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yaroslaw [1]
3 years ago
6

The equation of line a is y=3/4x-3 If line b runs parallel to line a and passes

Mathematics
1 answer:
katen-ka-za [31]3 years ago
5 0

Answer:

y = -4/3 x - 1/3

Step-by-step explanation:

y = -4/3 x + b

5 = -4/3 (-4) + b

15 = 16 + 3b

b = -1/3

y = -4/3 x - 1/3

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John has a piece of clay that weighs 12 1/2 pounds. He divides the clay into 5 equal pieces. How much does each piece weigh?
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2.44

Step-by-step explanation:

12 1/2 / 5 =2.44

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Nicholas wrote the steps below to simplify the fraction 20/30 find his error and correct it
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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}

7 0
3 years ago
Is this correct please correct me if i’m wrong.
Sever21 [200]

Answer: This is correct already.

Step-by-step explanation:

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