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dimulka [17.4K]
3 years ago
5

ILL MARK BRAINLIST

Mathematics
2 answers:
Alik [6]3 years ago
7 0

Answer:

vertical angles

Step-by-step explanation:

hope this helps

Mkey [24]3 years ago
5 0

Answer:

vertical angles

Step-by-step explanation:

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What is the value of the expression shown? 4(-2) + (-10) + 3(-8)
Vlad [161]

Answer:

-42

Step-by-step explanation:

1. 4(-2)

2. 3(-8)

3. -8 + -10 + -24

4. cobine like terms

5. -42

7 0
2 years ago
Ramapo College has a total of 25,000 students enrolling this year. 42% of
salantis [7]

Answer: im sorry im not sure but you can find the answer if you look it up

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
cos theta = 4/5, 0˚< theta < 90˚ use the information given to find sin2theta, cos2theta, and tan 2theta
NikAS [45]
First calculate \sin\theta.

\sin^2\theta+\cos^2\theta=1\\\\\sin^2\theta=1-\cos^2\theta\\\\\sin^2\theta=1-\left(\dfrac{4}{5}\right)^2\\\\\\\sin^2\theta=1-\dfrac{16}{25}\\\\\\\sin^2\theta=\dfrac{9}{25}\quad|\sqrt{(\dots)}\\\\\\\sin\theta=\sqrt{\dfrac{9}{25}}\\\\\\\boxed{\sin\theta=\dfrac{3}{5}}

We take positive value of sinθ because 0° < θ < 90°
Now we could calculate sin2θ, cos2θ and tan2θ:

\sin2\theta=2\sin\theta\cos\theta=2\cdot\dfrac{3}{5}\cdot\dfrac{4}{5}=\dfrac{2\cdot3\cdot4}{5\cdot5}=\dfrac{24}{25}\\\\\\\cos2\theta=\cos^2\theta-\sin^2\theta=\left(\dfrac{4}{5}\right)^2-\left(\dfrac{3}{5}\right)^2=\dfrac{16}{25}-\dfrac{9}{25}=\dfrac{7}{25}\\\\\\&#10;\tan2\theta=\dfrac{\sin2\theta}{\cos2\theta}=\dfrac{\frac{24}{25}}{\frac{7}{25}}=\dfrac{24\cdot25}{7\cdot25}=\dfrac{24}{7}=3\dfrac{3}{7}
5 0
3 years ago
Read 2 more answers
Please help. And give an explanation. I am so lost.
Verdich [7]

2 opposite interior angle (opposite to the exterior angle) added together are equal to the exterior angle

6x-1+85 = 18x

6x +84 = 18x

subtract 6x from both sides

84 = 12x

divide both sides by 12

x = 7

<em>hope this helps! </em>

8 0
3 years ago
Find x in the following two irregular similar polygons.
Levart [38]
The answer is C ....
7 0
3 years ago
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