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Bad White [126]
3 years ago
11

Please help me. i will give brainliest to the right answer<3

Mathematics
1 answer:
lord [1]3 years ago
3 0

Answer:

-2/3 and 5/9 are the only ratonal numbers resulting in a repeating decimal.

Mt Logic:

just covert the fractions to decimals. and the ones that are already repeating are not ratonal numbers.

Hope this helps!

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Find the midpoint of the segment with the following endpoints (-5,-1) and (1,9)
Bogdan [553]

Answer:

(x1,y1),(x2,y2)

(x1,y1)=(-5,-1)   (x2,y2)=(1,9)= (1-5/2, 9-1/2)

your answer:

(-2,4)

Step-by-step explanation:

4 0
3 years ago
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nadezda [96]
6the answer will be b 

7 0
3 years ago
Y=4x+8<br> Y=5x+1<br><br> I cant figure out this math problem
Crank
If solving for x:
Because both equations are equal to Y, you can set both equations equal to each other:
5x+1=4x+8
Then it's a matter of simplification (subtract the 4x to the left side and subtract the 1 over to the right):
x=7
If solving for Y:
Plug x=7 into either equation and solve:
Y=4(7)+8
Y=36
4 0
3 years ago
Plz Help!! <br><br> Factor the Polynomial: -6x² - 15x + 9
Salsk061 [2.6K]

Answer:

= -3(x+3)x(2x-1)

Step-by-step explanation:

5 0
4 years ago
Read 2 more answers
Simplify this equation please.
Sauron [17]
\dfrac{\csc^2\theta-3\csc\theta+2}{\csc^2\theta-1}

Identity:

\sin^2\theta+\cos^2\theta=1\implies1+\cot^2\theta=\csc^2\theta

So we can rewrite the denominator to get

\dfrac{\csc^2\theta-3\csc\theta+2}{\cot^2\theta}

Multiply numerator and denominator by \sin^2\theta. Several terms will cancel since \sin\theta\csc\theta=1. Also, \cot\theta=\dfrac{\cos\theta}{\sin\theta}. We get

\dfrac{1-3\sin\theta+2\sin^2\theta}{\cos^2\theta}

Factorize the numerator, and write \cos in terms of \sin in the denominator to factorize it further to get

\dfrac{(1-\sin\theta)(1-2\sin\theta)}{\cos^2\theta}=\dfrac{(1-\sin\theta)(1-2\sin\theta)}{1-\sin^2\theta}=\dfrac{(1-\sin\theta)(1-2\sin\theta)}{(1-\sin\theta)(1+\sin\theta)}


The 1-\sin\theta factors cancel, leaving you with

\dfrac{1-2\sin\theta}{1+\sin\theta}

which you could simplify a bit further by writing

\dfrac{1+\sin\theta-3\sin\theta}{1+\sin\theta}=1-\dfrac{3\sin\theta}{1+\sin\theta}
3 0
4 years ago
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