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3 years ago

12

Find the volume. Round to the nearest hundredth if necessary.

1 answer:

noname [10]3 years ago

6 0

Answer:

36 yd³

Step-by-step explanation:

The above solid shape given is a triangular prism.

The volume of triangular prism is given as ½ × base length of the triangle (b) × height of the triangle (h) × the length of the prism (l)

Base length of triangle (b) = 9 yd

Height of the triangle (h) = 2 yd

Length of the prism (l) = 4 yd

Volume = ½bhl

Volume = ½*9*2*4

Volume = 9*4

Volume of prism = 36 yd³

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Factor 4n 4 + n 3.<br><br> A) n(4n^3+1)<br> B) n^3(4n+1)<br> C) n^3(4n)

swat32

Honestly I think that it’s c but check that and get back to me

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3 years ago

Factor polnomial 4p^2−16p+12

iVinArrow [24]

Answer:

4(p−1)(p−3)

Step-by-step explanation:

Factor 4p2−16p+12

4p2−16p+12

=4(p−1)(p−3)

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3 years ago

Read 2 more answers

Given the function f(x)=3x-5, for which value in the domain does f(x)=-38?

victus00 [196]

$f(x) = -38\\\\\implies 3x -5 = -38\\\\\implies 3x = -38 +5 \\\\\implies 3x = -33\\\\\implies x = -\dfrac{33}{3}\\\\\implies x = -11$

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2 years ago

Dunya randomly chooses a number from 1 to 10. What is the probability she

kumpel [21]

Answer: 1/2

Step-by-step explanation:

The numbers that are less than 6 are 1, 2, 3, 4, and 5. Thus simply means that there are 5 numbers that are less than 6.

The probability that she chooses a number less than 6 will then be:

= 5/10

= 1/2

7 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