You can put this solution on YOUR website!
... Start with the given inequality.


... Distribute.


Distribute again.


Combine like terms on the left side.


Add to both sides.


Add to both sides.


Combine like terms on the left side.


Combine like terms on the right side.


Divide both sides by to isolate .


Reduce.


----------------------------------------------------------------------

Answer:

So the answer is

which in set-builder notation is

You can put this solution on YOUR website!
Start with the first equation.


Multiply EVERY term by 10 to make every number a whole number.

--------------------------

Move onto the second equation.


Multiply EVERY term by the LCD 23 to clear out the fraction.


Multiply and simplify


Multiply EVERY term by 10 to make every number a whole number.


So we have the system of equations:




Multiply the both sides of the first equation by 69.


Distribute and multiply.


Multiply the both sides of the second equation by 2.


Distribute and multiply.


So we have the new system of equations:




Now add the equations together. You can do this by simply adding the two left sides and the two right sides separately like this:





Group like terms.


Combine like terms.


Simplify.


Divide both sides by to isolate .


Reduce.


------------------------------------------------------------------


Now go back to the first equation.


Plug in .


Multiply.


Subtract from both sides.


Combine like terms on the right side.


Divide both sides by to isolate .


Reduce.

=========================================================================

Answer:


So the solutions are and


which form the ordered pair

You can put this solution on YOUR website!

Looking at the expression , we can see that the first coefficient is , the second coefficient is , and the last term is .


Now multiply the first coefficient by the last term to get .


Now the question is: what two whole numbers multiply to (the previous product) and add to the second coefficient ?


To find these two numbers, we need to list all of the factors of (the previous product).


Factors of :
1,2,4,8
-1,-2,-4,-8


Note: list the negative of each factor. This will allow us to find all possible combinations.


These factors pair up and multiply to .
1*(-8)
2*(-4)
(-1)*(8)
(-2)*(4)

Now let's add up each pair of factors to see if one pair adds to the middle coefficient :

First Number	Second Number	Sum
1	-8	1+(-8)=-7
2	-4	2+(-4)=-2
-1	8	-1+8=7
-2	4	-2+4=2

From the table, we can see that the two numbers and add to (the middle coefficient).


So the two numbers and both multiply to and add to


Now replace the middle term with . Remember, and add to . So this shows us that .


Replace the second term with .


Group the terms into two pairs.


Factor out the GCF from the first group.


Factor out from the second group. The goal of this step is to make the terms in the second parenthesis equal to the terms in the first parenthesis.


Combine like terms. Or factor out the common term

---------------------------------------------


Answer:


So factors to .


Note: you can check the answer by FOILing to get or by graphing the original expression and the answer (the two graphs should be identical).

You can put this solution on YOUR website!
It turns out that anything raised to the one half power is equivalent to taking the square root.

In other words


So...




So the solution is

You can put this solution on YOUR website!

Looking at we can see that the equation is in slope-intercept form where the slope is and the y-intercept is note: really looks like


Since this tells us that the y-intercept is .Remember the y-intercept is the point where the graph intersects with the y-axis

So we have one point




Now since the slope is comprised of the "rise" over the "run" this means


Also, because the slope is , this means:




which shows us that the rise is 1 and the run is 5. This means that to go from point to point, we can go up 1 and over 5



So starting at , go up 1 unit


and to the right 5 units to get to the next point



Now draw a line through these points to graph

So this is the graph of through the points and

You can put this solution on YOUR website!
To find the tangent line, we need the slope of the tangent line. To find that, we first need the first derivative of "y":


... Start with the given equation.


... Derive both sides with respect to "x"


... Derive the left and right sides. Note: remember, y is a function of "x", so use the chain rule.


... Subtract 2x from both sides.


... Divide both sides by 2y.


... Reduce


So the slope of any tangent line at the point (x,y) (on the circle) is



Now just plug in the values and to find the tangent slope at (-10,-12):





... Reduce



So the slope of the tangent line is


Now let's find the equation of the line that has a slope of and goes through (-10, -12):

If you want to find the equation of line with a given a slope of which goes through the point (-10,-12), 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 12 from both sides to isolate y


Combine like terms and to get


------------------------------------------------------------------------------------------------------------
Answer:


So the equation of the tangent line is

You can put this solution on YOUR website!
Here's some terminology:

Vertex: A point representing a location (city, town, etc...). In general, what the vertex represents isn't important.

Edge: A line connecting two vertices (plural for vertex)

Odd Vertex: A vertex with an odd number of edges connecting to it

Even Vertex: A vertex with an even number of edges connecting to it






So here's one way to draw the graph:




Take note that...

Vertex A: even vertex with 2 edges connecting to it

Vertex B: odd vertex with 3 edges connecting to it

Vertex C: even vertex with 2 edges connecting to it

Vertex D: odd vertex with 3 edges connecting to it

You can put this solution on YOUR website!
First, we need to find the length of diagonal:



If we cut the garden in half along the diagonal, we'll have this triangle set up (where "x" is the length of the diagonal):





Remember, the Pythagorean Theorem is where "a" and "b" are the legs of a triangle and "c" is the hypotenuse.


Since the legs are and this means that and


Also, since the hypotenuse is , this means that .


Start with the Pythagorean theorem.


Plug in , ,


Square to get .


Square to get .


Combine like terms.


Rearrange the equation.


Take the square root of both sides. Note: only the positive square root is considered (since a negative length doesn't make sense).


Take the square root of 900 to get 30.



So the diagonal is 30 meters long.



Now just multiply the length of the diagonal by the cost per linear meter to get:




So it will cost $360 to build the diagonal walkway.

You can put this solution on YOUR website!
"Two angles are complimentary." translates to . Solve for "y" to get



"The sum of the measure of the first angle and helf the second angle is 68 degrees." translates to


Start with the second equation.


Plug in


Distribute


Subtract 45 from both sides.


Combine like terms.


Multiply both sides by 2.


Multiply


------------------------


Go back to the first equation


Plug in


Subtract


So the solutions are and which means that the first and second angles are 46 and 44 degrees.

You can put this solution on YOUR website!
Start with the given equation.


Square both sides


Square the square roots to eliminate them


Add to both sides.


Subtract from both sides.


Combine like terms on the left side.


Combine like terms on the right side.


Divide both sides by to isolate .


----------------------------------------------------------------------

Answer:

So the answer is which approximates to .

You can put this solution on YOUR website!

Start with the given equation.


Subtract 5 from both sides.


Notice we have a quadratic equation in the form of where , , and


Let's use the quadratic formula to solve for z


Start with the quadratic formula


Plug in , , and


Square to get .


Multiply to get


Rewrite as


Add to to get


Multiply and to get .


or Break up the expression.


So the answers are or


which approximate to or (using a calculator)

You can put this solution on YOUR website!

First, let's draw a picture:




We can see that a triangle forms where the legs are "x" and 32 ft with a hypotenuse of 80 ft. To find the unknown length of the leg "x", we need to use the Pythagorean Theorem


Remember, the Pythagorean Theorem is where "a" and "b" are the legs of a triangle and "c" is the hypotenuse.


Since the legs are and this means that and


Also, since the hypotenuse is , this means that .


Start with the Pythagorean theorem.


Plug in , ,


Square to get .


Square to get .


Subtract from both sides.


Combine like terms.


Take the square root of both sides. Note: only the positive square root is considered (since a negative length doesn't make sense).


Evaluate the square root (using a calculator). Note: this value is approximate


Now round to the nearest whole number to get


So the kite is about 73 ft above Mike's head.

You can put this solution on YOUR website!
Since the square root is really a "second root", this means that . Since the square root is so commonly used, we've dropped the 2 (as we know that we're dealing with a square root when no number is present)


So


Now, recall that


In this case, , , and . So...





We can simplify further to get


So the answer is

You can put this solution on YOUR website!

Start with the given equation.


Subtract 1 from both sides.


Notice we have a quadratic equation in the form of where , , and


Let's use the quadratic formula to solve for x


Start with the quadratic formula


Plug in , , and


Negate to get .


Square to get .


Multiply to get


Rewrite as


Add to to get


Multiply and to get .


Simplify the square root


or Break up the expression.


or Reduce


So the answers are or


which approximate to or

You can put this solution on YOUR website!
Start with the given expression.


Factor 432 into . Note: 216 is a perfect cube


Factor into .


Break up the root.


Take the cube root of 216 to get 6


Take the cube root of to get


Rearrange the terms and multiply


So

You can put this solution on YOUR website!
Start with the given equation.


Take the log base 10 of both sides


Pull down the exponent


Divide both sides by .


Use the change of base formula to rewrite the right side


Subtract 1 from both sides.


Multiply EVERY term by -1 to isolate "x"


So the solution is which approximates to

You can put this solution on YOUR website!

Start with the given expression


Factor out the GCF


Now let's focus on the inner expression




------------------------------------------------------------



Looking at we can see that the first term is and the last term is where the coefficients are 5 and -11 respectively.

Now multiply the first coefficient 5 and the last coefficient -11 to get -55. Now what two numbers multiply to -55 and add to the middle coefficient -6? Let's list all of the factors of -55:



Factors of -55:
1,5,11,55

-1,-5,-11,-55 ...List the negative factors as well. This will allow us to find all possible combinations

These factors pair up and multiply to -55
(1)*(-55)
(5)*(-11)
(-1)*(55)
(-5)*(11)

note: remember, the product of a negative and a positive number is a negative number


Now which of these pairs add to -6? Lets make a table of all of the pairs of factors we multiplied and see which two numbers add to -6

First Number	Second Number	Sum
1	-55	1+(-55)=-54
5	-11	5+(-11)=-6
-1	55	-1+55=54
-5	11	-5+11=6

From this list we can see that 5 and -11 add up to -6 and multiply to -55


Now looking at the expression , replace with (notice adds up to . So it is equivalent to )




Now let's factor by grouping:


Group like terms


Factor out the GCF of out of the first group. Factor out the GCF of out of the second group


Since we have a common term of , we can combine like terms

So factors to


So this also means that factors to (since is equivalent to )



------------------------------------------------------------




So our expression goes from and factors further to


------------------
Answer:

So completely factors to

You can put this solution on YOUR website!
Start with the compound interest formula


Plug in , (this is the decimal form of 5.5% interest), and


Divide. Note: this value is approximate. So all future calculations will be approximate.


Add


Divide both sides by to isolate "P".


Rearrange the equation



Now simply plug in the given values of "t" to find P:

t=1:

Start with the given equation.


Plug in


Multiply


Raise 1.0001507 to the 365th power to get 1.0565420


Divide


So if you want to have $150,000 in the account in 1 year, you need to invest about $141,972.59 in the account.



I'll let you do the other value of t (it will follow the same procedure).

You can put this solution on YOUR website!
Since you CANNOT divide by zero, this means that we must take out values of x out of the domain that make the denominator equal to zero


Set the denominator equal to zero.


Subtract from both sides.


Combine like terms on the right side.


Divide both sides by to isolate .


So when the denominator is zero.


So we must make the restriction that


So the domain is x can be any real number BUT

You can put this solution on YOUR website!
Start with the given expression.


Rewrite each number as a square of a whole number


Take the square root of the squares (to eliminate the squares).


Reduce


So

You can put this solution on YOUR website!
I'm assuming that you want to solve for "x" right?



Start with the quadratic formula


Plug in , , and . Note: there are two different "a" terms here...


Negate to get .


Square to get .


Multiply to get


Multiply 2 and 2 to get 4


or Break up the "plus/minus" to form two equations.


So the solutions are or


If we knew the value of "a", then we could continue simplifying.

You can put this solution on YOUR website!
Start with the given inequality.


Multiply both sides by 5


Multiply.


Distribute.


Add to both sides.


Combine like terms on the right side.


Divide both sides by to isolate .


Reduce.


----------------------------------------------------------------------

Answer:

So the answer is


The answer in set-builder notation is

You can put this solution on YOUR website!
Start with the given inequality


Subtract y from both sides.


Add 7 to both sides.


Combine like terms.


Now let's graph


In order to graph , we need to graph the equation (just replace the inequality sign with an equal sign).
So lets graph the line (note: if you need help with graphing, check out this solver)

graph of


Now lets pick a test point, say (0,0). Any point will work, (just make sure the point doesn't lie on the line) but this point is the easiest to work with. Now evaluate the inequality with the test point



Substitute (0,0) into the inequality


Plug in and


Simplify




Since this inequality is true, we simply shade the entire region that contains (0,0)


Graph of with the boundary (which is the line in red) and the shaded region (in green)

You can put this solution on YOUR website!
In order to graph , we need to graph the equation (just replace the inequality sign with an equal sign).
So lets graph the line (note: if you need help with graphing, check out this solver)

graph of


Now lets pick a test point, say (0,0). Any point will work, (just make sure the point doesn't lie on the line) but this point is the easiest to work with. Now evaluate the inequality with the test point


Substitute (0,0) into the inequality

Plug in and

Simplify


Since this inequality is true, we simply shade the entire region that contains (0,0)

Graph of with the boundary (which is the line in red) and the shaded region (in green)

You can put this solution on YOUR website!
Let's solve the first inequality :


Start with the first inequality.


Multiply both sides by 6.


Multiply.


Add to both sides.


Combine like terms on the right side.


Divide both sides by to isolate .


---------------------------------------------------------------------


Now let's solve the second inequality :


Start with the second inequality.


Multiply both sides by 6.


Multiply.


Add to both sides.


Combine like terms on the right side.


Divide both sides by to isolate .


So our answer is or


Which in set-builder notation looks like

You can put this solution on YOUR website!
Start with the given expression.


Combine the roots.


Divide to get


Factor into


Factor into


Break up the square root using the identity .


Take the square root of to get .


Take the square root of to get .


Rearrange and combine the terms.

==================================================

Answer:


So simplifies to


In other words, where

You can put this solution on YOUR website!
a) The given statement is true, here's why...


... Start with the given expression.


... FOIL


... Combine like terms.


Let and to get





... Take half of "b" and square it to get . Add AND subtract this inside the parenthesis.


... Factor the first three terms in the parenthesis


... Distribute


Let to get





Let to get





So for ANY expression of the form you can rewrite it as



============================================================

b) The given statement is false (if you are only restricted to factor over the reals)


Here's a counter-example:


Let , , and . So the general expression becomes




-----------

Start with the given expression.


FOIL


Combine like terms.


Since you CANNOT factor over the reals, this means that CANNOT be written in the form of


Note: if you are allowed to factor over the complex numbers, then you can rewrite into

You can put this solution on YOUR website!
# 1

To factor , we can use synthetic division


First, let's find our test zero:

Set the given factor equal to zero

Solve for x.

so our test zero is -5


Now set up the synthetic division table by placing the test zero in the upper left corner and placing the coefficients of to the right of the test zero.

-5	|	6	31	4	-5
	|

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

-5	|	6	31	4	-5
	|
		6

Multiply -5 by 6 and place the product (which is -30) right underneath the second coefficient (which is 31)

-5	|	6	31	4	-5
	|		-30
		6

Add -30 and 31 to get 1. Place the sum right underneath -30.

-5	|	6	31	4	-5
	|		-30
		6	1

Multiply -5 by 1 and place the product (which is -5) right underneath the third coefficient (which is 4)

-5	|	6	31	4	-5
	|		-30	-5
		6	1

Add -5 and 4 to get -1. Place the sum right underneath -5.

-5	|	6	31	4	-5
	|		-30	-5
		6	1	-1

Multiply -5 by -1 and place the product (which is 5) right underneath the fourth coefficient (which is -5)

-5	|	6	31	4	-5
	|		-30	-5	5
		6	1	-1

Add 5 and -5 to get 0. Place the sum right underneath 5.

-5	|	6	31	4	-5
	|		-30	-5	5
		6	1	-1	0

Since the last column adds to zero, we have a remainder of zero. This means is a factor of

Now lets look at the bottom row of coefficients:

The first 3 coefficients (6,1,-1) form the quotient




So factors to


In other words,


I'll let you continue the factorization....

---

# 2


First lets find our test zero:

Set the denominator equal to zero

Solve for x.

so our test zero is -1


Now set up the synthetic division table by placing the test zero in the upper left corner and placing the coefficients of to the right of the test zero.(note: remember if a polynomial goes from to there is a zero coefficient for . This is simply because really looks like

-1	|	1	-2	1	0	-4
	|

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

-1	|	1	-2	1	0	-4
	|
		1

Multiply -1 by 1 and place the product (which is -1) right underneath the second coefficient (which is -2)

-1	|	1	-2	1	0	-4
	|		-1
		1

Add -1 and -2 to get -3. Place the sum right underneath -1.

-1	|	1	-2	1	0	-4
	|		-1
		1	-3

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

-1	|	1	-2	1	0	-4
	|		-1	3
		1	-3

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

-1	|	1	-2	1	0	-4
	|		-1	3
		1	-3	4

Multiply -1 by 4 and place the product (which is -4) right underneath the fourth coefficient (which is 0)

-1	|	1	-2	1	0	-4
	|		-1	3	-4
		1	-3	4

Add -4 and 0 to get -4. Place the sum right underneath -4.

-1	|	1	-2	1	0	-4
	|		-1	3	-4
		1	-3	4	-4

Multiply -1 by -4 and place the product (which is 4) right underneath the fifth coefficient (which is -4)

-1	|	1	-2	1	0	-4
	|		-1	3	-4	4
		1	-3	4	-4

Add 4 and -4 to get 0. Place the sum right underneath 4.

-1	|	1	-2	1	0	-4
	|		-1	3	-4	4
		1	-3	4	-4	0

Since the last column adds to zero, we have a remainder of zero. This means is a factor of

Now lets look at the bottom row of coefficients:

The first 4 coefficients (1,-3,4,-4) form the quotient




So factors to


In other words,


Now let's use the factor to factor


First lets find our test zero:

Set the denominator equal to zero

Solve for x.

so our test zero is 2


Now set up the synthetic division table by placing the test zero in the upper left corner and placing the coefficients of to the right of the test zero.

2	|	1	-3	4	-4
	|

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

2	|	1	-3	4	-4
	|
		1

Multiply 2 by 1 and place the product (which is 2) right underneath the second coefficient (which is -3)

2	|	1	-3	4	-4
	|		2
		1

Add 2 and -3 to get -1. Place the sum right underneath 2.

2	|	1	-3	4	-4
	|		2
		1	-1

Multiply 2 by -1 and place the product (which is -2) right underneath the third coefficient (which is 4)

2	|	1	-3	4	-4
	|		2	-2
		1	-1

Add -2 and 4 to get 2. Place the sum right underneath -2.

2	|	1	-3	4	-4
	|		2	-2
		1	-1	2

Multiply 2 by 2 and place the product (which is 4) right underneath the fourth coefficient (which is -4)

2	|	1	-3	4	-4
	|		2	-2	4
		1	-1	2

Add 4 and -4 to get 0. Place the sum right underneath 4.

2	|	1	-3	4	-4
	|		2	-2	4
		1	-1	2	0

Since the last column adds to zero, we have a remainder of zero. This means is a factor of

Now lets look at the bottom row of coefficients:

The first 3 coefficients (1,-1,2) form the quotient




So


Basically factors to


So


This means that then becomes





So all you have to do now is factor (I'll let you do that)

You can put this solution on YOUR website!
This question has been answered many times before. Check out this solution

You can put this solution on YOUR website!

1. P → Q
2. (P • Q) → R
3. P → (R → S)
4. (R • S) → T / P → T
-----------------------
5.   P                               Assumption
6.   Q                        1,5    Modus Ponens
7.   P • Q                    5,6    Conjunction 
8.   R                        2,7    Modus Ponens
9.   P • R                    5,8    Conjunction
10.  (P • R) -> S               3    Exportation
11.  S                       10,9    Modus Ponens
12.  R • S                   8,11    Conjunction
13.  T                       4,12    Modus Ponens
14.  P -> T                  5,13    Conditional Proof

You can put this solution on YOUR website!

1. (A v B) → (C • D) / A → C
----------------------------
2. A                              Assumption
3. C • D                    1,2   Modus Ponens
4. C                          3   Simplification 
5. A -> C                   2,4   Conditional Proof