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3 years ago

A store and a bank would both charge fees for _____.

Mathematics

1 answer:

-BARSIC- [3]3 years ago

6 0

I would say D but I am not sure 100 percent

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What is the distance between point A(−1, 3) and point B(−8, 3) ?

balu736 [363]

Answer:

7 units

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

Complete the statement with equal to, greater than, or less than.

enyata [817]

2 x 2/9 Will Be Less Than 2

5 0

2 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

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

5 0

3 years ago

Geraldine is asked to explain the limits on the range of an exponential equation using the function f(x) = 2^x. She makes these

GalinKa [24]

General Idea:

Domain is the values of x which gives a defined output to the function, and Range is the values of y that we get by substituting domain values of x. For the function f(x) = a^x, where a > 0, the domain will be all real numbers and range will be all values greater than 0. That is the graph will be very close to x-axis but will never touch it.

Applying the concept:

From the attached table, we can notice that as x increases infinitely, the y-values are continually doubled for each single increase in x and as x decreases infinitely, the y-values are continually halved for each single decrease in x.

Option 1: "Statement 1 is incorrect because the y-values are increased by 2, not doubled".

Option 1 is INCORRECT because Statement 1 is TRUE & CORRECT.

Option 2: "Statement 2 is incorrect because the y-values are doubled, not halved".

Option 2 is INCORRECT because Statement 2 is TRUE & CORRECT.

Option 3: "The conclusion is incorrect because the range is limited to the set of the integers"

Option 3 is INCORRECT because, the range will be always greater than 0, because irrespective of how many times y is halved for each single decrease in x, the y will still be a positive fraction and won't be a negative.

Option 4: "The conclusion is incorrect because the range is limited to the set of positive real numbers"

Option 4 is CORRECT because y value will be positive irrespective of whatever x we substitute in the function.

8 0

3 years ago

Read 2 more answers

NEED ANSWER ASAP- FOR FINALS

blagie [28]

Answer:

Width - 12ft Length - 30ft

Step-by-step explanation:

Width is 12ft

Length is 30ft

Depth is 6ft

The length, 30ft, is 2.5 times longer than the width which is 12 ft which confirms the two dimensions to be correct.

When you multiply all three of the dimensions of the swimming pool you will get 2940 cubic feet

8 0

3 years ago