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Leona [35]
2 years ago
14

Marc has 45 books on his bookshelf, and Corinna has 61. If Marc buys 4 new books each month and Corinna buys 2 new books each mo

nth, after how many months will Marc and Corinna have the same number of books?
Define the variable:_________________________________________



Write the equation: _________________________________________


Solve the equation in the space below. Show your work
Mathematics
1 answer:
luda_lava [24]2 years ago
7 0
Call x as the number of months that Marc and Corinna have the same number of book.
45 +4x= 61 +2x
4x- 2x= 61-45
2x= 16
x= 8
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tino4ka555 [31]
Axis of symmetry: x=1
Vertex: (1, -24)

Explanation:

6x²-12x-25=-7

6x²-12x-18=0

a=6, b=-12, c=-18

Formula of axis of symmetry: x=-b/2a

x=-b/2a

x= -(-12)/2(6)

x= 12/12

x= 1

Vertex: (1,?)

Sub x= 1 into the equation,
6(1)² - 12(1) - 18 = y

6 - 12 - 18 = y

y = -24



4 0
3 years ago
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Generate an equivalent expression using prime factorization for -60. show your workk ​
Darina [25.2K]

Answer:

60 70 80m9

Step-by-step explanation:

4 0
3 years ago
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+ ...) }

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3 years ago
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worty [1.4K]

Volume of piece candy is 113.04 cubic centimeter

<em><u>Solution:</u></em>

A student bought a piece of hard candy that is in the shape of a sphere

<em><u>The candy has a radius of 3 centimeters</u></em>

Radius = 3 cm

Use 3.14 for π

<em><u>Volume of sphere is given as:</u></em>

V = \frac{4}{3} \times \pi \times r^3

Where, "r" is the radius of sphere

<em><u>Therefore, volume of candy is given as:</u></em>

V = \frac{4}{3} \times 3.14 \times 3^3\\\\V = 4 \times 3.14 \times 3^2\\\\V = 12.56 \times 9\\\\V = 113.04

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