F(x) = 2(x+3)2 -7 a= h= k= - id23275062392, sxbenn

Ответ:

dunk36

18.01.2022

Mathematics

Step-by-step explanation:

In vertex form, the equation is

y = a(x-h)^2 + k

So just read off the values!

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Ответ:

brennarfa

23.04.2022

Mathematics

$2\sin 2\theta - 3 \cos2\theta - 3\sin\theta + 3 = \sin\theta(4\cos\theta + 6\sin\theta - 3)$ left-hand side $\begin{aligned} \text{lhs} & = 2\sin 2\theta - 3 \cos2\theta - 3\sin\theta + 3 \\ & = 2\boxed{\sin 2\theta} - 3 \boxed{\cos2\theta} - 3\sin\theta + 3 \end{aligned}$ use identities: $\begin{aligned} \sin 2\theta & = 2\sin\theta\cos\theta \\ \cos 2\theta & = 1 - 2\sin^2\theta \end{aligned}$ (using that form of cosine double angle because there's one 1 cosine on the right-hand side. if we used another form, there'd be too many cosines) $\begin{aligned} \text{lhs} & = 2(2\sin\theta\cos\theta)- 3(1 - 2\sin^2\theta) - 3\sin\theta + 3 \\ & = 4\sin\theta\cos\theta - 3 + 6\sin^2\theta - 3\sin\theta + 3 \\ & =4\sin\theta\cos\theta - \boxed{3} + 6\sin^2\theta - 3\sin\theta + \boxed{3} \end{aligned}$ those +3 and -3 cancel each other out (add to become 0) $\begin{aligned} \text{lhs} & =4\sin\theta\cos\theta + 6\sin^2\theta - 3\sin\theta \end{aligned}$ factor out $\sin\theta$ out of everything $\begin{aligned} \text{lhs} & =4\sin\theta\cos\theta + 6\sin^2\theta - 3\sin\theta \\ & = \sin\theta\left(4\cos\theta + 6\sin\theta - 3\right) \\ & = \text{rhs} \end{aligned}$

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F(x) = 2(x+3)2 -7

a=

h=

k=

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