How to solve these three math problems?

Question

I need help with these three problems:

1. How do I draw a line for the following equation: (2x)^2(x^2)^3y=(-5/2)x

2. The variables x and y vary invrsely. when x is 10, y is 7. If x is 5, then y is ______?

3. What team should be added to x^2-(2/3)x so that the result is a perfect square trinomial?

I dont understand how to do these three problems. Please dont post any comments regarding "not doing my own homework" or something similar because These three problemsa re among the 100 problems I had to do. I finished all of them except for these three. Please Help me and explain how to do these problems because I have to show my work.

Thanx in advance!!!!

Answer 1

I don't have the time to get into these questions, but what I WILL tell you is that there can be multiple answers to question number 2.  When variables vary inversely, it just means that as one goes up, the other goes down.  The question is proportion.  We don't know if they are HALVING "x" or subtracting 5 from "x" in that problem.  EITHER WAY, x becomes 5 from 10.  How they arrived at 5 from 10 is ambiguous.  If they vary inversely, with y being 7 when x is 10, x-5=5, so y+5 (the opposite of what you did to x) is 12.  If x/2 is 5, then 2y (the opposite of what you did to x) is 14.

Get me?

Answer 2

1. How do I draw a line for the following equation: (2x)^2(x^2)^3y=(-5/2)x

This is not a linear equation, so you cannot draw a line for it.

Solve for y in terms of x:

(2x)^2(x^2)^3y=(-5/2)x

4x^2 * x^6 * y = -5x/2

y = -5x / 8x^8

y = -5 / 8x^7

2. The variables x and y vary invrsely. when x is 10, y is 7. If x is 5, then y is ______?

y = k / x

Solve for k from the first data pair: (x, y) = (10, 7)

7 = k/10

k = 70

Solve for y for the second data pair:

y = k/x

y = 70/5

y = 14 Answer

3. What team should be added to x^2-(2/3)x so that the result is a perfect square trinomial?

A perfect square trinomial is of the form a^2 + 2ab + b^2 whose factors are (a + b)(a + b) or (a + b)^2

In x^2-(2/3)x,

a^2 = x^2 which means that a = x

2ab = -(2/3)x

Solve for b:

2ab = -(2/3)a

b = -(2/3)a / 2a

b = -(2/2)(a/a)(1/3)

b = -(1/3)

Therefore, b^2 = 1/9

Add this to x^2-(2/3)x and you'll have: x^2-(2/3)x + 1/9, whose factors are (x - 1/9)^2.

God bless you!