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3 years ago

Write an expression for the sequence of operations described below.

Mathematics

1 answer:

aniked [119]3 years ago

4 0

Answer:

It should be 9+h+2

Step-by-step explanation:

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The line with a slope of 1/3 and a y- intercept of -15 intersects the line passing through which of these pairs of points? Selec

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3 years ago

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3 years ago

5 +4.2 + 5 – 2. 2 - 1

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3 years ago

Read 2 more answers

PLEASE HELP I DO NOT UNDERSTAND AT ALL ITS PRECALC PLEASE SERIOUS ANSWERS

Elina [12.6K]

You want to end up with $A\sin(\omega t+\phi)$ . Expand this using the angle sum identity for sine:

$A\sin(\omega t+\phi)=A\sin(\omega t)\cos\phi+A\cos(\omega t)\sin\phi$

We want this to line up with $2\sin(4\pi t)+5\cos(4\pi t)$ . Right away, we know $\omega=4\pi$ .

We also need to have

$\begin{cases}A\cos\phi=2\\A\sin\phi=5\end{cases}$

Recall that $\sin^2x+\cos^2x=1$ for all $x$ ; this means

$(A\cos\phi)^2+(A\sin\phi)^2=2^2+5^2\implies A^2=29\implies A=\sqrt{29}$

Then

$\begin{cases}\cos\phi=\frac2{\sqrt{29}}\\\sin\phi=\frac5{\sqrt{29}}\end{cases}\implies\tan\phi=\dfrac{\sin\phi}{\cos\phi}=\dfrac52\implies\phi=\tan^{-1}\left(\dfrac52\right)$

So we end up with

$2\sin(4\pi t)+5\cos(4\pi t)=\sqrt{29}\sin\left(4\pi t+\tan^{-1}\left(\dfrac52\right)\right)$

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3 years ago

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Find the limit. Use l'Hospital's Rule if appropriate. If there is a more elementary method, consider using it. lim x→0+ (2x + 1)

ElenaW [278]

$\displaystyle\lim_{x\to0^+}(2x+1)\cot x=\lim_{x\to0^+}\frac{2x+1}{\tan x}$

As $x\to0$ , the numerator approaches 1 while $\tan x\to0$ , but since $x\to0$ from above we have $\tan x>0$ , which suggests the limit is $+\infty$ .

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3 years ago