Which number supports the given conjecture?

Mathematics

1 answer:

taurus [48]3 years ago

5 0

The correct answer is B.

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This is for 20 points!!!!!!!

3241004551 [841]

All really need to do is pick one number. so, we will go with 6 and 87.

6 times 12 is 72 then you subtract that from $87.

That would be 15

To double check, you can use another one. 9 and 123.

9 times 12 is 108. subtract that from 123 and you get 15.

Your answer is B) $15

Hope I helped :)

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4 years ago

What number represents the same amount as 2 hundreds +12 tens +6 ones?

Svetradugi [14.3K]

The answer is 326. this is simple.

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3 years ago

Read 2 more answers

The asymptote of the function f(x) = 3^x+1 - 2 is ____. Its y-intercept is ____.

xxMikexx [17]

We are given the function f(x) = 3^x+1 - 2 and is asked for the asymptote and y-intercept. In this case, we transform the function to

y +2 = 1/3^(x+1)-1
y = 1-2/3^(x+1)-1/(3^(x+1)-1)
let 3^(x+1)-1 = 0
1/3^(x+1) = 0
x = infinity
when x = infinity
y = 0 - 2
y = -2

The y-intercept is the value where x = 0. substituting,

y = 3^(0+1) - 2 = 1

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4 years ago

If an airplane flew at 432 mph for 2.5 hours how far did it fly

grin007 [14]

The plane flew 1,080 miles in 2.5 hours

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3 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 :

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