sorry. another algebra question.?

Question

1. find the reciprocal of -8/7.
what is reciprocal? is it the same thing as additive inverse?

2. x[(2)(x)] is (2x)(x^2) right?

3. simplify:
[6(x-3)+3x]-2{3[2(2y-4)]-5}

thank you soo much!

Answer 1

find the reciprocal of -8/7

Reciprocal of x is 1/x

So, here reciprocal of -8/7 will be -7/8


2) x[(2)(x)] is (2x)(x^2)

No,

x[(2)(x)] = x * 2 * x = 2 x^2

3) [6(x-3)+3x]-2{3[2(2y-4)]-5}

[(6x-18)+3x]-2{3[4y-8)]-5}

[6x-18+3x]-2{[12y-24]-5}

[9x -18] -2{12y-24 - 5}

9x -18 -2{12y - 29}

9x -18 - 24y + 58

9x - 24y + 40

Answer 2

1.
The reciprocal is just flipping the numerator and denominator. So, the reciprocal of -8/7 would be -7/8

2. No,
x[(2)(x)] = x * 2 * x = 2x^2

3.
[6(x-3)+3x]-2{3[2(2y-4)]-5} =
[6x - 18 + 3x] - 2{3(4y - 8) - 5} =
[9x - 18] - 2{12y - 24 - 5} =
[9x - 18] - 24y + 48 + 10 =
9x -24y + 40

Answer 3

1)-7/8. Reciprocal means you just interchange the numerator and the denominator of the fraction.

2)No. The first equation is equal to 2x^2. The second equation is equal to 2x^3. So, naturaly, they are different.

36x-18+3x-2(3[4y-8]-5)
9x-18-2(12y-24-5)
9x-18-24y+58
9x-24y+40

Answer 4

1. Reciprocal of - 8/7
= 1 / (- 8/7)
= 1 * - 7/8
= - 7/8

2. x(2 * x)
= x^2 * 2x or 2x^2: You're right.

3. (6[x - 3]) - 2(3[2{2y - 4}] - 5)
= (6x - 18) - 2(3[4y - 8] - 5)
= (6x - 18) - 2(12y - 24 - 5)
= (6x - 18) - 2(12y - 29)
= 6x - 18 - 24y + 58
= 6x - 24y + 40

Answer 5

1)
The reciprocal is -7/8 because
(-8/7) * (-7/8) = 1

I belive it's also called the multiplicative inverse.

2) No, it's 2x^2

x * 2 * x = 2(x * x) = 2x^2

3)
[6(x-3)+3x]-2{3[2(2y-4)]-5}
= [6x-18+3x]-2{3[4y - 8]-5}
= [6x-18+3x]-2{12y - 24]-5}
= (3x - 18) - 2(12y - 29)
= (3x - 18) - (24y - 58)
= 3x - 18 - 24y + 58
= 3x - 24y + 40

Answer 6

Doctor Q is right for the first 2
3. 6x-18+3x-2(3[4y-8]-5)
9x-18-2(12y-24-5)
9x-18-24y+58
9x-24y+40

Answer 7

1. -7/8
2. 2x^2

Answer 8

1) -7/8

2) 2x^2

3) Solve yourself it's easy....