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Y_Kistochka [10]
3 years ago
5

Describe how to transform the quantity of the fifth root of x to the seventh power, to the third powerinto an expression with a

rational exponent. Make sure you respond with complete sentences.
Mathematics
1 answer:
geniusboy [140]3 years ago
7 0
(\sqrt[5]{x^{7}})^{3}=(x^{\frac{7}{5}})^{3}=x^{\frac{7\cdot3}{5}}=x^{\frac{21}{5}}

The root is equivalent to a fractional power with that number as the denominator. Otherwise, the rules of exponents apply.
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bija089 [108]

Answer:

0.9544    

Step-by-step explanation:

P(-2 < Z < 2) means that Z has mean 0 and standard deviation 2.

P(−2 < Z < 2) = F(2) - F(-2)

Using the Z - table,

F(2) = 0.9772

and F(-2) = 0.0228

Thus,

P(−2 < Z < 2) = 0.9772 - 0.0228 = 0.9544

This means that data within two standard deviation is 95%.

8 0
3 years ago
(ab^2 x -b^4a^3)^2<br><br> What is the answers
Umnica [9.8K]

Answer:

\left(ab^2x-b^4a^3\right)^2:\quad a^2b^4x^2-2a^4b^6x+a^6b^8

Step-by-step explanation:

Given the expression

\left(ab^2\:x\:-b^4a^3\right)^2

solving the expression

\left(ab^2\:x\:-b^4a^3\right)^2

\mathrm{Apply\:Perfect\:Square\:Formula}:\quad \left(a-b\right)^2=a^2-2ab+b^2

a=ab^2x,\:\:b=b^4a^3

so the expression becomes

=\left(ab^2x\right)^2-2ab^2xb^4a^3+\left(b^4a^3\right)^2

=a^2b^4x^2-2a^4b^6x+a^6b^8

Thus,

\left(ab^2x-b^4a^3\right)^2:\quad a^2b^4x^2-2a^4b^6x+a^6b^8

3 0
3 years ago
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irina1246 [14]

Answer:

1. ASA

2. AAS

3. AAS

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5 0
3 years ago
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zloy xaker [14]

Answer:

  26

Step-by-step explanation:

The two angles shown are a linear pair. Their sum is 180°.

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  7z = 182 . . . . . . . . . divide by °, add 2

  z = 26 . . . . . . . . . . divide by 7

3 0
3 years ago
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Ivenika [448]

the balance would probably be $37

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