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3 years ago

Find the slope. (4, 10) and (-2,-5)

Mathematics

1 answer:

marta [7]3 years ago

7 0

Answer:

Slope in fraction form = 5/2

Slope in decimal form = 2.5

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Work Shown:

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

m = (-5-10)/(-2-4)

m = (-15)/(-6)

m = (-3*5)/(-3*2)

m = 5/2 slope in fraction form

m = 2.5 slope in decimal form

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Explanation:

The first step is the slope formula where (x1,y1) and (x2,y2) are the two points the line goes through.

I used (x1,y1) = (4,10) and (x2,y2) = (-2,-5)

The slope formula basically says "subtract the y values, then subtract the x values in the same order. Then divide the two differences".

Or you can think of it as

slope = rise/run

rise = y2-y1 = change in y

run = x2-x1 = change in x

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If AC = x + 5 and AB = 3x - 1... what is the length of BC?

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If B is the midpoint of AC, then |AB| = |AC|.
|AB| = 3x + 2
|BC| = 5x - 10
Therefore we have the equation:
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4 years ago

Read 2 more answers

(5) Find the Laplace transform of the following time functions: (a) f(t) = 20.5 + 10t + t 2 + δ(t), where δ(t) is the unit impul

Aloiza [94]

Answer

(a) $F(s) = \frac{20.5}{s} - \frac{10}{s^2} - \frac{2}{s^3}$

(b) $F(s) = \frac{-1}{s + 1} - \frac{4}{s + 4} - \frac{4}{9(s + 1)^2}$

Step-by-step explanation:

(a) $f(t) = 20.5 + 10t + t^2 +$ δ(t)

where δ(t) = unit impulse function

The Laplace transform of function f(t) is given as:

$F(s) = \int\limits^a_0 f(s)e^{-st} \, dt$

where a = ∞

=> $F(s) = \int\limits^a_0 {(20.5 + 10t + t^2 + d(t))e^{-st} \, dt$

where d(t) = δ(t)

=> $F(s) = \int\limits^a_0 {(20.5e^{-st} + 10te^{-st} + t^2e^{-st} + d(t)e^{-st}) \, dt$

Integrating, we have:

=> $F(s) = (20.5\frac{e^{-st}}{s} - 10\frac{(t + 1)e^{-st}}{s^2} - \frac{(st(st + 2) + 2)e^{-st}}{s^3} )\left \{ {{a} \atop {0}} \right.$

Inputting the boundary conditions t = a = ∞, t = 0:

$F(s) = \frac{20.5}{s} - \frac{10}{s^2} - \frac{2}{s^3}$

(b) $f(t) = e^{-t} + 4e^{-4t} + te^{-3t}$

The Laplace transform of function f(t) is given as:

$F(s) = \int\limits^a_0 (e^{-t} + 4e^{-4t} + te^{-3t} )e^{-st} \, dt$

$F(s) = \int\limits^a_0 (e^{-t}e^{-st} + 4e^{-4t}e^{-st} + te^{-3t}e^{-st} ) \, dt$

$F(s) = \int\limits^a_0 (e^{-t(1 + s)} + 4e^{-t(4 + s)} + te^{-t(3 + s)} ) \, dt$

Integrating, we have:

$F(s) = [\frac{-e^{-(s + 1)t}} {s + 1} - \frac{4e^{-(s + 4)}}{s + 4} - \frac{(3(s + 1)t + 1)e^{-3(s + 1)t})}{9(s + 1)^2}] \left \{ {{a} \atop {0}} \right.$

Inputting the boundary condition, t = a = ∞, t = 0:

$F(s) = \frac{-1}{s + 1} - \frac{4}{s + 4} - \frac{4}{9(s + 1)^2}$

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help, please.. . Let f(x) = 2x2 – 2x + 7 and g(x) = x2 + 9x – 6. Find f(x) – g(x).. . x2 – 11x + 13. . x2 – 11x + 1. . x2 + 7x +

AlladinOne [14]

F(x) = 2x² - 2x + 7

g(x) = x² + 9x - 6

f(x) - g(x) = 2x² - 2x + 7 - (x² + 9x - 6)

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The 1st option.

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3 years ago

Find the slope. (4, 10) and (-2,-5)​

Find the slope. (4, 10) and (-2,-5)