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3 years ago

e. Noah earned v dollars over the summer. Mai earned m dollars, which is 45 dollars more than Noah did. What's the answer?

Mathematics

1 answer:

Veseljchak [2.6K]3 years ago

6 0

Answer:

v=m-45

Step-by-step explanation:

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In a school, the ratio of soccer players to volleyball players is 4 : 1. The number of soccer players is how many times the numb

soldier1979 [14.2K]

Your question appears to be phrased incorrectly. If the ratio of soccer players to football players is 4:1 then for every 1 football player there are 4 soccer players. For example, if there are 3 football players, there would be 12 soccer players.

The equivalent expression would be S = 4F

The statement: "there are 4 more soccer players than football players" is not the same thing. It simply means that we add 4 to the total of football players to find out how many soccer players there are.

The equivalent expression would be S = F + 4

That being said: An equivalent ratio to 4:1 would be 8:2 , 12:3, 16:4, ...

Think of the ratio as a fraction. 4:1 = 4/1

4/1 = 8/2 = 12/3 = 16/4 ..., etc.

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3 years ago

(45pts) An elevator holds at most 1,400 pounds. A team of movers and their cart have a combined weight of 1,200 pounds. If each

emmainna [20.7K]

Let $x$ be the number of boxes the team brings with them. Their weight combined with the boxes can't exceed the capacity of 1400. Assuming the elevator runs fine at that exact weight, you want to find the number of boxes, each of which contributes 40 pounds. This is given by the equation

$1400=1200+40x$

Solving for $x$ , you have

$1400=1200-40x$
$200=40x$
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So the team can bring *at most* 5 boxes at a time.

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4 years ago

(x^2y+e^x)dx-x^2dy=0

klio [65]

It looks like the differential equation is

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

Check for exactness:

$\dfrac{\partial\left(x^2y+e^x\right)}{\partial y} = x^2 \\\\ \dfrac{\partial\left(-x^2\right)}{\partial x} = -2x$

As is, the DE is not exact, so let's try to find an integrating factor µ(x, y) such that

$\mu\left(x^2y + e^x\right) \,\mathrm dx - \mu x^2\,\mathrm dy = 0$

*is* exact. If this modified DE is exact, then

$\dfrac{\partial\left(\mu\left(x^2y+e^x\right)\right)}{\partial y} = \dfrac{\partial\left(-\mu x^2\right)}{\partial x}$

We have

$\dfrac{\partial\left(\mu\left(x^2y+e^x\right)\right)}{\partial y} = \left(x^2y+e^x\right)\dfrac{\partial\mu}{\partial y} + x^2\mu \\\\ \dfrac{\partial\left(-\mu x^2\right)}{\partial x} = -x^2\dfrac{\partial\mu}{\partial x} - 2x\mu \\\\ \implies \left(x^2y+e^x\right)\dfrac{\partial\mu}{\partial y} + x^2\mu = -x^2\dfrac{\partial\mu}{\partial x} - 2x\mu$

Notice that if we let µ(x, y) = µ(x) be independent of y, then ∂µ/∂y = 0 and we can solve for µ :

$x^2\mu = -x^2\dfrac{\mathrm d\mu}{\mathrm dx} - 2x\mu \\\\ (x^2+2x)\mu = -x^2\dfrac{\mathrm d\mu}{\mathrm dx} \\\\ \dfrac{\mathrm d\mu}{\mu} = -\dfrac{x^2+2x}{x^2}\,\mathrm dx \\\\ \dfrac{\mathrm d\mu}{\mu} = \left(-1-\dfrac2x\right)\,\mathrm dx \\\\ \implies \ln|\mu| = -x - 2\ln|x| \\\\ \implies \mu = e^{-x-2\ln|x|} = \dfrac{e^{-x}}{x^2}$

The modified DE,

$\left(e^{-x}y + \dfrac1{x^2}\right) \,\mathrm dx - e^{-x}\,\mathrm dy = 0$

is now exact:

$\dfrac{\partial\left(e^{-x}y+\frac1{x^2}\right)}{\partial y} = e^{-x} \\\\ \dfrac{\partial\left(-e^{-x}\right)}{\partial x} = e^{-x}$

So we look for a solution of the form F(x, y) = C. This solution is such that

$\dfrac{\partial F}{\partial x} = e^{-x}y + \dfrac1{x^2} \\\\ \dfrac{\partial F}{\partial y} = e^{-x}$

Integrate both sides of the first condition with respect to x :

$F(x,y) = -e^{-x}y - \dfrac1x + g(y)$

Differentiate both sides of this with respect to y :

$\dfrac{\partial F}{\partial y} = -e^{-x}+\dfrac{\mathrm dg}{\mathrm dy} = e^{-x} \\\\ \implies \dfrac{\mathrm dg}{\mathrm dy} = 0 \implies g(y) = C$

Then the general solution to the DE is

$F(x,y) = \boxed{-e^{-x}y-\dfrac1x = C}$

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We are asked in this problem the surface area of the prism given the dimensions: 9mm by 12mm by 8mm. Surface area is the sum of the areas of each face of the prism. The formula for surface area is 2*lw+2*lh+ 2*wh. In this case, upon substitution, SA = 2*9*12+2*9*8+2*12*8 equal to a total of 552 mm2.

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Andreas93 [3]

Answer:

Domain= how many boxes there are from each circle to the other, and the range is basically the height. In this case, Domain= 8, Range=2

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