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3 years ago

Charlie's eraser has a mass of 9 grams. How many milligrams is the eraser?

Mathematics

2 answers:

Vadim26 [7]3 years ago

8 0

Answer:

9000 milligrams

Step-by-step explanation:

One gram is the equivalent of 1000 milligrams.

So, 9 grams is 9*1000 milligrams, which is 9000 milligrams.

lozanna [386]3 years ago

5 0

Answer:

9000 milligrams

Step-by-step explanation:

There are 1000 milligrams in one gram, So 9 grams is 9000 milligrams.

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If g(x)=2(x-4), find the value of x if g(x)=20

kherson [118]

Answer:

3) 14

Step-by-step explanation:

in the given equation replace the "g(x)" with the provided "20"

20=2(x-4) solve as usual

divide each side by 2

10=x-4

add 4 to each side

14=x

the value of x is 14

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3 years ago

6 = a/4 + 2 What’s the answer for the variable a?

Vikki [24]

Answer:

a = 16

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3 years ago

Read 2 more answers

Consider the following initial-value problem. (x + y)2 dx + (2xy + x2 − 2) dy = 0, y(1) = 1 Let ∂f ∂x = (x + y)2 = x2 + 2xy + y2

IRISSAK [1]

$(x+y)^2\,\mathrm dx+(2xy+x^2-2)\,\mathrm dy=0$

Suppose the ODE has a solution of the form $F(x,y)=C$ , with total differential

$\dfrac{\partial F}{\partial x}\,\mathrm dx+\dfrac{\partial F}{\partial y}\,\mathrm dy=0$

This ODE is exact if the mixed partial derivatives are equal, i.e.

$\dfrac{\partial^2F}{\partial y\partial x}=\dfrac{\partial^2F}{\partial x\partial y}$

We have

$\dfrac{\partial F}{\partial x}=(x+y)^2\implies\dfrac{\partial^2F}{\partial y\partial x}=2(x+y)$

$\dfrac{\partial F}{\partial y}=2xy+x^2-2\implies\dfrac{\partial^2F}{\partial x\partial y}=2y+2x=2(x+y)$

so the ODE is indeed exact.

Integrating both sides of

$\dfrac{\partial F}{\partial x}=(x+y)^2$

with respect to $x$ gives

$F(x,y)=\dfrac{(x+y)^3}3+g(y)$

Differentiating both sides with respect to $y$ gives

$\dfrac{\partial F}{\partial y}=2xy+x^2-2=(x+y)^2+\dfrac{\mathrm dg}{\mathrm dy}$

$\implies x^2+2xy-2=x^2+2xy+y^2+\dfrac{\mathrm dg}{\mathrm dy}$

$\implies\dfrac{\mathrm dg}{\mathrm dy}=-y^2-2$

$\implies g(y)=-\dfrac{y^3}3-2y+C$

$\implies F(x,y)=\dfrac{(x+y)^3}3-\dfrac{y^3}3-2y+C$

so the general solution to the ODE is

$F(x,y)=\dfrac{(x+y)^3}3-\dfrac{y^3}3-2y=C$

Given that $y(1)=1$ , we find

$\dfrac{(1+1)^3}3-\dfrac{1^3}3-2=C\implies C=\dfrac13$

so that the solution to the IVP is

$F(x,y)=\dfrac{(x+y)^3}3-\dfrac{y^3}3-2y=\dfrac13$

$\implies\boxed{(x+y)^3-y^3-6y=1}$

5 0

2 years ago

How do I solve this system by graphing?

guajiro [1.7K]

You can first put into y=mx+b form so for the first one it would be y=-1/2x+3 3=y-intercept or where it'll touch the y-axis and -1/2 is the slope you can use the same approach for the next equation

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3 years ago