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Sergeeva-Olga [200]

3 years ago

9

Use the work shown below to write the equation for a line that passes through the points (−5, 0) and (−1, −8). 1. Use slope form

ula to find slope: 2. Substitute one point and slope into slope-intercept form to find the y-intercept: What is the equation of the line in slope-intercept form? y = –10x – 2 y = –1x – 8 –8 = –2x – 1 y = –2x – 10

2 answers:

atroni [7]3 years ago

3 0

Answer:

<h2>y = -2x - 10</h2>

Step-by-step explanation:

The formula of a slope:

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

We have two points (-5, 0) and (-1, -8). Substitute:

$m=\dfrac{-8-0}{-1-(-5)}=\dfrac{-8}{4}=-2$

The slope-intercept form of an equation of a line:

$y=mx+b$

m - slope

b - y-intercpet

Put the value of the slope and coordinates of the point (-5, 0) to the equation of a line:

$0=-2(-5)+b$

$0=10+b$ <em>subtract 10 from both sides</em>

$-10=b\to b=-10$

Finally:

$y=-2x-10$

faltersainse [42]3 years ago

3 0

Answer:

The answer is y = -2x - 10

Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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2 years ago

Define the double factorial of n, denoted n!!, as follows:n!!={1⋅3⋅5⋅⋅⋅⋅(n−2)⋅n} if n is odd{2⋅4⋅6⋅⋅⋅⋅(n−2)⋅n} if n is evenand (

tekilochka [14]

Answer:

Radius of convergence of power series is $\lim_{n \to \infty}\frac{a_{n}}{a_{n+1}}=\frac{1}{108}$

Step-by-step explanation:

Given that:

n!! = 1⋅3⋅5⋅⋅⋅⋅(n−2)⋅n n is odd

n!! = 2⋅4⋅6⋅⋅⋅⋅(n−2)⋅n n is even

(-1)!! = 0!! = 1

We have to find the radius of convergence of power series:

$\sum_{n=1}^{\infty}[\frac{8^{n}n!(3n+3)!(2n)!!}{2^{n}[(n+9)!]^{3}(4n+3)!!}](8x+6)^{n}\\\\\sum_{n=1}^{\infty}[\frac{8^{n}n!(3n+3)!(2n)!!}{2^{n}[(n+9)!]^{3}(4n+3)!!}]2^{n}(4x+3)^{n}\\\\\sum_{n=1}^{\infty}[\frac{8^{n}n!(3n+3)!(2n)!!}{[(n+9)!]^{3}(4n+3)!!}](x+\frac{3}{4})^{n}\\$

Power series centered at x = a is:

$\sum_{n=1}^{\infty}c_{n}(x-a)^{n}$

$\sum_{n=1}^{\infty}[\frac{8^{n}n!(3n+3)!(2n)!!}{2^{n}[(n+9)!]^{3}(4n+3)!!}](8x+6)^{n}\\\\\sum_{n=1}^{\infty}[\frac{8^{n}n!(3n+3)!(2n)!!}{2^{n}[(n+9)!]^{3}(4n+3)!!}]2^{n}(4x+3)^{n}\\\\\sum_{n=1}^{\infty}[\frac{8^{n}4^{n}n!(3n+3)!(2n)!!}{[(n+9)!]^{3}(4n+3)!!}](x+\frac{3}{4})^{n}\\$

$a_{n}=[\frac{8^{n}4^{n}n!(3n+3)!(2n)!!}{[(n+9)!]^{3}(4n+3)!!}]\\\\a_{n+1}=[\frac{8^{n+1}4^{n+1}n!(3(n+1)+3)!(2(n+1))!!}{[(n+1+9)!]^{3}(4(n+1)+3)!!}]\\\\a_{n+1}=[\frac{8^{n+1}4^{n+1}(n+1)!(3n+6)!(2n+2)!!}{[(n+10)!]^{3}(4n+7)!!}]$

Applying the ratio test:

$\frac{a_{n}}{a_{n+1}}=\frac{[\frac{32^{n}n!(3n+3)!(2n)!!}{[(n+9)!]^{3}(4n+3)!!}]}{[\frac{32^{n+1}(n+1)!(3n+6)!(2n+2)!!}{[(n+10)!]^{3}(4n+7)!!}]}$

$\frac{a_{n}}{a_{n+1}}=\frac{(n+10)^{3}(4n+7)(4n+5)}{32(n+1)(3n+4)(3n+5)(3n+6)+(2n+2)}$

Applying n → ∞

$\lim_{n \to \infty}\frac{a_{n}}{a_{n+1}}= \lim_{n \to \infty}\frac{(n+10)^{3}(4n+7)(4n+5)}{32(n+1)(3n+4)(3n+5)(3n+6)+(2n+2)}$

The numerator as well denominator of $\frac{a_{n}}{a_{n+1}}$ are polynomials of fifth degree with leading coefficients:

$(1^{3})(4)(4)=16\\(32)(1)(3)(3)(3)(2)=1728\\ \lim_{n \to \infty}\frac{a_{n}}{a_{n+1}}=\frac{16}{1728}=\frac{1}{108}$

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2 years ago

What is 4(b + 2), if b = 7?

AlladinOne [14]

Answer:

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Step-by-step explanation:

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8 0

3 years ago

Read 2 more answers

A satellite dish has cross-sections shaped like parabolas. The receiver is located 13 inches from the base along the axis of sym

horsena [70]

Answer:

Depth = 3.3 inches

Step-by-step explanation:

Given that the shape of the satellite looks like a parabola

The equation of parabola is given as follows

$x^2=4\times a\times y$

Where

a= 13

Therefore

$x^2=4\times 13\times y$

$x^2=52\times y$

Lets take (13 , y) is a

Now by putting the values in the above equation we get

$13^2=52\times y$

$y=\dfrac{13^2}{52}=3.25$

y=3.25 in

Therefore the depth of the satellite at the nearest integer will be 3.3 inches.

Depth = 3.3 inches

7 0

3 years ago