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Nataly_w [17]
2 years ago
11

Rewrite using exponential notation \root(6)((4\pi )^(3))

Mathematics
1 answer:
Nezavi [6.7K]2 years ago
5 0

~\hspace{7em}\textit{rational exponents} \\\\ a^{\frac{ n}{ m}} \implies \sqrt[ m]{a^ n} ~\hspace{10em} a^{-\frac{ n}{ m}} \implies \cfrac{1}{a^{\frac{ n}{ m}}} \implies \cfrac{1}{\sqrt[ m]{a^ n}} \\\\[-0.35em] ~\dotfill\\\\ \sqrt[6]{(4\pi )^3}\implies (4\pi )^{\frac{3}{6}}\implies (4\pi )^{\frac{1}{2}}

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Translate the following into an algebraic expression: A number is 30% of 20% of the number x.
bezimeni [28]

Answer: 0.06x

Step-by-step explanation:

  • An algebraic expression is an expression consist of integer constants, variables, and algebraic operations.

The given statement:  A number is 30% of 20% of the number x.

The required algebraic expression would be:

30% of 20% of x

=\dfrac{30}{100}\times \dfrac{20}{100}\times x  [we divide a percentage by 100 to convert it into decimal]

=\dfrac{6}{100}\times x\\\\=0.06x

Hence, the required algebraic expression would be :

0.06x

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3 years ago
Function or not a Function​
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Answer:

Function; Not a function

Step-by-step explanation:

Write the ordered pair in the first set (x, y):     (1, 11), (2, 7) , (3, 2)

Write the ordered pair in the second set (x, y):     (1, 3), (1, 4),  (2, 4) , (3, 2)

The first set in a function because no ordered pair has the same x value.

The second set in not a function because the x value is the same in two of the ordered pairs.

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Find the area of the trapezoid by decomposing it into other shapes.
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Please help.... <br> don’t understand
BlackZzzverrR [31]

5.\\a_0=24\\a_1=a_0+7\to a_1=24+7=31\\a_2=a_1+7\to a_2=31+7=38\\a_3=a_2+7\to a_3=38+7=45\\a_4=a_3+7\to a_4=45+7=52\\\vdots\\a_n=a_{n-1}+7\\\\Answer:\ \left\{\begin{array}{ccc}a_0=24\\a_{n}=a_{n-1}+7\end{array}\right

6..\\a_0=-5\\a_1=a_0\cdot(-6)\to a_1=-5(-6)+30\\a_2=a_1\cdot(-6)\to a_2=30(-6)=-180\\a_3=a_2\cdot(-6)\to a_3=-180(-6)=1080\\a_4=a_3\cdot(-6)\to a_4=1040(-6)=-6480\\\vdots\\a_n=(-6)a_{n-1}\\\\Answer:\ \left\{\begin{array}{ccc}a_0=-5\\a_n=-6a_{n-1}\end{array}\right

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