Anyone knows what’s is the answer for 5 ?

Mathematics

1 answer:

Andreas93 [3]3 years ago

7 0

Answer:

f(x) = |x - 4| + 3

Step-by-step explanation:

Identify (from the graph) the coordinates of the vertex: (4, 3). That +4 tells us that the vertex of the basic function y = |x| has been translated 4 units to the right; that +3 tells us that the vertex has also been translated 3 units up. Consider y = |x-a| + b. Here, b = +3 and a=+4. Thus, the correct function is f(x) = |x - 4| + 3.

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Alik [6]

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4 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:

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

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