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3 years ago

6) Sally wrote the number 30, 048 in expanded form.

Mathematics

1 answer:

vichka [17]3 years ago

6 0

Answer:

the answer is G) Change 30 to 3,000

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Find the y value for point F such that DF and EF form a 1:3 ratio. (5 points) Segment DE is shown. D is at negative 3, negative

SVETLANKA909090 [29]

I drew the segment and used Pythagorean theorem to solve for its measure. The line formed is the hypotenuse of the imaginary right triangle.

Among the choices only -1.33 and -1.25 is a feasible choice. But I am leaning towards -1.25 as the y-value of point F based on my diagram. Please see attachment.

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3 years ago

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A candy store makes a 9-pound mixture of gummy candy, jelly beans, and hard candy. The cost of gummy candy is $2.00 per pound, j

Musya8 [376]

Let

x--------> the amount of gummy candy in pounds

y--------> the amount of jelly beans in pounds

z--------> the amount of hard candy in pounds

we know that

$x+y+z=9$ --------> equation $1$

$2x+3y+3z=23$ --------> equation $2$

$x=2y$ --------> equation $3$

Substitute equation $3$ in equation $1$ and equation $2$

$[2y]+y+z=9$ -----> $3y+z=9$ ------> equation $4$

$2[2y]+3y+3z=23$ -----> $7y+3z=23$ ------> equation $5$

using a graphing tool ------> Solve the system of equations

see the attached figure

the solution is the point $(3,2)$

$z=3\ pounds\\y=2\ pounds$

Find the value of x

$x=2y$

$x=2*2=4\ pounds$

therefore

the answer is

the amount of gummy candy is $4\ pounds$

the amount of jelly beans is $2\ pounds$

the amount of hard candy is $3\ pounds$

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Which of the following is the graph of y=1/2X ?

ivanzaharov [21]

Where’s the graph?????

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(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 :

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