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2 years ago

A line through (2, 15) and (6, 7)

Mathematics

1 answer:

erastovalidia [21]2 years ago

5 0

Answer:

y = -2x + 19

Step-by-step explanation:

Let the equation of the line is y = mx + b

Here, m = slope of the line

b = y - intercept

slope of a line passing through (x₁, y₁) and (x₂, y₂) is given by

$m=\frac{y_2-y_1}{x_2-x_1}$

slope of line passing through (2, 15) and (6, 7) will be

$m=\frac{15-7}{2-6}$

= (-2)

Equation of the line will be,

y = -2x + b

Since line passes through point (2, 15)

15 = -2 (2) + b

b = 15 + 4

b = 19

Therefore, equation will be,

y = -2x + 19

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Which pair of ratios makes a proportion?

Nana76 [90]

Answer:

D. 1/4 and 9/36

Step-by-step explanation:

If two ratios make a proportion then it can be shown as:

a/b = c/d ⇒ ad = bc

Let's verify the answer choices with this method

A. 1/4 and 5/25

1*25 = 4*5
25 = 20, incorrect, not a proportion

B. 1/4 and 7/24

1*24 = 4*7
24 = 28, incorrect, not a proportion

C. 1/4 and 10/48

1*48 = 4*10
48 = 40, incorrect, not a proportion

D. 1/4 and 9/36

1*36 = 4*9
36 = 36, correct, this is a proportion

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1 year ago

What is the output of a function when the input is 10?

Blizzard [7]

You did not give us the function

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2 years ago

Two points on a line are ( 3, 8 )and ( -3, 2 ). What is the slope of the line?

professor190 [17]

Answer:

Step-by-step explanation:

m=(y2-y1)/(x2-x1)

m=(2-8)/(-3-3)

m=-6/-6

m=1

8 0

2 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

2 years ago

If you can expplain

lana [24]

Answer:

I can explain. The answer is A

Step-by-step explanation:

Think of it this way: when you write a mixed number, the fraction is out of 100. If the denominator is 8, find 1/8 of 100 by doing 100 divided by 1/8. This gives you 12.5. The entire fraction is 2/8, so we need to multiply 12.5 by 2, since it's only 1/8. 12.5 x 2 gives us 25. In the decimal, this would be equal to .25. A is the only answer that represents the mixed number as a decimal correctly.

6 0

2 years ago

Read 2 more answers

A line through (2, 15) and (6, 7)​

A line through (2, 15) and (6, 7)