3 years ago

Write the following equation in standard form: y + 4 = -3(x + 2)

Mathematics

1 answer:

Irina18 [472]3 years ago

5 0

Answer:

-3x-y=10

Step-by-step explanation:

y+4=-3(x+2)

y+4=-3x-6

y=-3x-6-4

y=-3x-10

-3x-y=10

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Read 2 more answers

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

Elena L [17]

Answer:

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

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

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

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

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