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3 years ago

Brainliest, what is the total measure of angles 8 and 5 of angle 7 equals 61

Mathematics

1 answer:

Nata [24]3 years ago

8 0

Angle 6 = angle 7 = 61⁰, as vertical angles
Angle 5 = angle 8, as vertical angles

angle 6+angle 7 + angle 5 + angle 8 =360
61+61 + angle 5 + angle 8 =360
angle 5 + angle 8 = 360 - 61*2 = 238⁰

Answer D. 238⁰

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Consider this sphere inside the cylinder. Which statements are true? Check all that apply. NEED HELP ASAP

-Dominant- [34]

Option A: The height of the cylinder is equal to the diameter of the sphere.

Option C: The radius of the sphere is half the height of the cylinder.

Option E: The volume of the sphere is two-thirds the volume of the cylinder.

Solution:

The sphere is inside the cylinder.

Let r be the radius of the sphere.

Option A: The height of the cylinder is equal to the diameter of the sphere.

The sphere is fully occupied the cylinder.

If we draw the vertical line through enter of the sphere, which is equal to the height of cylinder.That is h = d. It is true.

Option B: The height of the cylinder is two times the diameter of the sphere.

That is h = 2d. From the above option, we know that h = d.

So, it is not true.

Option C: The radius of the sphere is half the height of the cylinder.

we know that diameter = 2 × radius (d = 2r)

From option A, we have h = d.

Substitute d = 2r, we get

⇒ h = 2r

Divide by 2 on both sides, we get

$$\Rightarrow \frac{h}{2}=r$

Therefore, it is true.

Option D: The diameter of the sphere is equal to the radius of the cylinder.

It is not true, because diameters of both cylinder and sphere are equal.

Option E: The volume of the sphere is two-thirds the volume of the cylinder.

Radius of cylinder and sphere = r

Height of cylinder h = 2r (by option C)

Volume of cylinder = Volume of sphere

$$\Rightarrow \pi r^2 h = \frac{4}{3} \pi r^3$ (formula)

$$\Rightarrow \pi r^2 (2r) = \frac{2\times2}{3} \pi r^3$

$$\Rightarrow 2 \pi r^3= \frac{2}{3} \times 2\pi r^3$

Hence it is true.

Option A, option C and option E are true.

6 0

3 years ago

Read 2 more answers

Factor: 128 + 2d^3 <br><br> cubes lol

BARSIC [14]

The factorization if the given expression 128 + 2d³ is 2(64 + d³).

<h3>Factorization</h3>

128 + 2d³

There are two parameters in the expression;

128
2d³

The common factors between 128 and 2d³ is 2

So,

128 + 2d³

= 2(64 + d³)

Learn more about factorization:

brainly.com/question/25829061

#SPJ1

4 0

2 years ago

Im giving extra points and brainliest

Veseljchak [2.6K]

Answer: the c

Step-by-step explanation:

5 0

2 years ago

Read 2 more answers

John used AABC to write a proof of the Centroid Theorem. He began by drawing medians AK and CL,

kumpel [21]

Answer:

Hello your question is poorly written attached below is the complete question

answer : attached below

Step-by-step explanation:

To Prove: Z is located 2/3 of the distance from each vertex of ΔABC to the midpoint of the opposite side. we will apply ; property of bisecting a line , equality theorem , transitive property and similarity theorem

Attached below is the proof

8 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