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3 years ago

5

The school store sells Midville High T-shirts for $12 each and Midville High gym shorts for $9 a pair. One afternoon they sold a

total of 53 T-shirts and pairs of shorts for a total of $570 . How many T-shirts did they sell and how many pairs of shorts did they sell?

1 answer:

Oksi-84 [34.3K]3 years ago

7 0

Answer:

The store sold 21 T-shirts and 18 pairs of shorts.

Step-by-step explanation:

x = the number of T-shirts sold

y = the number of pairs of shorts sold

x + y = 53

12x + 9y = 570

x = 21 T-shirts

y = 18 shorts

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3 years ago

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The amount of material removed is the volume of the region within the sphere bounded by the cylinder. Consider a sphere of radius $2a$ centered at the origin; this sphere has equation

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$\displaystyle\iiint_{\mathcal D}\mathrm dV=\int_{\theta=0}^{\theta=2\pi}\int_{r=0}^{r=a}\int_{\zeta=-\sqrt{4a^2-r^2}}^{\zeta=\sqrt{4a^2-r^2}}r\,\mathrm d\zeta\,\mathrm dr\,\mathrm d\theta$

The integral with respect to $\zeta$ is symmetric about $\zeta=0$ , so we instead compute twice the integral from $\zeta=0$ to $\zeta=\sqrt{4a^2-r^2}$ , and we can immediately compute the integral with respect to $\theta$ :

$=\displaystyle4\pi\int_{r=0}^{r=a}r\sqrt{4a^2-r^2}\,\mathrm dr$

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$=\displaystyle-2\pi\int_{r=0}^{r=a}-2r\sqrt{4a^2-r^2}\,\mathrm dr=-2\pi\int_{s=4a^2}^{s=3a^2}\sqrt s\,\mathrm ds$

$=-2\pi\cdot\dfrac23s^{3/2}\bigg|_{s=4a^2}^{s=3a^2}$

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4 years ago

3.6/= 1.2/2 help please

LUCKY_DIMON [66]

I think the question is supposed to be 3.6/y = 1.2/2

STEP 1: Divide 1.2 by 2 to get 0.6

3.6/y = 0.6

STEP 2: Multiply each term by y and simplify

3.6 = 0.6y

STEP 3: Since y is on the right side of the equation, switch the sides so it is on the left side of the equation

0.6y = 3.6

STEP 4: Divide each term by 0.6 and simplify

y = 6

SO, THE ANSWER IS...

y = 6

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3 years ago