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NARA [144]
3 years ago
6

Solve for x 16x = 20(12)

Mathematics
1 answer:
marysya [2.9K]3 years ago
4 0

Answer: X=15

Step-by-step explanation: Isolate the variable by dividing each side by factors that don't contain the variable.

Hope this helps!! :)

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What is the answer to this eqaution 20 ÷ 15/16
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Answer:

21 1/3

Step-by-step explanation:

20 ÷ 15/16=

20/1 ÷ 15/16=

20/1*16/15=

4/1*16/3=

64/3=

21 1/3

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Mark put his dog on a diet. The dog's total weight change for the first two weeks was -3/4 pound. The dog lost the same amount o
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The change in the first week is .375 pounds =. Divide 3/4 by 2
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30 POINTS PLEASE ANSWER CORRECTLY AND CLEARLY THANK YOU!
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B

Step-by-step explanation:

The investment grows by 12 percent every year. Therefore, if you multiply the 1500 by 0.12 you should get the value of the amount of extra money made in one year. If you substitute 2 into t you get the value increased by year 2 and so on.

5 0
3 years ago
Read 2 more answers
Use multiplication or division of power series to find the first three nonzero terms in the Maclaurin series for each function.
Lunna [17]

Answer:

The first three nonzero terms in the Maclaurin series is

\mathbf{ 5e^{-x^2} cos (4x)  }= \mathbf{ 5 ( 1 -9x^2 + \dfrac{115}{6}x^4+ ...) }

Step-by-step explanation:

GIven that:

f(x) = 5e^{-x^2} cos (4x)

The Maclaurin series of cos x can be expressed as :

\mathtt{cos \ x = \sum \limits ^{\infty}_{n =0} (-1)^n \dfrac{x^{2n}}{2!} = 1 - \dfrac{x^2}{2!}+\dfrac{x^4}{4!}-\dfrac{x^6}{6!}+...  \ \ \ (1)}

\mathtt{e^{-2^x} = \sum \limits^{\infty}_{n=0}  \ \dfrac{(-x^2)^n}{n!} = \sum \limits ^{\infty}_{n=0} (-1)^n \ \dfrac{x^{2n} }{x!} = 1 -x^2+ \dfrac{x^4}{2!}  -\dfrac{x^6}{3!}+... \ \ \  (2)}

From equation(1), substituting x with (4x), Then:

\mathtt{cos (4x) = 1 - \dfrac{(4x)^2}{2!}+ \dfrac{(4x)^4}{4!}- \dfrac{(4x)^6}{6!}+...}

The first three terms of cos (4x) is:

\mathtt{cos (4x) = 1 - \dfrac{(4x)^2}{2!}+ \dfrac{(4x)^4}{4!}-...}

\mathtt{cos (4x) = 1 - \dfrac{16x^2}{2}+ \dfrac{256x^4}{24}-...}

\mathtt{cos (4x) = 1 - 8x^2+ \dfrac{32x^4}{3}-... \ \ \ (3)}

Multiplying equation (2) with (3); we have :

\mathtt{ e^{-x^2} cos (4x) = ( 1- x^2 + \dfrac{x^4}{2!} ) \times ( 1 - 8x^2 + \dfrac{32 \ x^4}{3} ) }

\mathtt{ e^{-x^2} cos (4x) = ( 1+ (-8-1)x^2 + (\dfrac{32}{3} + \dfrac{1}{2}+8)x^4 + ...) }

\mathtt{ e^{-x^2} cos (4x) = ( 1 -9x^2 + (\dfrac{64+3+48}{6})x^4+ ...) }

\mathtt{ e^{-x^2} cos (4x) = ( 1 -9x^2 + \dfrac{115}{6}x^4+ ...) }

Finally , multiplying 5 with \mathtt{ e^{-x^2} cos (4x) } ; we have:

The first three nonzero terms in the Maclaurin series is

\mathbf{ 5e^{-x^2} cos (4x)  }= \mathbf{ 5 ( 1 -9x^2 + \dfrac{115}{6}x^4+ ...) }

7 0
3 years ago
Estimate:________<br>74,529<br>- 38453​
kiruha [24]

Answer:

35,000?

Step-by-step explanation:

3 0
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