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3 years ago

13

A fair spinner has eight equal sections.

1 answer:

yanalaym [24]3 years ago

3 0

Answer:

Your Answer is B!

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704÷46 what the answer tell me im in the

olasank [31]

15?im not sure...but I think thats right

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3 years ago

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They make a base of 50 gallons (20%). They make a “heavy batch of 150 gallons (80%) They mix them both. What is the percent of

SSSSS [86.1K]

Answer #1: x= 250

50 = x times 20/100

X= 100 times 50/ 20

X=250

Answer #2: same explanation but the answer is X= 120

If your talking about merging the percentage of the answer, I can’t help you with that

4 0

3 years ago

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Find the coordinates of a point that is in Quadrant IV and is √45 units away from the point (-4, 1).

exis [7]

Answer:

15-3

Step-by-step explanation:

15x3=45 or it could be 3-15

6 0

3 years ago

In the two functions fand g, fog and go fare both equivalent to x. Which of the following statements must be TRUE?

garik1379 [7]

F Must be the inverse of g

8 0

1 year ago

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

3 years ago