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Romashka-Z-Leto [24]
2 years ago
14

(x^3)^-5 solution please! 1/x^15 x^15 x^8 1/x^8

Mathematics
1 answer:
Sati [7]2 years ago
4 0
(x^3)^{-5}=\dfrac1{(x^3)^5}=\dfrac1{x^{3\times5}}=\dfrac1{x^{15}}
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How to write (1.5 x 10^-4) x (2.5 x 10^3) in scientific notation
maria [59]
<h2>Answer:</h2>

The answer to your question is: (4 x 10^-1)

7 0
3 years ago
Activity 4: Performance Task
Nookie1986 [14]

An arithmetic progression is simply a progression with a common difference among consecutive terms.

  • <em>The sum of multiplies of 6 between 8 and 70 is 390</em>
  • <em>The sum of multiplies of 5 between 12 and 92 is 840</em>
  • <em>The sum of multiplies of 3 between 1 and 50 is 408</em>
  • <em>The sum of multiplies of 11 between 10 and 122 is 726</em>
  • <em>The sum of multiplies of 9 between 25 and 100 is 567</em>
  • <em>The sum of the first 20 terms is 630</em>
  • <em>The sum of the first 15 terms is 480</em>
  • <em>The sum of the first 32 terms is 3136</em>
  • <em>The sum of the first 27 terms is -486</em>
  • <em>The sum of the first 51 terms is 2193</em>

<em />

<u>(a) Sum of multiples of 6, between 8 and 70</u>

There are 10 multiples of 6 between 8 and 70, and the first of them is 12.

This means that:

\mathbf{a = 12}

\mathbf{n = 10}

\mathbf{d = 6}

The sum of n terms of an AP is:

\mathbf{S_n = \frac n2(2a + (n - 1)d)}

Substitute known values

\mathbf{S_{10} = \frac{10}2(2*12 + (10 - 1)6)}

\mathbf{S_{10} = 390}

<u>(b) Multiples of 5 between 12 and 92</u>

There are 16 multiples of 5 between 12 and 92, and the first of them is 15.

This means that:

\mathbf{a = 15}

\mathbf{n = 16}

\mathbf{d = 5}

The sum of n terms of an AP is:

\mathbf{S_n = \frac n2(2a + (n - 1)d)}

Substitute known values

\mathbf{S_{16} = \frac{16}2(2*15 + (16 - 1)5)}

\mathbf{S_{16} = 840}

<u>(c) Multiples of 3 between 1 and 50</u>

There are 16 multiples of 3 between 1 and 50, and the first of them is 3.

This means that:

\mathbf{a = 3}

\mathbf{n = 16}

\mathbf{d = 3}

The sum of n terms of an AP is:

\mathbf{S_n = \frac n2(2a + (n - 1)d)}

Substitute known values

\mathbf{S_{16} = \frac{16}2(2*3 + (16 - 1)3)}

\mathbf{S_{16} = 408}

<u>(d) Multiples of 11 between 10 and 122</u>

There are 11 multiples of 11 between 10 and 122, and the first of them is 11.

This means that:

\mathbf{a = 11}

\mathbf{n = 11}

\mathbf{d = 11}

The sum of n terms of an AP is:

\mathbf{S_n = \frac n2(2a + (n - 1)d)}

Substitute known values

\mathbf{S_{16} = \frac{11}2(2*11 + (11 - 1)11)}

\mathbf{S_{11} = 726}

<u />

<u>(e) Multiples of 9 between 25 and 100</u>

There are 9 multiples of 9 between 25 and 100, and the first of them is 27.

This means that:

\mathbf{a = 27}

\mathbf{n = 9}

\mathbf{d = 9}

The sum of n terms of an AP is:

\mathbf{S_n = \frac n2(2a + (n - 1)d)}

Substitute known values

\mathbf{S_{9} = \frac{9}2(2*27 + (9 - 1)9)}

\mathbf{S_{9} = 567}

<u>(f) Sum of first 20 terms</u>

The given parameters are:

\mathbf{a = 3}

\mathbf{d = 3}

\mathbf{n = 20}

The sum of n terms of an AP is:

\mathbf{S_n = \frac n2(2a + (n - 1)d)}

Substitute known values

\mathbf{S_{20} = \frac{20}2(2*3 + (20 - 1)3)}

\mathbf{S_{20} = 630}

<u>(f) Sum of first 15 terms</u>

The given parameters are:

\mathbf{a = 4}

\mathbf{d = 4}

\mathbf{n = 15}

The sum of n terms of an AP is:

\mathbf{S_n = \frac n2(2a + (n - 1)d)}

Substitute known values

\mathbf{S_{15} = \frac{15}2(2*4 + (15 - 1)4)}

\mathbf{S_{15} = 480}

<u>(g) Sum of first 32 terms</u>

The given parameters are:

\mathbf{a = 5}

\mathbf{d = 6}

\mathbf{n = 32}

The sum of n terms of an AP is:

\mathbf{S_n = \frac n2(2a + (n - 1)d)}

Substitute known values

\mathbf{S_{32} = \frac{32}2(2*5 + (32 - 1)6)}

\mathbf{S_{32} = 3136}

<u>(g) Sum of first 27 terms</u>

The given parameters are:

\mathbf{a = 8}

\mathbf{d = -2}

\mathbf{n = 27}

The sum of n terms of an AP is:

\mathbf{S_n = \frac n2(2a + (n - 1)d)}

Substitute known values

\mathbf{S_{27} = \frac{27}2(2*8 + (27 - 1)*-2)}

\mathbf{S_{27} = -486}

<u>(h) Sum of first 51 terms</u>

The given parameters are:

\mathbf{a = -7}

\mathbf{d = 2}

\mathbf{n = 51}

The sum of n terms of an AP is:

\mathbf{S_n = \frac n2(2a + (n - 1)d)}

Substitute known values

\mathbf{S_{51} = \frac{51}2(2*-7 + (51 - 1)*2)}

\mathbf{S_{51} = 2193}

Read more about arithmetic progressions at:

brainly.com/question/13989292

4 0
2 years ago
Read 2 more answers
M/9+2/3=7/3<br><br> Solve for m.
Aleks04 [339]
<span>M/9+2/3=7/3
</span><span>M/9 =7/3 - 2/3
M/9 = 5/3
M = 5/3 x 9
M = 15</span>
3 0
3 years ago
I need help with all 4 <br> For # 31 - 34, solve the equation for the unknown variable.
Sati [7]
31 is g = -19
32 is x = -9
34 is w = -11
31:
g-(-7) = -12
g+7 = -12
g = -19
32:
-5x = 45 divide by -5
x = -9
33 looks like it has a print error so I can’t figure that out, I don’t know where the |-2| is supposed to be
34:
(w-1)/4 = -3 multiply by 4
w - 1 = -12
w = -11
3 0
3 years ago
Ryan is looking for the sum of the cube of a number, n, and 16 What is the sum Ryan is looking for if n=2?
mihalych1998 [28]

Answer: The sum =14

Step-by-step explanation:

Given that n=2

And

sum = cube of a number, n, and 16 which can be expressed mathematically as

n^3 + 6

2^3 + 6

8 + 6= 14

6 0
3 years ago
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