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posledela
3 years ago
11

Which is a perfect square?

Mathematics
2 answers:
Leni [432]3 years ago
7 0

Answer:

I believe it's 6^2.

Step-by-step explanation: 6^2 is the same as 6 times 6. Here's a tip, if you see a 2 as an exponent, you can also say x squared. X to the power of 2 is the same as x squared.

Lena [83]3 years ago
5 0

Answer:

6^{2}

Step-by-step explanation:

6^{1} = 6        \sqrt{6} = 2.45 (NOT a perfect square)

6^{2} = 36      \sqrt{36} = 6 (perfect square)

6^{3} = 216     \sqrt{216} = 14.7 (NOT a perfect square)

6^{5} = 7776   \sqrt{7776} = 88.18 (NOT a perfect square)

If the answer ends with a decimal, they are not perfect squares.

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Use substitution to solve the<br> following system of equations.<br> - 4y - 5x = 30 AND x= y + 3
IRISSAK [1]

Answer:

The solution to the system of equations is y = -5 and x = -2.

Step-by-step explanation:

The question tells us to use substitution to solve the system.  This means that the given value for x (in terms of y) should be substituted into the other equation.  This is modeled below:

-4y - 5x = 30

-4y - 5(y+3) = 30

Next, we should use the distributive property to simplify the left side of the equation.

-4y -5y - 15 = 30

The next step is to combine like terms on the left side of the equation.

-9y - 15 = 30

Then, we can add 15 to both sides of the equation.

-9y = 45

Finally, we can divide both sides of the equation by -9.

y = -5

To find the value for x, we substitute in the value we just found for y into either of our original equations.

x = y + 3

x = -5 + 3

x = -2

Therefore, the correct answer is y = -5 and x = -2.

Hope this helps!

3 0
3 years ago
Prove that<br>{(tanθ+sinθ)^2-(tanθ-sinθ)^2}^2 =16(tanθ+sinθ)(tanθ-sinθ)
USPshnik [31]

First, expand the terms inside the bracket you will get

(( \tan {}^{2} (x)  + 2 \tan(x)  \sin(x)  +  \sin {}^{2} (x)  - ( \tan {}^{2} (x)  - 2 \tan(x)  +  \sin {}^{2} (x) ) {}^{2}  = 16( \tan(x)  +  \sin(x) )( \tan(x)  -  \sin(x) )

( 4 \tan(x)  \sin(x) ) {}^{2}  = 16( \tan(x)  +  \sin(x) )( \tan(x)  -  \sin(x) )

16 \tan {}^{2} (x)  \sin {}^{2} (x)  = 16( \tan(x)  +  \sin(x) )( \tan(x)  -  \sin(x) )

16 \tan {}^{2} (x) (1 -  \cos {}^{2} (x) ) = 16 (\tan(x)  +  \sin(x) )( \tan(x)  -  \sin(x) )

16( \tan {}^{2} (x)  -   \frac{  \sin {}^{2} (x) \cos {}^{2} ( {x}^{} )  }{ \cos {}^{2} (x) }

16( \tan {}^{2} (x)  -  \sin {}^{2} (x) ) = 16( \tan(x)  +  \sin(x) )( \tan(x)  -  \sin(x) )

16( \tan(x)  +  \sin(x) )( \tan(x)  -  \sin(x)  = 16( \tan(x)  +  \sin(x) )( \tan(x)  -  \sin(x) )

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2 years ago
Xy=22 as a proportion please! :)
Elina [12.6K]
I’m sorry if I’m wrong but I do think your answer is X=22y^-1
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Step-by-step explanation:

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