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barxatty [35]
3 years ago
8

PLEASE PLEASE PLEASE HELP!! Explanations on how to set it up would be great!

Mathematics
1 answer:
shusha [124]3 years ago
3 0
AD = BC
x+2 = 2x - 3
2x -x = 2+3
x = 5

AB =  DC
2y +1 = 3x - 2
2y + 1 = 3(5) - 2
2y + 1 = 13
2y = 13 - 1
2y = 12
y = 6

answer: x = 5 and y = 6
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area of a square prism is 2×a square +4ah

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Please help! I’ll give brainlest. Idk the answer bruh it’s for a grade :/
slamgirl [31]

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A

Step-by-step explanation:

Process of elimination.

It can't be B or D cause those say that the temp go down while in your graph it goes up, and C is saying that it rises and becomes constant while in your graph it's the other way around.

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3 years ago
Read 2 more answers
Photo has question! Need answer ASAP
galben [10]

Answer:

My best guess would be B and C since those are the only numbers that aren't multiples of 5, and all the values of a(n) are.

7 0
2 years ago
I need help proving this ASAP
Ket [755]

Answer:

See explanation

Step-by-step explanation:

We want to show that:

\tan(x +  \frac{3\pi}{2} )  =  -   \cot \: x

One way is to use the basic double angle formula:

\frac{ \sin(x +  \frac{3\pi}{2} ) }{\cos(x +  \frac{3\pi}{2} )}  =  \frac{ \sin(x)  \cos( \frac{3\pi}{2} )  +   \cos(x)  \sin( \frac{3\pi}{2}) }{\cos(x)  \cos( \frac{3\pi}{2} )   -    \sin(x)  \sin( \frac{3\pi}{2}) }

\frac{ \sin(x +  \frac{3\pi}{2} ) }{\cos(x +  \frac{3\pi}{2} )}  =  \frac{ \sin(x) ( 0)  +   \cos(x) (  - 1) }{\cos(x) (0)   -    \sin(x) (  - 1) }

We simplify further to get:

\frac{ \sin(x +  \frac{3\pi}{2} ) }{\cos(x +  \frac{3\pi}{2} )}  =  \frac{ 0  -   \cos(x) }{0 +    \sin(x) }

We simplify again to get;

\frac{ \sin(x +  \frac{3\pi}{2} ) }{\cos(x +  \frac{3\pi}{2} )}  =  \frac{- \cos(x) }{ \sin(x) }

This finally gives:

\frac{ \sin(x +  \frac{3\pi}{2} ) }{\cos(x +  \frac{3\pi}{2} )}  =  -  \cot(x)

6 0
3 years ago
-45*5x+3x=666<br><br> A) -2<br> B) -3<br> C) 2<br> D) 3
Ivahew [28]

Answer: -3

Step-by-step explanation: Explanation is below

:) Hopes this helps!

3 0
3 years ago
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