutions

Answer 82858 by jim_thompson5910(28598)   on 2007-12-02 00:16:12 (Show Source):
You can put this solution on YOUR website!
 Solved by pluggable solver: Solving a System of Linear Equations by Elimination/Addition Lets start with the given system of linear equations In order to solve for one variable, we must eliminate the other variable. So if we wanted to solve for y, we would have to eliminate x (or vice versa). So lets eliminate x. In order to do that, we need to have both x coefficients that are equal but have opposite signs (for instance 2 and -2 are equal but have opposite signs). This way they will add to zero. So to make the x coefficients equal but opposite, we need to multiply both x coefficients by some number to get them to an equal number. So if we wanted to get 1 and 1 to some equal number, we could try to get them to the LCM. Since the LCM of 1 and 1 is 1, we need to multiply both sides of the top equation by 1 and multiply both sides of the bottom equation by -1 like this: Multiply the top equation (both sides) by 1 Multiply the bottom equation (both sides) by -1 So after multiplying we get this: Notice how 1 and -1 add to zero (ie ) Now add the equations together. In order to add 2 equations, group like terms and combine them Notice the x coefficients add to zero and cancel out. This means we've eliminated x altogether. So after adding and canceling out the x terms we're left with: Divide both sides by to solve for y Reduce Now plug this answer into the top equation to solve for x Plug in Multiply Subtract from both sides Combine the terms on the right side Multiply both sides by . This will cancel out on the left side. Multiply the terms on the right side So our answer is , which also looks like (, ) Notice if we graph the equations (if you need help with graphing, check out this solver) we get graph of (red) (green) (hint: you may have to solve for y to graph these) and the intersection of the lines (blue circle). and we can see that the two equations intersect at (,). This verifies our answer.

 Coordinate-system/113842: Determine which two equations represent parallel lines. (a) y = 2x – 6 (b) y = –2x + 6 (c) y = 2x + 1 (d) y = -1/2 x – 6 1 solutions Answer 82855 by jim_thompson5910(28598)   on 2007-12-01 22:30:49 (Show Source): You can put this solution on YOUR website!Since A) and C) have equal slopes of 2 (the slope is the number in front of the x), these two lines are parallel.
Coordinate-system/113839: graph 3x + y = 3
1 solutions

Answer 82853 by jim_thompson5910(28598)   on 2007-12-01 22:29:45 (Show Source):
You can put this solution on YOUR website!
 Solved by pluggable solver: Graphing Linear Equations Start with the given equation Subtract from both sides Multiply both sides by Distribute Multiply Rearrange the terms Reduce any fractions So the equation is now in slope-intercept form () where (the slope) and (the y-intercept) So to graph this equation lets plug in some points Plug in x=-2 Multiply Add So here's one point (-2,9) Now lets find another point Plug in x=-1 Multiply Add So here's another point (-1,6). Add this to our graph Now draw a line through these points So this is the graph of through the points (-2,9) and (-1,6) So from the graph we can see that the slope is (which tells us that in order to go from point to point we have to start at one point and go down -3 units and to the right 1 units to get to the next point), the y-intercept is (0,)and the x-intercept is (,0) . So all of this information verifies our graph. We could graph this equation another way. Since this tells us that the y-intercept (the point where the graph intersects with the y-axis) is (0,). So we have one point (0,) Now since the slope is , this means that in order to go from point to point we can use the slope to do so. So starting at (0,), we can go down 3 units and to the right 1 units to get to our next point Now draw a line through those points to graph So this is the graph of through the points (0,3) and (1,0)

Coordinate-system/113841: Find the slope of the line passing through the points (5, 3) and (6, 7)
1 solutions

Answer 82852 by jim_thompson5910(28598)   on 2007-12-01 22:28:55 (Show Source):
You can put this solution on YOUR website!
 Solved by pluggable solver: Finding the slope Slope of the line through the points (5, 3) and (6, 7) Answer: Slope is

 Coordinate-system/113840: Graph using the intercept method: 2x – 5y = 101 solutions Answer 82851 by jim_thompson5910(28598)   on 2007-12-01 22:28:16 (Show Source): You can put this solution on YOUR website! Start with the given equation Let's find the x-intercept To find the x-intercept, let y=0 and solve for x: Plug in Simplify Divide both sides by 2 Reduce So the x-intercept is (note: the x-intercept will always have a y-coordinate equal to zero) ------------------ Start with the given equation Now let's find the y-intercept To find the y-intercept, let x=0 and solve for y: Plug in Simplify Divide both sides by -5 Reduce So the y-intercept is (note: the y-intercept will always have a x-coordinate equal to zero) ------------------------------------------ So we have these intercepts: x-intercept: y-intercept: Now plot the two points and Now draw a line through the two points to graph graph of through the points and
 Polynomials-and-rational-expressions/113829: 10. Solve the following system both by graphing and then check result algebraically. y = 8 2y = x^2 1 solutions Answer 82837 by jim_thompson5910(28598)   on 2007-12-01 21:45:24 (Show Source): You can put this solution on YOUR website!First let's graph In order to graph , simply draw a horizontal line through the y=8 Graph of Now let's move onto the second equation Start with the given equation Divide both sides by 2 to solve for y Now let's graph In order to do so, let's make a table by plugging in x-values (you get to choose which ones) to find the y-values  x y -5.00000 12.50000 -4.00000 8.00000 -3.00000 4.50000 -2.00000 2.00000 -1.00000 0.50000 0.00000 0.00000 1.00000 0.50000 2.00000 2.00000 3.00000 4.50000 4.00000 8.00000 5.00000 12.50000  Now let's plot these points and connect them to get this graph Graph of Now let's plot the two graphs together: Graph of (red) (green) From the graph, we can see that the two lines intersect at x=-4 and x=4 which means the solutions are (-4,8) or (4,8) ===================================== Now let's solve algebraically: Start with the given system Take the 2nd equation and plug in Multiply Take the square root of both sides So we then get or So this means our solutions are (-4,8) or (4,8) (remember y is always 8 since the first equation is )
 Polynomials-and-rational-expressions/113827: 9. Evaluate the following: a. log1/2 4 b. 1og8 64 c. log32 8 d. 1og3 81 e. pi^(log_pi 2pi)1 solutions Answer 82836 by jim_thompson5910(28598)   on 2007-12-01 21:43:52 (Show Source): You can put this solution on YOUR website!Remember if we have , we can write it as a) Start with the given equation. Just set the original expression equal to y. Note: the entire is the base of the log. Rewrite the original expression using the property I listed above Rewrite as Multiply the exponents Rewrite as Since the bases are equal, the exponents are equal. So which means So ------------------------------------- b) Start with the given equation. Just set the original expression equal to y Rewrite the original expression using the property: <===> Rewrite as Since the bases are equal, the exponents are equal. So So --------------------------------------- c) Start with the given equation. Just set the original expression equal to y Rewrite the original expression using the property: <===> Rewrite as and as Multiply the exponents Since the bases are equal, the exponents are equal. So which means So --------------------------------------- d) Start with the given equation. Just set the original expression equal to y Rewrite the original expression using the property: <===> Rewrite as Since the bases are equal, the exponents are equal. So So --------------------------------------- e) Start with the given expression Break up the logarithm using the identity . Think of as 2 times . Break up the exponent using the identity Evaluate to get 1 Since we cannot simplify the expression any further, we cannot simplify the entire expression any further.
 Polynomials-and-rational-expressions/113826: 8. Solve and Check the Following Equations: C. log6 x + log6 (x – 2) = log6 151 solutions Answer 82835 by jim_thompson5910(28598)   on 2007-12-01 21:41:30 (Show Source): You can put this solution on YOUR website! Start with the given equation Combine the logs using the identity Distribute Since the base of the logs are equal, the arguments (the stuff inside the logs are equal). So So let's solve Start with the given equation Subtract 15 from both sides Factor the left side (note: if you need help with factoring, check out this solver) Now set each factor equal to zero: or or Now solve for x in each case So our possible solutions are or However, since you cannot take the log of a negative number, the only solution is ------------------------------------- Check: Let's check the solution Start with the equation found on the third step. Plug in Square and multiply Subtract. Since both sides of the equation are equal, this solution is verified. So the solution is verified
 Polynomials-and-rational-expressions/113823: 8. Solve and Check the Following Equations: A. 2^(x^2+x) = 64 1 solutions Answer 82833 by jim_thompson5910(28598)   on 2007-12-01 21:35:55 (Show Source): You can put this solution on YOUR website! Start with the given equation Rewrite as Since the bases are equal, the exponents are equal. So Now let's solve for x Subtract x from both sides Factor the left side (note: if you need help with factoring, check out this solver) Now set each factor equal to zero: or or Now solve for x in each case So our solutions are or ------------------------ Check: Let's check the solution Start with the given equation Plug in Square -3 to get 9 Add Raise 2 to the sixth power to get 64. Since both sides of the equation are equal, this solution is verified. ---- Let's check the solution Start with the given equation Plug in Square 2 to get 4 Add Raise 2 to the sixth power to get 64. Since both sides of the equation are equal, this solution is verified. So this verifies our solutions and
 Polynomials-and-rational-expressions/113822: 7. Graph and it's inverse on the same set of coordinate axes. Label all units, intercepts, and asymptotes. Determine the domain and range of each function.1 solutions Answer 82831 by jim_thompson5910(28598)   on 2007-12-01 21:32:19 (Show Source): You can put this solution on YOUR website! Graph of (red) and it's inverse (green) So we can see from the graph of we can see... f: Domain: all real numbers, Range=, Asymptotes: no vertical, horizontal at Intercepts: x-int=none, y-int=(0,1) And we can see from the graph of we can see... : Domain: , Range=all real numbers, Asymptotes: vertical at , no horizontal, Intercepts: x-int=(1,0) y-int=none Notice how everything switched from the graphs to
Polynomials-and-rational-expressions/113821: 6. Determine the behavior of

Domain :
x – intercept(s) :
Hole(s) :
y – intercept :
Vertical Asymptote(s) :
Oblique Asymptote :

1 solutions

Answer 82830 by jim_thompson5910(28598)   on 2007-12-01 21:29:13 (Show Source):
You can put this solution on YOUR website!
Domain:

Set the denominator equal to zero. Remember, dividing by 0 is undefined. So if we find values of x that make the
denominator zero, then we must exclude them from the domain.

Combine like terms on the right side

Since makes the denominator equal to zero, this means we must exclude from our domain

So our domain is:

which in plain English reads: x is the set of all real numbers except

So our domain looks like this in interval notation

x-intercept(s):

To find the x-intercepts, set f(x) (which is y) equal to zero

This means the numerator is equal to zero. Remember the denominator cannot equal zero

Factor the left side (note: if you need help with factoring, check out this
solver)

Now set each factor equal to zero:

or

or Now solve for x in each case

So our solutions are or
This means the x-intercepts are (-3,0) and (2,0)

Hole(s):

Factor

Since the expression does not simplify further, there are no exceptions to make about the two expressions. So in this case there are no holes in this function

y-intercept(s):

To find the y-intercept, plug in x=0

Simplify

Divide

So the y intercept is (0,2)

Vertical Asymptote(s):

Since the value x=3 is excluded from the domain, and there are no holes, there is one vertical asymptote at x=3

So the vertical asymptote is

Oblique Asymptote:

To find the oblique asymptote, divide using synthetic division

First lets find our test zero:

Set the denominator equal to zero

Solve for x.

so our test zero is 3

Now set up the synthetic division table by placing the test zero in the upper left corner and placing the coefficients of the numerator to the right of the test zero.
 3 | 1 1 -6 |

Start by bringing down the leading coefficient (it is the coefficient with the highest exponent which is 1)
 3 | 1 1 -6 | 1

Multiply 3 by 1 and place the product (which is 3) right underneath the second coefficient (which is 1)
 3 | 1 1 -6 | 3 1

Add 3 and 1 to get 4. Place the sum right underneath 3.
 3 | 1 1 -6 | 3 1 4

Multiply 3 by 4 and place the product (which is 12) right underneath the third coefficient (which is -6)
 3 | 1 1 -6 | 3 12 1 4

Add 12 and -6 to get 6. Place the sum right underneath 12.
 3 | 1 1 -6 | 3 12 1 4 6

Since the last column adds to 6, we have a remainder of 6. This means is not a factor of
Now lets look at the bottom row of coefficients:

The first 2 coefficients (1,4) form the quotient

So the oblique asymptote is the line

Notice if we graph and it's asymptotes, we can visually verify our answers:

Graph of with the oblique asymptote (green) and the vertical asymptote (blue)