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2 years ago

9. Jenny is selling t-shirts to raise money for the cheerleading team. She sells long sleeve t-shirts

Mathematics

1 answer:

Lorico [155]2 years ago

8 0

Answer:

43 long sleeve t-shirts

Step-by-step explanation

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The area of square pond is 5625meter square and 2 meter width path is made around the pond

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Answer:

Step-by-step explanation:

good luck hoped it helped

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What did the exponent say to the base? Riddle

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3 years ago

Monica bought a sweater for $48.00. A week later the sweater went on sale for 25% off the original price. If Monica needed to pa

IRINA_888 [86]

Answer:

A $12.84

Step-by-step explanation:

Sweater was originally $48.00 before tax. $48.00 x 7% tax rate (.07) = $3.36 in tax

So the sweater originally would've cost $48.00 + $3.36 = $51.36

The sale price was 25% off the $48.00. $48.00 x 25% (.25) = $12.00

Then you subtract the $12.00 from the $48.00 to get the new price of $36.00

Now that's before tax, so $36.00 x 7% tax rate (.07) = $2.52 in tax

So the sweater would cost on sale $36.00 + $2.52 = $38.52

To figure out how much she would've saved you subtract the sale price & tax from the original price & tax $51.36 - $38.52 = $12.84

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1 year ago

Find the value of kk for which the constant function x(t)=kx(t)=k is a solution of the differential equation 5t3dxdt+2x−2=05t3dx

uranmaximum [27]

Given the differential equation

$5t^3 \frac{dx}{dt} +2x-2=0$

The solution is as follows:

$5t^3 \frac{dx}{dt} +2x-2=0 \\ \\ \Rightarrow5t^3 \frac{dx}{dt} =2-2x \\ \\ \Rightarrow \frac{5}{2-2x} dx= \frac{1}{t^3} dt \\ \\ \Rightarrow \int {\frac{5}{2-2x} } \, dx = \int {\frac{1}{t^3}} \, dt \\ \\ \Rightarrow- \frac{5}{2} \ln(2-2x)=- \frac{1}{2t^2} +A \\ \\ \Rightarrow\ln(2-2x)= \frac{1}{5t^2} +B\\ \\ \Rightarrow2-2x=Ce^{\frac{1}{5t^2}} \\ \\ \Rightarrow 2x=Ce^{\frac{1}{5t^2}}+2 \\ \\ \Rightarrow x=De^{\frac{1}{5t^2}}+1$

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3 years ago

A submarine started at 750 meters below sea level. It rose 50 meters per hour over a 4-hour period. Which expression represents

Cerrena [4.2K]

Answer:

$y(t) = 750 -50t$

Step-by-step explanation:

We want to model the position of the submarine as a function of time.

Notice that we have a constant amount. The initial depth: 750 meters

Then we have a factor that varies over time. Every hour the submarine ascends 50 meters. therefore as t increases the depth y(t) of the submarine decreases. Then the factor is:

-50t.

Where t represents the time in hours.

Then the equation that represents the new position of the submarine in relation to the level of the sea:

$y(t) = 750 -50t$

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