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# Recent problems solved by 'jsmallt9'

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 Functions/717316: or questions 4–5, is the relation a function? 4. {(4, 4), (13, 2), (–5, –12), (–7, 12)} (1 point) yes no 5. {(–3, –2), (12, 7), (8, 9), (12, 15)} (1 point) yes no 1 solutions Answer 440234 by jsmallt9(3296)   on 2013-02-21 19:08:15 (Show Source): You can put this solution on YOUR website!4. Yes. Each x-coordinate is different. 5. No. The ordered pairs (12, 7) and (12, 15) have the same x-coordinates but different y-coordiates.
 Radicals/717301: What is the extraneous solution of sqrt(x-3)= (x-5) also sqrt of(x+4)=sqrt(3x)1 solutions Answer 440232 by jsmallt9(3296)   on 2013-02-21 19:04:47 (Show Source): You can put this solution on YOUR website!First we find solutions. Then we determine which, if any, are extraneous. Square both sides: Squaring the left side is as easy as it looks. Squaring an expression like the right side is where a lot of mistakes are made. It is not !! To square a two-term expression like (x-5) correctly we must use FOIL on (x-5)(x-5) or use the pattern. I prefer using the pattern: which simplifies to: Now that x is out of the square root we can solve for it. This is a quadratic equation so we want one side to be zero. Subtracting x and adding 3 we get: Next we factor (or use the Quadratic Formula). This factors fairly easily: From the Zero Product Property: x-7 = 0 or x-4 = 0 Solving these we get: x = 7 or x = 4 Now we check. Use the original equation: Checking x = 7: Check! Checking x = 4: Check fails! So x = 7 is actually a solution and x = 4 is extraneous. Square both sides: x+4 = 3x Subtract x: 4 = 2x Divide by 2: 2 = x Check. Checking x = 2: Check! x = 2 is the solution to this equation. There are no extraneous solutions.
 Divisibility_and_Prime_Numbers/717356: there are four diffrent digits which,when inserted in the blank space in the numbers 4blank5,make the number divisble by 3. write them. i dont understand can u help me 1 solutions Answer 440227 by jsmallt9(3296)   on 2013-02-21 18:48:21 (Show Source): You can put this solution on YOUR website!If a number is divisible by 3 then the sum of the digits of that number are also divisible by 3. So we want the 4 and the 5 and the "four different digits" to add up to a number divisible by 3. Probably the easiest way to do this is to pick three of the digits and then see what 4th digit would make the sum of the digits be divisible by 3. Let's say the 3 of the 4 digits are 1, 2 and 3 and let's call the one missing digit "n". So the digits are 4123n5. Now let's add the digits: 4+1+2+3+n+5. This simplifies to 15+n. So what digit (or digits) would make 15+n divisible by 3? I'll leave it up to you to figure it out. Hint: There are 4 digits that would make 15+n divisible by 3. One of them is the same as one of the digits already in the number. So there are three possible 4th digits (when you pick 1, 2 and 3 to start with) which would make the number divisible by 3 and would be different from the others.
 logarithm/717358: I am having some trouble with solving a few problems on my Algebra homework. Idea:Write each expression as a single Logarithm. Question: log^5 4 + log^5 3 1 solutions Answer 440222 by jsmallt9(3296)   on 2013-02-21 18:34:50 (Show Source): You can put this solution on YOUR website!log^5 4 + log^5 3??? Logarithms are especially difficult to post since it is not possible to type subscripts like the bases of logarithms. Please use some English to explain what it is your logarithms are (like "base 5 log of (x+1)") or teach yourself the syntax used on Algebra.com to display logs. If you click on the "Show source" link above you will be able to see what I typed to get to display.
 Divisibility_and_Prime_Numbers/717353: it says write the missing digit to make each number divisible by 9 what do i do 88,blank,12 what do i do 1 solutions Answer 440218 by jsmallt9(3296)   on 2013-02-21 18:29:52 (Show Source): You can put this solution on YOUR website!A number is divisible by 9 if the sum of the digits is divisible by 9. If we call the missing digit "n" then the sum of the digits, 8+8+n+1+2, must add up to a number which is divisible by 9. Simplifying 8+8+n+1+2 we get 19+n. "n" must be a valid digit, 0 through 9. Which one must it be if 19+n has to be divisible by 9? There's only one possible answer. I'll leave it up to you to finish. Just try different digits for "n" until 19+n is divisible by 9.
 Quadratic-relations-and-conic-sections/717354: x2+y2+4x-6y=11 is this an hyperbola?1 solutions Answer 440215 by jsmallt9(3296)   on 2013-02-21 18:22:40 (Show Source): You can put this solution on YOUR website!Since there are no xy terms and since the coefficients of the squared terms are equal, this is the equation of a circle.
 logarithm/717352: How do you solve the log of √6/3 with the base of 3/2 = -1/2?1 solutions Answer 440213 by jsmallt9(3296)   on 2013-02-21 18:19:58 (Show Source): You can put this solution on YOUR website!There are no unknowns/variables in what you posted. What are you/we supposed to solve for?
 Exponents/717186: Explain the Law of Exponents that states a to the -n power = 1 divided by a to the n power?1 solutions Answer 440175 by jsmallt9(3296)   on 2013-02-21 15:40:01 (Show Source): You can put this solution on YOUR website!You probably understand the positive exponents: , , , , ... To help us understand negative exponents, let's think about the opposite of the above. Instead of starting with a and repeatedly multiplying by a, think about starting with a high power of a and repeatedly dividing by a: , , , ... As we did this the exponents kept going down by 1. Now, let's keep going! The exponent went down by 1 again! again, the exponent went down by 1 etc.
 Polynomials-and-rational-expressions/717184: What is the LMC of the pair of polynomials? 2x^2-18 and 5x^3+30x^2+45x= 1 solutions Answer 440166 by jsmallt9(3296)   on 2013-02-21 15:23:56 (Show Source): You can put this solution on YOUR website!To find the LCM of some polynomials you need to see their factors. So we will factor them. First the greatest common factor: Next, factoring patterns. The second factor of the first polynomial is a difference of squares which will factor according to the pattern . The second factor of the other polynomial fits a different pattern: or (a+b)(a+b) with the "a" being x and the "b" being 3. Using these patterns to further factor we get: If you have trouble with LCM's I find that it can be helpful to write the factors in a special way: ```2x^2-18 = 1 * 2 * (x+3) * (x-3) 5x^3+30x^2+45x = 1 * (x+3) * 5 * x * (x+3) ======================================================== LCM = 1 * 2 * (x+3) * (x-3) * 5 * x * (x+3) = 10x(x+3)(x-3)(x+3) ```The first lines display the polynomials and their factors. The bottom line is the LCM. Notice how I used spacing and the Commutative Property (which allows me to rearrange the order of factors of the polynomials in any way I choose. I arranged the factors so the the factors in each column were all the same. I lined up the factors of 1, of 2, of (x+3), etc. Taking 1 factor from each column, the LCM is the product of these factors. Sometimes you want an LCM in factored form (like when you are finding a lowest common denominator (LCD) which is just an LCM of denominators). Sometimes you want it in simplified form. If you want this LCM in simplified form then we just multiply out the factors. We can multiply in any order so I am going to choose an order that looks easy. First, (x+3)(x-3). We already know how this works out from when we factored 2x^2=18: LCM = Next I'll use FOIL to multiply the last two factors: LCM = Simplifying... LCM = And finally the 10x (with the Distributive Property): LCM = P.S. There are additional uses for arranging the factors in a table like the one above:This table will handle as many polynomials as you need. Just add a line for each additional polynomial.If you are looking for an LCD, not only can you find the LCD but you can see what you need to multiply each fraction by to make the denominator become the LCD. For example, if the problem was to add: the LCD would be as was found earlier. And from the table you can literally see which factors of the LCD a certain denominator is missing. For 2x^2-18 we can see that it is "missing" factors of 5, x and a second (x+3). And for the other denominator we can see that, compared to the LCM/LCD, it is "missing" factors of 2 and (x-3). We now now how to change the fractions so their denominators become the LCD: etc.This table of factors can also be used to find the greatest common factor (GCF) between polynomials. In this case, instead of using factors from every column you use only the factors from "full" columns. IOW from columns that have no empty spaces. In this case there are only two "full" columns: the 1's and the first (x+3)'s. So the GCF for 2x^2-18 and 5x^3+30x^2+45x is 1*(x+3) or just (x+3)
 absolute-value/717064: Please help me solve this equation: 3|x + 5|+ 4 = 71 solutions Answer 440102 by jsmallt9(3296)   on 2013-02-21 10:44:50 (Show Source): You can put this solution on YOUR website!I just solved a problem very much like this one. The steps to its solution should help you figure out this problem. Click here to see the other solution.
 absolute-value/717066: Please answer this equation: 4 + 3 |5 - x| = 1541 solutions Answer 440101 by jsmallt9(3296)   on 2013-02-21 10:41:20 (Show Source): You can put this solution on YOUR website! First, isolate the absolute value. Subtracting 4: and divide by 3: With the absolute value isolated, we now rewrite the equation as two equivalent equations without absolute value. If the absolute value of something is 50 then it must be either a 50 or a -50: or With the x out of the absolute value, we can now solve for it. Subtracting 5 in each equation: or Dividing (or multiplying) by -1: or
 Probability-and-statistics/717065: from the following data find the missing frequency when mean is 15.38. ______________________________________________________ MARKS: 10 12 14 16 18 20 FREQUENCY: 3 7 12 20 ? 5 ________________________________________________________1 solutions Answer 440096 by jsmallt9(3296)   on 2013-02-21 10:33:49 (Show Source): You can put this solution on YOUR website!If "n" is the missing frequency then: Simplifying... Eliminate the fraction by multiplying each side by (47+n): Subtracting 15.38n: Subtracting 702: Dividing by 2.62: Since "n" was a frequency, it should be a whole number. The reason we did not get a whole number for n must be eitherWe made a mistake. Check the work to see if we did. Or...The average we were given is a rounded-off number. In this case, "n" must be 8.
 Matrices-and-determiminant/717092: x+3y-z=0 2x-y+4z=0 x-11y+14z=0 please solve the system of equations using matrix inverse method1 solutions Answer 440094 by jsmallt9(3296)   on 2013-02-21 10:12:17 (Show Source): You can put this solution on YOUR website!With all the equations with a zero on the right side, we know that a solution (perhaps the only solution) to this system will be (0, 0, 0). But we will go ahead and find an inverse anyway. The coefficient matrix for this system is: ```1 3 -1 2 -1 4 1 -11 14```(Note: To save time I am not going to put the matrices in brackets. So you'll just have to deal with just the numbers.) To find the inverse of this matrix:"Append" the identity matrix to the coefficient matrix. This should give you a matrix with the left half being the original coefficient matrix and the right half being the identity matrix.Use elementary row operations to change this expanded matrix so the the left half is the identity matrix.The right half of this modified, expanded matrix will be the inverse.Let's see this in action: 1. Append the identity matrix: ```1 3 -1 1 0 0 2 -1 4 0 1 0 1 -11 14 0 0 1```2. Use elementary row operations to modify this expanded matrix. (Some notes before I start:You probably want to check my arithmeticThe exact steps I use and the order in which I do them are not the only way to accomplish this step. All that is important is that you use proper elementary row operations correctly and that you end up with the right matrixBetween each matrix I will describe how I am getting from the one above to the one below. In these descriptions I will refer to row 1 as r1, row 2 as r2 and row 3 as r3. ```Replace r2 with r2 + -2*r1: 1 3 -1 1 0 0 0 -7 6 -2 1 0 1 -11 14 0 0 1 Replace r3 with r3 + -r1: 1 3 -1 1 0 0 0 -7 6 -2 1 0 0 -14 15 -1 0 1 Replace r3 with r3 + -2*r2: 1 3 -1 1 0 0 0 -7 6 -2 1 0 0 0 3 3 -2 1 Replace r2 with r2 + -2*r3: 1 3 -1 1 0 0 0 -7 0 -8 5 -2 0 0 3 3 -2 1 Replace r2 with (-1/7)*r2 and replace r3 with (1/3)*r3: 1 3 -1 1 0 0 0 1 0 8/7 -5/7 2/7 0 0 1 1 -2/3 1/3 Replace r1 with r1 + -3*r2: 1 0 -1 -17/7 -15/7 -6/7 0 1 0 8/7 -5/7 2/7 0 0 1 1 -2/3 1/3 Replace r1 with r1 + -r3: 1 0 0 -10/7 -59/21 -11/21 0 1 0 8/7 -5/7 2/7 0 0 1 1 -2/3 1/3```The identity matrix has now "moved" from the right half to the left half. 3. The right half is now the inverse of our original coefficient matrix. So: ```-10/7 -59/21 -11/21 8/7 -5/7 2/7 1 -2/3 1/3 is the inverse of: 1 3 -1 2 -1 4 1 -11 14```Note: Since the inverse does not have a row of all zeros, then there will be just one solution to this system, the (0, 0, 0) we've know of since the start. To use this inverse to find a solution to the system, write the system in matrix form and then left-multiply (remember, matrix multiplication is not commutative!) both sides by the inverse matrix. The system, in matrix form is: ``` 1 3 -1 x 0 2 -1 4 * y = 0 1 -11 14 z 0 Left-multiply each side by the inverse: -10/7 -59/21 -11/21 1 3 -1 x -10/7 -59/21 -11/21 0 8/7 -5/7 2/7 * 2 -1 4 * y = 8/7 -5/7 2/7 * 0 1 -2/3 1/3 1 -11 14 z 1 -2/3 1/3 0```On the left side the multiplying first two matrices, if we found the inverse correctly, will result in the identity matrix. And that identity matrix times the variable matrix will result in the variable matrix. On the right side we end up with a 3x1 column matrix with the x value of the solution in the first row, the y value in the second row and the z value in the third row (all zeros as we've known from the start). P.S. I suggest you multiply the coefficient matrix and its inverse. If you do not get the identity matrix as a result then one or more of the calculations made while we found the inverse must be in error.
 Linear-systems/717033: What is the solution to the system of equations y = 2x and y = -x + 3?1 solutions Answer 440085 by jsmallt9(3296)   on 2013-02-21 08:46:56 (Show Source): You can put this solution on YOUR website!With not just one but both equations already "solved for y", this system is an ideal candidate for using the Substitution Method. The first equation says the y = 2x. Substituting 2x for the y in the other equation we get: 2x = -x + 3 Now we solve for x. Adding x to each side: 3x = 3 Dividing by 3 we get: x = 1 Now we use this value for x and either of the original equations to find the y value: y = 2(1) So y = 2 The solution is the point (1, 2)
 Triangles/717073: if triangle abc is congruent to triangle tuv which statement is incorrect? angle c is congruent to angle v line bc is congruent to line uv abc is congruent to utv line ab is congruent to line tu1 solutions Answer 440084 by jsmallt9(3296)   on 2013-02-21 08:42:32 (Show Source): You can put this solution on YOUR website!When you're told that triangle ABC is congruent to triangle TUV, you are told more than just the fact that the triangles are congruent. The letters of the names of triangles represent vertices of the triangle. And usually the order of the letters is unimportant. But when a statement says that two triangles are congruent the order of the letters means something important. The first letters in the names of the triangle are corresponding vertices; the second letters are corresponding vertices and the third letters are corresponding vertices. Since corresponding parts of congruent triangles are congruent we know that if triangle abc is congruent to triangle tuv then:Angle A is congruent to angle TAngle B is congruent to angle UAngle C is congruent to angle VSide AB (which is between the first two vertices) is congruent to side TUSide BC (which is between the second and third vertices) is congruent to side UVSide AC (which is between the first and third vertices) is congruent to side TVExcept for "ABC is congruent to UTV" you should now what the answers are. For "ABC is congruent to UTV" (assuming this means "triangle ABC is congruent to triangle UTV") then this is false because the corresponding parts don't match up. If it had been "triangle BCA is congruent to triangle UVT", then it would be true because the same letters correspond with each other: B and U, C and V and A and T P.S. Points should be named with capital/upper-case letters. And vertices of a triangle are just special points so they also should be named with capital letters.
 logarithm/717077: log under root 3 1/31 solutions Answer 440083 by jsmallt9(3296)   on 2013-02-21 08:20:32 (Show Source): You can put this solution on YOUR website!What does "log under root" mean?? You'll have to learn how to read your problems correctly so you can then describe them correctly when you post. Find out how to read your problem correctly and re-post it in a way we can understand.
 Inverses/717078: Asked: inverse function of x = y^2 - y already done: y = x^2 - x But how do I get a function with y=... out of this? Is this possible?1 solutions Answer 440081 by jsmallt9(3296)   on 2013-02-21 08:16:08 (Show Source): You can put this solution on YOUR website!Isn't already in "y = ..." form? I think you're done!
 Polynomials-and-rational-expressions/716670: using synthetic division solve (5x^2 - 12x - 8)/(x + 3)1 solutions Answer 439954 by jsmallt9(3296)   on 2013-02-20 13:58:02 (Show Source): You can put this solution on YOUR website! is just an expression and expressions cannot be "solved". But they can be simplified. In this case we can divide the numerator by the denominator. Synthetic division is designed for division by an expression of the form: x - n where "n" is some number. If we are to use synthetic division to divide by x+3 we must imagine (or actually rewrite) it as a subtraction: x - (-3): ```-3 | 5 -12 -8 ----- -15 81 ------------- 5 -27 73 ```The remainder is in the lower right corner. It is 73. The rest of the bottom row is the quotient. The "5 -27" translates into 5x-27. So after dividing: we get: (Note what happens with the remainder. We handle in just the same way as in the "good old days" before variables: We make a fraction with the remainder on top and the divisor on the bottom!)
 Polynomials-and-rational-expressions/716674: a.list all Zeroes b.use synthetic division f(x)=x^3-4x^2+8x-51 solutions Answer 439951 by jsmallt9(3296)   on 2013-02-20 13:47:44 (Show Source): You can put this solution on YOUR website! a. To find the zeros of a polynomial of degree 2 or more we factor the polynomial. (Although in the special case of Quadratics we can use the Quadratic Formula instead of factoring.) The greatest common factor is 1 (which we rarely factor out). There are too many terms for any of the factoring patterns or for trinomial factoring. I don't see how to factor the polynomial by grouping so we are left with factoring by trial and error of the possible rational roots. The possible rational roots of a polynomial are all the ratios, positive and negative, that can be formed using a factor of the constant term (at the end) in the numerator over a factor of the leading coefficient (at the front). The contant term in our polynomial is 5. (Actually it is -5 but since we are going to use all positive and negative ratios it doesn't matter if we use the "-".) The factors of 5 are just 1 and 5. The leading coefficient (in front of ) is 1 which just has 1's for factors. This makes the possible rational roots of our polynomial: +1/1 or +5/1 which reduce to: +1 or +5 Now we see if any of these actually are roots/zeros of the polynomial. Since powers of 1 are very simple we can easily check that mentally. But since the problem tells you to use synthetic division, we will do so: ```1 | 1 -4 8 -5 ---- 1 -3 5 ----------------- 1 -3 5 0 ```The remainder, in the lower right corner is zero. This means that (x-1) is a factor and that 1 is a root/zero of the polynomial. Not only that but the rest of the bottom row tells us what the other factor will be. "1 -3 5" translates into . So f(x) in (partially) factored form is: The remaining zeros of the polynomial will be zeros of the second factor. That factor will not factor further, no matter which factoring techniques we use! But it is quadratic so we can use the formula: Simplifying... which is short for: or With the negative inside the square root, these roots/zeros will be complex numbers. So we should express them in standard a + bi form: or or or or or These two, plus the zero of 1 we found earlier, are the three zeros of your polynomial function.
 Trigonometry-basics/716705: f(x)=2sin(1/2x+5pi/6) Find five exact value points on the graph1 solutions Answer 439944 by jsmallt9(3296)   on 2013-02-20 13:19:49 (Show Source): You can put this solution on YOUR website!Since all multiples of are special angles and since the constant part of the argument, , is already a multiple of , it will probably be easiest if we choose values for x that would make the other term in the argument, , also be a multiple of . By doing so we would then be able to add the fractions and end up with an argument that is also a multiple of . Perhaps you can already tell what values x should have that would turn into a multiple of . If not, then just pick a multiple of , set equal to it and solve for x. Let's be easy on ourselves and choose itself: Multiplying both sides by 6 we get: Dividing by 3 we get: Perhaps you can now see that if x is any multiple of , then would end up being some multiple of . I'll get you started on a table of values: ``` x (1/2)x (1/2)x+5pi/6 sin((1/2)x+5pi/6) 2sin((1/2)x+5pi/6) f(x) pi/3 pi/6 6pi/6 = pi 0 0 0 -pi/3 -pi/6 4pi/6 = 2pi/3 sqrt(3)/2 sqrt(3) sqrt(3) 0 0 5pi/6 1/2 1 1 ```I'll leave it up to you to find two more. P.S. We did not have to make x be a value that made the argument of sin be a multiple of pi/6. We just did that because it made things easier. Any x that would make (1/2)x+5pi/6 turn out to be any of the special angles would work.