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3 years ago

Write these values in order, starting with the smallest. 72%, 0.6, 21/25, 0.75, 1/2

Mathematics

1 answer:

sweet-ann [11.9K]3 years ago

3 0

Answer:

1/2 , .6 , 72%, .75, 21/25

Step-by-step explanation:

.5, .6, .72, .75, .84

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(05.02 MC)

tankabanditka [31]

Answer:

there is no value to x in this system because you don't have a number for the variable y

Step-by-step explanation:

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3 years ago

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A tiny but horrible alien is standing at the top of the Empire State Building (which is

Leto [7]

Answer:

87.7 degrees.

Step-by-step explanation:

In triangle ABC, attached.

The height of the building |AB|=443 meters

The distance of the agent across the street , |BC|=18 meters

We want to determine the angle at C.

Now,

$Tan C=\dfrac{|AB|}{|BC|} \\C=arctan (\dfrac{|AB|}{|BC|} )\\=arctan (\dfrac{443}{18} )\\=87.67^\circ\\\approx 87.7^\circ $(correct to the nearest tenth)$

The agent should sfoot his laser gun at an angle of 87.7 degrees.

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4 years ago

Molly opened a bank account. She deposited $86.50 into her account every month for 20 months. She used $24.50 every month to pay

harkovskaia [24]

First, let's multiply 86.50 by 20 since she deposited $86.50 each month for 20 months.

86.50 * 20 = $1,730

Now, we have to divide this by 24.50 * 20 to find how much out of her $1,730 that she used on singing lessons.

24.50 * 20 = $490

Subtract how much she used for singing lessons from her total amount.

1,730 - 490 = $1,240

Finally, we multiply the amount she has left by 1/4 to find out how much she spent on her summer camp. We can also do this by multiplying 1,240 by 0.25 since 0.25 = 1/4.

1,240 * 0.25 = $310

So, Molly spent $310.00 on her summer camp.
Hope this helped!

7 0

3 years ago

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A geometric sequence has a first term of x-y and a common ratio of x+y. what if the third term of the sequence?

kotegsom [21]

First term t(1)=(x-y)
common ratio =(x+y) (don't omit the parentheses)
then the nth term:
t(n)=(x-y)(x+y)^(n-1)
check:
t(1)=(x-y)(x+y)^(1-1)=(x-y)(x+y)^0=(x-y)(1)=(x-y)
t(3)=(x-y)(x+y)^(3-1)=(x-y)(x+y)^2

When n is small, t(n) can also be found by repeated multiplication of the common ratio (x+y)

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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 :

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