PLEASE HELP ME!!! (assume that no variable equals 0)

Mathematics

1 answer:

irina1246 [14]3 years ago

8 0

Answer:

9a - 2b

I hope this helps!

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Find the remainder when f(x) is divided by (x - k). <br><br> f(x) = 8x4 + 7x3 + 5x2 - 5x + 35; k = 4

Slav-nsk [51]

Using the remainder theorem, evaluating the function for k will be the same value as the remainder.
8(4^4) + 7(4^3) + 5(4^2) -5(4) + 35 =
2048 + 448 + 80 + - 20 + 35 =
2591

6 0

3 years ago

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(12a^(2) - 3)/(2)*(2a + 1)^(-2)*((6)/((2a + 1)))^(-1)

Elena L [17]

Answer:

$\frac{144a^{4}+144a^{3} - 36 - 12a} {2a+1}\\$

Step-by-step explanation:

$\frac{(12a^2-3)}{2*(2a+1)^{-2}}*\frac{6}{(2a+1)^{-1} }\\ = \frac{(12a^2-3)}{2 * \frac{1}{(2a+1)^{2} }} * \frac{6}{\frac{1}{2a+1} }\\ =\frac{(12a^2-3)(2a+1)^{2}}{2} * \frac{6}{2a+1}\\=\frac{(12a^2-3)(2a+1)^{2}}{2} * \frac{6}{2a+1}$

$= \frac{(12a^2-3)(4a^{2} + 1 + 2 (2a)(1))}{2} * \frac{6}{2a+1}\\= \frac{(12a^2-3)(4a^{2} + 1 + 4a)}{2} * \frac{6}{2a+1}\\= \frac{(12a^2-3)(4a^{2} + 1 + 4a)}{1} * \frac{3}{2a+1}\\= \frac{3(12a^2-3)(4a^{2} + 1 + 4a)} {2a+1}\\\\= \frac{3(12a^2(4a^{2} + 1 + 4a) - 3 (4a^{2}+1+4a)} {2a+1}\\= \frac{3(48a^{4}+12a^{2}+48a^{3} - 12a^{2} -12 - 4a)} {2a+1}\\= \frac{144a^{4}+36a^{2}+144a^{3} - 36a^{2} - 36 - 12a)} {2a+1}$