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2 years ago

https://lh5.googleusercontent.com/nNcMGuld4KbRoBooDf5Liva-NAJyj-IEknnSniAdiBKPNhKWtKqEw2SwN2uQUidgVX1Vh4rDHA=w739

Mathematics

2 answers:

fgiga [73]2 years ago

7 0

Answer:

The answer is F: 108

Step-by-step explanation:

144 - 108 = 36

108 ÷ 3 = 36

Marta_Voda [28]2 years ago

6 0

Answer:

giytg8h0ygrubigrfbk

Step-by-step explanation:

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What is the perimeter of a square, if the area is 9<br> square units?

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2 years ago

A line with a slope of 3 passes through the point (2, 5). Write an equation for this line in point-slope form.

Phantasy [73]

Answer:

slope = 3

point (2, 5)

b = y - m*x

b = 5 -3*2

b = -1

Then we enter the slope and the value of b into this equation:

y = mx + b

y = 3x -1

Source: https://www.1728.org/distance.htm

Step-by-step explanation:

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3 years ago

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During an experiment, Juan rolled a six-sided number cube 18 times. The number two occurred four times. Juan claimed the experim

dem82 [27]

Because u would have to find the undercorse of 010-1 witch makes the out of part by 6

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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

If eat 1 apple and then i eat another how many apples have i eaten in total

stealth61 [152]

2 apples. Because you ate one already and you eat another one that equals 2

8 0

2 years ago

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