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2 years ago

Can someone pls help me

Mathematics

1 answer:

Yuri [45]2 years ago

6 0

Answer:

I'm not sure, but I'll try

Step-by-step explanation:

(Side)^2 = (Hypotenuse)^2 - (Side)^2

(Side)^2 = (9)^2 - ( √ 65 )^2

(Side)^2 = (81) - (65)

(Side)^2 = 16

Take root of answer

Side = √16

Side = 4

Answer = 4

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hi people i suck at math and is in 10th grade i have a question if you start at 5,2 and move 3 units left and 2 units up what do

irina [24]

Answer:

(2,4)

Step-by-step explanation:

left( - ) and right( + ) (x axes)

up( + ) and down( - ) (y axes)

(5,2) just add

5 - 3 = 2

2 + 2 = 4

Answer :

(2,4)

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3 years ago

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Dimas [21]

Step-by-step explanation:

v= 4/3πr³

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3 years ago

Which sum or difference identity would you use to verify that cos (180° - q) = -cos q?

Phantasy [73]

Answer:

$\cos (a-b)=\cos a \cos b+\sin a \sin b$

Step-by-step explanation:

Given : $\cos (180^{\circ}-q)=-\cos q$

We have to write which identity we will use to prove the given statement.

Consider $\cos (180^{\circ}-q)=-\cos q$

Take left hand side of given expression $\cos (180^{\circ}-q)$

We know

$\cos (a-b)=\cos a \cos b+\sin a \sin b$

Comparing , we get, a= 180° and b = q

Substitute , we get,

$\cos (180^{\circ}-q)=\cos 180^{\circ} \cos (q)+\sin q \sin 180^{\circ}$

Also, we know $\sin 180^{\circ}=0$ and $\cos 180^{\circ}=-1$

Substitute, we get,

$\cos (180^{\circ}-q)=-1\cdot \cos (q)+\sin q \cdot 0$

Simplify , we get,

$\cos (180^{\circ}-q)=-\cos (q)$

Hence, use difference identity to prove the given result.

7 0

3 years ago

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Help plz who ever helps they are nice

slava [35]

Im mean I’d be down to help

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2 years ago

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What is the prime factorization of 616?

Leni [432]

A would be the correct answer:)

Hope this helps!

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3 years ago

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