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2 years ago

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Fre-e points 915 9482 0043 Gm1fgZ join

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UNO [17]2 years ago

7 0

Answer:

What is thiissssssssssssssssssss

Minchanka [31]2 years ago

6 0

Answer:

Thanks, but join what?

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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

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

8 0

3 years ago

Read 2 more answers

Which angle is an obtuse angle?<br> A. 310°<br> B. 80°<br> O c. 260°<br> O D. 170°

Murljashka [212]

Step-by-step explanation:

obtuse angle is an angle whose measure is more than 90 degree but less than 180 degree..

So,

170°

3 0

3 years ago

Read 2 more answers

If it takes 5 sewing machines 5 minutes to make 5 shirts, how long will it take for 100 sewing machines to make 100 shirts?

alekssr [168]

SOLUTION:

If it takes 5 sewing machines 5 minutes to make 5 shirts, then that means:

100 machines will make 100 shirts in 5 minutes as well.

Hope this helps! :)
Have a lovely day! <3

3 0

3 years ago

How do i solve 16x+9=9y-2x

Misha Larkins [42]

Answer:

18x+9=9y

Step-by-step explanation:

16x+9=9y-2x

+2x +2x

18x+9=9y

4 0

3 years ago

Read 2 more answers

Prove by mathematical induction that

postnew [5]

For $n=1$ , on the left we have $\cos\theta$ , and on the right,

$\dfrac{\sin2\theta}{2\sin\theta}=\dfrac{2\sin\theta\cos\theta}{2\sin\theta}=\cos\theta$

(where we use the double angle identity: $\sin2\theta=2\sin\theta\cos\theta$ )

Suppose the relation holds for $n=k$ :

$\displaystyle\sum_{n=1}^k\cos(2n-1)\theta=\dfrac{\sin2k\theta}{2\sin\theta}$

Then for $n=k+1$ , the left side is

$\displaystyle\sum_{n=1}^{k+1}\cos(2n-1)\theta=\sum_{n=1}^k\cos(2n-1)\theta+\cos(2k+1)\theta=\dfrac{\sin2k\theta}{2\sin\theta}+\cos(2k+1)\theta$

So we want to show that

$\dfrac{\sin2k\theta}{2\sin\theta}+\cos(2k+1)\theta=\dfrac{\sin(2k+2)\theta}{2\sin\theta}$

On the left side, we can combine the fractions:

$\dfrac{\sin2k\theta+2\sin\theta\cos(2k+1)\theta}{2\sin\theta}$

Recall that

$\cos(x+y)=\cos x\cos y-\sin x\sin y$

so that we can write

$\dfrac{\sin2k\theta+2\sin\theta(\cos2k\theta\cos\theta-\sin2k\theta\sin\theta)}{2\sin\theta}$

$=\dfrac{\sin2k\theta+\sin2\theta\cos2k\theta-2\sin2k\theta\sin^2\theta}{2\sin\theta}$

$=\dfrac{\sin2k\theta(1-2\sin^2\theta)+\sin2\theta\cos2k\theta}{2\sin\theta}$

$=\dfrac{\sin2k\theta\cos2\theta+\sin2\theta\cos2k\theta}{2\sin\theta}$

(another double angle identity: $\cos2\theta=\cos^2\theta-\sin^2\theta=1-2\sin^2\theta$ )

Then recall that

$\sin(x+y)=\sin x\cos y+\sin y\cos x$

which lets us consolidate the numerator to get what we wanted:

$=\dfrac{\sin(2k+2)\theta}{2\sin\theta}$

and the identity is established.

8 0

3 years ago