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san4es73 [151]
3 years ago
5

What is 0.58 as a fraction

Mathematics
2 answers:
zvonat [6]3 years ago
7 0
0.58
= 58/100
= 29/50
wlad13 [49]3 years ago
3 0
Here is the answer: 58 over 100, hope it helps.
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Original price: $35<br> Percent of discount:<br> Sale price: $31.50
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Answer:

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Step-by-step explanation:

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3 years ago
On a floor plan for Rosedale Middle School, 1 inch represents 10 feet. If Sarah's classroom is 2 inches by 3 inches on the floor
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Answer:

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Step-by-step explanation:

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2 years ago
Solve for x in the following angle relationship. Please also identify the type of angles given:
Nuetrik [128]

9514 1404 393

Answer:

  x= 10, corresponding

Step-by-step explanation:

Corresponding angles are in the same direction from their respective intersections. Here, the angles are both northwest of the point where the lines cross. The angles are corresponding.

__

Corresponding angles are congruent, so we have ...

  10x +10 = 110

  10x = 100 . . . . . subtract 10

  x = 10 . . . . . . . . divide by 10

6 0
2 years ago
Andy Dixon bought 3 paintbrushes for $1.75 each and 7 tubes of oil paint for $2.49 each.how much did he spend in all?
AnnZ [28]
The answer would be $22.68!

If Andy bought 3 paintbrushes for $1.75 each, then we just multiply how much they cost by the number that the bought.

1.75 x 3 = 5.25

If Andy bought 7 tubes of oil paint, we just do the same thing we did in the first step! Multiply how much they cost by the number that the bought.

2.49 x 7 = 
<span>17.43</span>
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
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