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Veseljchak [2.6K]
2 years ago
10

-16 = -5a -21. Plz help. Show work if possible. Pre-Algebra is hard. :(

Mathematics
1 answer:
Leona [35]2 years ago
5 0

Answer:

a = -1

Step-by-step explanation:

-16 = -5a - 21 (Given)

5 = -5a (Added 21 on both sides)

-1 = a (Divided -5 on both sides)

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In this problem we consider an equation in differential form Mdx+Ndy=0. (4x+2y)dx+(2x+8y)dy=0 Find My= 2 Nx= 2 If the problem is
zheka24 [161]

Answer:

f(x,y)=2x^2+4y^2+2xy=C_1\\\\Where\\\\y(x)=\frac{1}{4} (-x\pm \sqrt{-7x^2+C_1} )

Step-by-step explanation:

Let:

M(x,y)=4x+2y\\\\and\\\\N(x,y)=2x+8y

This is and exact equation, because:

\frac{\partial M(x,y)}{\partial y} =2=\frac{\partial N}{\partial x}

So, define f(x,y) such that:

\frac{\partial f(x,y)}{\partial x} =M(x,y)\\\\and\\\\\frac{\partial f(x,y)}{\partial y} =N(x,y)

The solution will be given by:

f(x,y)=C_1

Where C1 is an arbitrary constant

Integrate \frac{\partial f(x,y)}{\partial x} with respect to x in order to find f(x,y):

f(x,y)=\int\ {4x+2y} \, dx =2x^2+2xy+g(y)

Where g(y) is an arbitrary function of y.

Differentiate f(x,y) with respect to y in order to find g(y):

\frac{\partial f(x,y)}{\partial y} =2x+\frac{d g(y)}{dy}

Substitute into \frac{\partial f(x,y)}{\partial y} =N(x,y)

2x+\frac{dg(y)}{dy} =2x+8y\\\\Solve\hspace{3}for\hspace{3}\frac{dg(y)}{dy}\\\\\frac{dg(y)}{dy}=8y

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g(y)=\int\ {8y} \, dy =4y^2

Substitute g(y) into f(x,y):

f(x,y)=2x^2+4y^2+2xy

The solution is f(x,y)=C1

f(x,y)=2x^2+4y^2+2xy=C_1

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y(x)=\frac{1}{4} (-x\pm \sqrt{-7x^2+C_1} )

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3 years ago
Cherise has a bag of marbles for her little sister. It contains 10 orange marbles, 7 yellow marbles, and 3 green marbles. She dr
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(7/20) (7/20) (7/20) = 343/8000

We are given
x = 2 and 3
u = 2.5
s = 0.5

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z = (2 - 2.5) / 0.5 and (3 - 2.5)/ 0.5
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Rearrange the following and ad .005, 2,002.181, 795.41, 14.0,184
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Please help me out with this!!<br> BRAINLIEST AVAILABLE!!
lapo4ka [179]

Answer:

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Step-by-step explanation:

Question One

The first and third frames look to me to be the same. I'll treat them that way.

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2y + 6 = 2(x + 3)         Remove the brackets

2y + 6 = 2x + 6           Subtract 6 from both sides

2y = 2x                       Divide by 2

y = x

Now solve these two equations.

so x^2 = x                  

x > 0

1 solution is x = 0 from which y = 0. This won't work. x must be greater than 0. So the other is

x(x) = x                           Divide both sides by x            

x = 1                            

y = x^2                           Put x = 1 into x^2

y = 1^2                           Solve

y = 1                      

The second solution is

(1,1)

xy = 1*1

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Question Two

square root(k + 2) - x = 0

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k + 2 = 9^2                    Square 9

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