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2 years ago

I NEED HELP :( PLZ ASAP

Mathematics

2 answers:

Vilka [71]2 years ago

6 0

Answer:

what

Step-by-step explanation:

VikaD [51]2 years ago

4 0

We need the question :(

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Maria Addai has been offered a future payment of $990 two years from now. If she can earn 8.2 percent compounded annually on her

IgorLugansk [536]

Answer:

845.6306

Step-by-step explanation:

Firstly this is annuity based

Let, investment at beginning of year = x

Then value at year 1 end = x + (8.2% $\times$ x)

Value at end of year 2 = (x + 0.082x) + (8.2% $\times$ (x + 0.082x))

Now this value = $990

Therefore,

990 = (x + 0.082x) + ((x + 0.082x) $\times$ 8.2%)

990 = x + 0.082x + 0.082x + 0.006724x = 1.170724x

x = 990/1.170724 = 845.6306

3 0

2 years ago

Read 2 more answers

Solve the equations to find the number and type of solutions

luda_lava [24]

Answer:

This has one real solution, x=4

Step-by-step explanation:

8 - 4x = 0

Add 4x to each side

8 - 4x+4x = 0+4x

8 =4x

Divide each side by 4

8/4 = 4x/4

2 =x

This has one real solution, x=4

6 0

3 years ago

Read 2 more answers

(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 :

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