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3 years ago

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

Help me :}

nadezda [96]

6the answer will be b

7 0

3 years ago

Y=4x+8 Y=5x+1 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!! 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