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antoniya [11.8K]

3 years ago

14

Can I get some help on this pls

1 answer:

defon3 years ago

6 0

2.) 3
3.) 3723
4.) 1512
5.)4692
6.)5525
7.) 760
8.) 696

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Hour how much does davon charge per hour for a skateboard rental HELPP I BEG

Setler79 [48]

Answer:

$7

Davon charges $7 per hour of rental

6 0

3 years ago

This is due in a hour

Dimas [21]

Formula- A = π r²
a= area
r= radius

diameter of the circle- 3.75 x 2 = 7.5

hope this helps :)

6 0

3 years ago

The height, h, in feet of the tip of the hour hand of a wall clock varies from 9 feet to 10 feet. Which of the following equatio

marshall27 [118]

Y=Acos(p)+m, A=amplitude, p=period, m=midline, in this case:

A=1/2, p=360(t/12)=30t, m=(10-9)/2+9=9.5 so

h(t)=(1/2)cos(30t)+9.5

4 0

3 years ago

Read 2 more answers

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

The 100th term of 70, 100, 130

fgiga [73]

Answer:

3040

Step-by-step explanation:

given arithmetic progression is

70,100,130,...

here

first term (a)=70

common difference (d)=100-70=30

number of term n=100

using the formula of arithmetic progression

an=a+(n-1)d

a100=70+(100-1)30

a100=70+99×30

a100=70+2970

a100=3040

6 0

2 years ago