Please help me with this math problem

Mathematics

2 answers:

Lynna [10]3 years ago

7 0

Answer:

18vw²y^9 + 24v^6w^5

= 6vw²( 3y^9 + 4v^5w³)

Hope this helps.

eduard3 years ago

6 0

Answer:

6vw^2(3y^9 + 4v^5w^3)

Step-by-step explanation:

If you were to multiply the equation 6vw^2(3y^9 + 4v^5w^3) then the result would be

18vw^2y^9 + 24v^6w^5

Hope this helps!

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mihalych1998 [28]

Answer:

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Step-by-step explanation:

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3 years ago

Let f(x) = .............

galben [10]

Answer:

Step-by-step explanation:

g(3) = -2(3)^2 -4 = 2(9) -4 = 18-4 = 14

|f(2)| = | 2^2 -3| = | 4-3| = 1

(g(3) +2) = (14+2) = 16

4* |f(2)| = 4*1 = 4

16/4 = 4

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3 years ago

Write an equation in standard form of the hyperbola described.

marishachu [46]

Check the picture below, so the hyperbola looks more or less like so, so let's find the length of the conjugate axis, or namely let's find the "b" component.

$\textit{hyperbolas, horizontal traverse axis } \\\\ \cfrac{(x- h)^2}{ a^2}-\cfrac{(y- k)^2}{ b^2}=1 \qquad \begin{cases} center\ ( h, k)\\ vertices\ ( h\pm a, k)\\ c=\textit{distance from}\\ \qquad \textit{center to foci}\\ \qquad \sqrt{ a ^2 + b ^2} \end{cases} \\\\[-0.35em] ~\dotfill$

$\begin{cases} h=0\\ k=0\\ a=2\\ c=4 \end{cases}\implies \cfrac{(x-0)^2}{2^2}-\cfrac{(y-0)^2}{b^2} \\\\\\ c^2=a^2+b^2\implies 4^2=2^2+b^2\implies 16=4+b^2\implies \underline{12=b^2} \\\\\\ \cfrac{(x-0)^2}{2^2}-\cfrac{(y-0)^2}{12}\implies \boxed{\cfrac{x^2}{4}-\cfrac{y^2}{12}}$