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Mnenie [13.5K]
3 years ago
7

jillian bought 1/2 pound of turkey and 1 1/3 pound of cheddar cheese at the deli. both items are on sale for $6 per pound. how m

uch does each turkey cost>
Mathematics
1 answer:
Neko [114]3 years ago
4 0
Divide you cost by the weight and than you should have your answer

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Please help <br><br>Simplify 3 - (2- 8).<br><br>O -1<br>O 7<br>O -7<br>O 9​
Galina-37 [17]

Answer:

3-(2-8)

3-(-6)

3+6

9

d).9

Step-by-step explanation:

5 0
2 years ago
The difference between 3 times a number x and 2 is 19. What is the value of x?
Ber [7]

Answer:

the value of x is 7.

Step-by-step explanation:

2+19=21

21÷3=7

Check:

3×7=21

21-2=19

Therefore, 7 is the answer.

6 0
3 years ago
Convert the rectangular equation to polar form.
N76 [4]
To convert from rectangular coordinate (x, y) to polar coordinate (r, θ).
r is given by the square root of the sum of the squares of the rectangular coordinate.
r= \sqrt{x^2+y^2}
and θ is given by the arctan of the ratio of y to x.
\theta=\tan^{-1} \frac{y}{x}
Example:
To convert rectangular coordinate (3, 4) to polar coordinate
r= \sqrt{3^2+4^2} = \sqrt{9+16} = \sqrt{25} =5 \\ \theta=\tan^{-1}( \frac{4}{3}) =53.13^o
Therefore, rectangular coordinate (3, 4) = polar coordinate (5, 53.13°)
3 0
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 <em>µ(x, y)</em> 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 <em>µ(x, y)</em> = <em>µ(x)</em> be independent of <em>y</em>, then <em>∂µ/∂y</em> = 0 and we can solve for <em>µ</em> :

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 <em>F(x, y)</em> = <em>C</em>. 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 <em>x</em> :

F(x,y) = -e^{-x}y - \dfrac1x + g(y)

Differentiate both sides of this with respect to <em>y</em> :

\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
Write 235 using powers of 10
vaieri [72.5K]

Answer:

(10^2)*2, 10*3, 10/2

Step-by-step explanation:

Thats my best guess to your very vague question

8 0
2 years ago
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