2.06: similar triangles practice quiz

We need to simplify it (or just divide). Since the coefficient of the area (on the x^2) is the same as that on the width, we know that that same coefficient on the length is 1.

This gets us to the basic frame (1x+/-y).

To find the value of y, we need to pay attention to the "-43x+63" and "-7" aspects of the area and width, respectively. To get "63," the "-7" was multiplied by the y — by dividing 63 by -7, we know that the value of y is -9 (the numbers both have to be negative to multiply to a positive number).

We are left with the length (x-9). Put together, this means (x-9)(4x-7) is the area. Multiplying, that makes (4x^2-7x-36x+63), or (4x^2-43x+63). Since this is the area given to us, we know our answer is correct. For this question, the answer is A.

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Divide $9x^4-2-6x-x^2$ by $3x-1$ . First, put the first equation in order by exponents. We get $\frac{9x^4-x^2-6x-2}{3x-1}$ . Since the exponent on the upper equation goes up to x^4, and we are dividing by a simple x, we know that the first exponent in our answer will be x^3. Since our coefficient needs to have a product of 9 when multiplied by 3, it is 3. The first part of our answer is $(3x^3)$ . Since there is no exponent of 3 for x in the upper equation, and since the "x^2" that has to be the next term in the equation due to it being present in each answer choice, we know that it as to be a "+x^2" — the "3x^3 that result from the "-1" (in the lower equation) being multiplied by 3x^3 have to be cancelled out by a "-3x^3", and if the sign for the term "x^2" is negative, we end up with tw0 "3x^3" that add up to "6x^3" instead of cancelling each other out.

Now, we have $(3x^3-x^2)$ . We can immediately rule out C. Moving on.

We now have $(3x-1)(3x^3+x^2)=9x^4-x^2$ . We can rule out answer choice B because it is incomplete - we are missing the second part of the upper equation, "-6x-2." Both of the remaining answers include "-2" as the next term, whether with an x or without.

Honestly, I haven't done algebra in a few years — while I know there's a way to deduce the rest of the equation, let's solve the equation using the two remaining answer choices and see which one is correct.

A: $(3x-1)(3x^3+x^2-2-\frac{4}{3x-1} )$

$=9x^4+3x^3-6x -3x^3-x^2+2-4$ (FOIL) (3x-1 times $-\frac{4}{3x-1}$ = -4)

$=9x^4+3x^3-3x^3-x^2-6x+2-4$ (in order of exponents)

$=9x^4-x^2-6x+2-4$ (simplify)

$=9x^4-x^2-6x-2$ (simplify)

D: $(3x-1)(3x^3+x^2-2x-\frac{4}{3x-1} )$

$=9x^4+3x^3-6x^2-\frac{4}{1} -3x^3-x^2+3x+\frac{4}{1}$ (FOIL)

$=9x^4+3x^3-3x^3-7x^2-4+4$ (in order of exponents)

$=9x^4-7x^2$ (simplify)

A is exactly the long fraction we started with ( $9x^4-2-6x-x^2$ ), just in a different order! This means that answer A, when multiplied by (3x-1), equals the same thing which, if divided by (3x-1), yields answer A. Because of the rules of multiplication/division (xy=z, z/x=y, z/y=z), this means that we have the proper set of numbers. Answer A is correct.

I hope this helps!!!!! Let me know if there's anything else I can help with :)

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2 years ago