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3 years ago

What is the slope of (12,-18) and (11,12)? Please I’m very confused.

Mathematics

1 answer:

Lesechka [4]3 years ago

6 0

Answer:

-30

Step-by-step explanation:

$\boxed{slope = \frac{y1 - y2}{x1 - x2} }$

where (x₁, y₁) is the first coordinate and (x₂, y₂) is the second coordinate.

Slope

$= \frac{12 - ( - 18)}{11 - 12}$

$= \frac{12 + 18}{ - 1}$

= -30

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3 years ago

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Answer:

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4 years ago

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ss7ja [257]

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Step-by-step explanation:

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$\implies \textsf {3x - 2 = 4 (Given)}$

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2 years ago

Read 2 more answers

Find the distance between the points (3, 8) and (0,4).

Bingel [31]

Answer:

Step-by-step explanation:

(3,8) (0,4)

X1 = 3 X2 = 0

Y1 = 8 Y2 = 4

$D=\sqrt{(X1-X2)^{2}+ (Y1-Y2)^{2} }$

$D= \sqrt{(3-0)^{2} + (8-4)^{2} }$

Do what's in the parentheses so...

3-0= 3
8-4= 4

Now plug it in!

$D= \sqrt{(3)^{2}+(4)^{2} }$

Now you are going to finish the parentheses so...

(3)^2= 9
(4)^2= 16

Plug that in so that you have this...

$\sqrt{9+16}$

Add 9+16 to get...

$\sqrt{25}$

Then you are going to find the number or numbers that make this a perfect square...

$\sqrt{25}= 5$

So 5 is your answer

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3 years ago

1. (a) Solve the differential equation (x + 1)Dy/dx= xy, = given that y = 2 when x = 0. (b) Find the area between the two curves

erastova [34]

(a) The differential equation is separable, so we separate the variables and integrate:

$(x+1)\dfrac{dy}{dx} = xy \implies \dfrac{dy}y = \dfrac x{x+1} \, dx = \left(1-\dfrac1{x+1}\right) \, dx$

$\displaystyle \frac{dy}y = \int \left(1-\frac1{x+1}\right) \, dx$

$\ln|y| = x - \ln|x+1| + C$

When x = 0, we have y = 2, so we solve for the constant C :

$\ln|2| = 0 - \ln|0 + 1| + C \implies C = \ln(2)$

Then the particular solution to the DE is

$\ln|y| = x - \ln|x+1| + \ln(2)$

We can go on to solve explicitly for y in terms of x :

$e^{\ln|y|} = e^{x - \ln|x+1| + \ln(2)} \implies \boxed{y = \dfrac{2e^x}{x+1}}$

(b) The curves y = x² and y = 2x - x² intersect for

$x^2 = 2x - x^2 \implies 2x^2 - 2x = 2x (x - 1) = 0 \implies x = 0 \text{ or } x = 1$

and the bounded region is the set

$\left\{(x,y) ~:~ 0 \le x \le 1 \text{ and } x^2 \le y \le 2x - x^2\right\}$

The area of this region is

$\displaystyle \int_0^1 ((2x-x^2)-x^2) \, dx = 2 \int_0^1 (x-x^2) \, dx = 2 \left(\frac{x^2}2 - \frac{x^3}3\right)\bigg|_0^1 = 2\left(\frac12 - \frac13\right) = \boxed{\frac13}$

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3 years ago