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Y_Kistochka [10]
3 years ago
10

How do I solve this??

Mathematics
1 answer:
Liula [17]3 years ago
7 0

2(2p + m) = 3 - 5m

Distributive

4p + 2m = 3 - 5m

2m + 5m = 3 - 4p

7m = 3 - 4p

m = (3 - 4p)/ 7

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What is the area of the parallelogram shown above
adoni [48]

The area of the parallelogram is 55 square feet

<h3>How to determine the area of the parallelogram?</h3>

From the figure, we have the following dimensions:

Height = 5 ft

Base = 11 ft

The area of the parallelogram is calculated as:

Area = Base * Height

This gives

Area = 5ft * 11ft

Evaluate the product

Area = 55 square feet

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brainly.com/question/76387

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8 0
2 years ago
A new car worth ​$18 comma 000 is depreciating in value by ​$3 comma 000 per year. after how many years will the​ car's value be
ehidna [41]
Its 5 years
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8 0
3 years ago
12^1/2(2^1/2+3^1/2)(2^1/2-3^1/2)
krek1111 [17]
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4 0
3 years ago
Read 2 more answers
Show that the line integral is independent of path by finding a function f such that ?f = f. c 2xe?ydx (2y ? x2e?ydy, c is any p
Juli2301 [7.4K]
I'm reading this as

\displaystyle\int_C2xe^{-y}\,\mathrm dx+(2y-x^2e^{-y})\,\mathrm dy

with \nabla f=(2xe^{-y},2y-x^2e^{-y}).

The value of the integral will be independent of the path if we can find a function f(x,y) that satisfies the gradient equation above.

You have

\begin{cases}\dfrac{\partial f}{\partial x}=2xe^{-y}\\\\\dfrac{\partial f}{\partial y}=2y-x^2e^{-y}\end{cases}

Integrate \dfrac{\partial f}{\partial x} with respect to x. You get

\displaystyle\int\dfrac{\partial f}{\partial x}\,\mathrm dx=\int2xe^{-y}\,\mathrm dx
f=x^2e^{-y}+g(y)

Differentiate with respect to y. You get

\dfrac{\partial f}{\partial y}=\dfrac{\partial}{\partial y}[x^2e^{-y}+g(y)]
2y-x^2e^{-y}=-x^2e^{-y}+g'(y)
2y=g'(y)

Integrate both sides with respect to y to arrive at

\displaystyle\int2y\,\mathrm dy=\int g'(y)\,\mathrm dy
y^2=g(y)+C
g(y)=y^2+C

So you have

f(x,y)=x^2e^{-y}+y^2+C

The gradient is continuous for all x,y, so the fundamental theorem of calculus applies, and so the value of the integral, regardless of the path taken, is

\displaystyle\int_C2xe^{-y}\,\mathrm dx+(2y-x^2e^{-y})\,\mathrm dy=f(4,1)-f(1,0)=\frac9e
8 0
3 years ago
Simplfy (x)^2(2xy^3)^5 plz help
Deffense [45]

Answer: 32x^7y^15

Step-by-step explanation:

(x)^2(2xy^3)^5

= x^2 * 32x^5y^15

= 32x^7y^15

5 0
2 years ago
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