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3 years ago

13

Which one answer fast !

2 answers:

max2010maxim [7]3 years ago

6 0

Answer:

A, 1/4

0.25 simplified is 1/4, as shown in the diagram.

Montano1993 [528]3 years ago

5 0

Answer:

A is the correct answer to your question

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I’ll give brainliest please help!

vodka [1.7K]

Answer:

ummm they have biography for them but i don't know they full answer

Step-by-step explanation:

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3 years ago

X / 8 = -2 <br> I don't understand it.

Alenkinab [10]

Answer:

x=-16

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

Find the equation for the line that passes through the point (-4,5) and that is perpendicular to the line with the equation-3/2x

IgorLugansk [536]

Answer:

y = -4/3x - 1/3

Step-by-step explanation:

-3/2x+2y=-7

2y = 3/2x - 7

y = 3/4x - 7/2

y = -4/3x + b

5 = -4/3(-4) + b

5 = 16/3 + b

-1/3 = b

4 0

2 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

4 years ago

1. 6 + 48 / 8* 7 = ?

gladu [14]

E.48 because 6+48/8(7) so 6+(6)(7) then do 6+42=48

8 0

3 years ago