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vodka [1.7K]
3 years ago
10

Can someone help me graph y = 2x + 4

Mathematics
1 answer:
katen-ka-za [31]3 years ago
8 0
Y-intercept (0,4)
Your welcome :)
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What is 13.5/10 simplified
andrezito [222]

Answer:

27/20

Step-by-step explanation:

\frac{13.5}{10} = \frac{13.5 * 2}{10 * 2} = \frac{27}{20}

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A number cube labeled 1 to 6 is rolled 450 times. Which is the best prediction for the number of times that the number cube will
Vedmedyk [2.9K]
Haha, I don't think those answers belong to the question asked :P

The probability of rolling less than 3 is 2/6=1/3

So if you rolled 450 times you should expect to roll less than 3:

450(1/3)=150 times.
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3 years ago
Hurry!! Worth 10 points
Darina [25.2K]

Answer:

its D

Step-by-step explanation:

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3 years ago
18 miles is what percent of 24 miles?
Gekata [30.6K]

Answer:

3/4

Cause there are 3 sixes in 18 and 4 sixes in 24

4 0
3 years ago
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
3 years ago
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