All categories

3 years ago

What is the answer to this problem shown?

Mathematics

1 answer:

GrogVix [38]3 years ago

3 0

There's no problem there??

You might be interested in

Please help me for the love of God if i fail I have to repeat the class

Elena-2011 [213]

$\theta$ is in quadrant I, so $\cos\theta>0$ .

$x$ is in quadrant II, so $\sin x>0$ .

Recall that for any angle $\alpha$ ,

$\sin^2\alpha+\cos^2\alpha=1$

Then with the conditions determined above, we get

$\cos\theta=\sqrt{1-\left(\dfrac45\right)^2}=\dfrac35$

and

$\sin x=\sqrt{1-\left(-\dfrac5{13}\right)^2}=\dfrac{12}{13}$

Now recall the compound angle formulas:

$\sin(\alpha\pm\beta)=\sin\alpha\cos\beta\pm\cos\alpha\sin\beta$

$\cos(\alpha\pm\beta)=\cos\alpha\cos\beta\mp\sin\alpha\sin\beta$

$\sin2\alpha=2\sin\alpha\cos\alpha$

$\cos2\alpha=\cos^2\alpha-\sin^2\alpha$

as well as the definition of tangent:

$\tan\alpha=\dfrac{\sin\alpha}{\cos\alpha}$

Then

1. $\sin(\theta+x)=\sin\theta\cos x+\cos\theta\sin x=\dfrac{16}{65}$

2. $\cos(\theta-x)=\cos\theta\cos x+\sin\theta\sin x=\dfrac{33}{65}$

3. $\tan(\theta+x)=\dfrac{\sin(\theta+x)}{\cos(\theta+x)}=-\dfrac{16}{63}$

4. $\sin2\theta=2\sin\theta\cos\theta=\dfrac{24}{25}$

5. $\cos2x=\cos^2x-\sin^2x=-\dfrac{119}{169}$

6. $\tan2\theta=\dfrac{\sin2\theta}{\cos2\theta}=-\dfrac{24}7$

7. A bit more work required here. Recall the half-angle identities:

$\cos^2\dfrac\alpha2=\dfrac{1+\cos\alpha}2$

$\sin^2\dfrac\alpha2=\dfrac{1-\cos\alpha}2$

$\implies\tan^2\dfrac\alpha2=\dfrac{1-\cos\alpha}{1+\cos\alpha}$

Because $x$ is in quadrant II, we know that $\dfrac x2$ is in quadrant I. Specifically, we know $\dfrac\pi2$ , so $\dfrac\pi4$ . In this quadrant, we have $\tan\dfrac x2>0$ , so

$\tan\dfrac x2=\sqrt{\dfrac{1-\cos x}{1+\cos x}}=\dfrac32$

8. $\sin3\theta=\sin(\theta+2\theta)=\dfrac{44}{125}$

6 0

4 years ago

15X^2+20x factor greatest common monomial

velikii [3]

Both 15 and 20 can be divided evenly by 5, so 5 is the greatest common monomial: 5(3x^2 + 4x) or 5x(3x + 4)

4 0

4 years ago

<img src="https://tex.z-dn.net/?f=f%28x%29%20%3D%20%28x%20-%20%20%5Cfrac%7B2%7D%7B9%7D%20%29%28x%20%2B%20%20%5Cfrac%7B1%7D%7B2%7

Aleonysh [2.5K]

$f(x) = (x - \dfrac{2}{9} )(x + \dfrac{1}{2} )$

$\text {When } f(x) = 0 :$

$(x - \dfrac{2}{9} )(x + \dfrac{1}{2} ) = 0$

$(x - \dfrac{2}{9} ) = 0 \text { or } (x + \dfrac{1}{2} ) = 0$

$x = \dfrac{2}{9} \text { or } x = -\dfrac{1}{2}$

3 0

4 years ago

PLZ HELP WILL GIVE BRAINLEST HELP ASAP

mestny [16]

Answer:

176 feet

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

Which of the following represents the graph of f(x) = 2(x + 3)?

Yuliya22 [10]