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nadezda [96]
3 years ago
14

Find the slope!!! Quickly please

Mathematics
2 answers:
Artyom0805 [142]3 years ago
7 0

Answer:

2

Step-by-step explanation:

Slope is (y1 - y2) / (x1 - x2) for any 2 respective x's and y's.

For any of the points here, you get 2.

VLD [36.1K]3 years ago
6 0

Answer:

the slope is m= 2

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What is the slope of the line that is perpendicular to a line whose equation 3y=-4x+2?
Zarrin [17]
The slopes of lines perpendicular to each other are opposite reciprocals. So, if you are given that the slope of a line is 3 and need to find the slope of a line perpendicular to that line, you'd flip that number around and negate it, leaving you with -1/3.

To find the slope of the given line, first get it into slope-intercept form (y - mx + b, where m is the slope and b is the y-intercept).

    3y = -4x + 2
    y = -4/3x + 2/3

The slope is -4/3. To find the slope of a perpendicular line, find its opposite reciprocal. It is 3/4.

Answer:
3/4 (the first option)
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3 years ago
Sid brought $200 to the amusement park . the cost of admission is $16.25 , and the cost of a lunch is $6.50 . After Sid purchase
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Let x represent the # of people in sid’s group: 200 - (16.25x + 6.50x) = 18.25
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3 years ago
Venny Had 5 friends she ate 3 of them. The rest ran away. How much friends dose she have now?
nadya68 [22]
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4 0
3 years ago
Which is the denominator and which is the numerator?
Mashcka [7]
Hey there!

<span>The numerator is the top part of the fraction and the denominator is the bottom part of the fraction. </span>

For example, in the fraction \frac{1}{2} , 1 would be the numerator and 2 would be the denominator. 

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7 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+ ...) }

7 0
3 years ago
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