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kow [346]
4 years ago
15

Emma is training for a 10-kilometer race. She wants to beat her last 10-kilometer time, which was 1 hour 10 minutes. Emma has al

ready run for 55 minutes. Which inequality can be used to find how much longer she can run and still beat her previous time? (1 hr = 60 min)

Mathematics
2 answers:
Sauron [17]4 years ago
4 0

Answer:

70 > I+55

Step-by-step explanation:

The last inequality expresses better this situations.

Emma wants to beat her last record, which was 70 minutes. So, she needs to run in less than 70 minutes.

However, she already ran 55 minutes, the remaining time would be the variable I, so, the sum of them must be less than 70 minutes. So, there are two options to represent this:

I + 55 < 70 or 70 > I + 55.

And if this helped you, feel free to make this the brainliest answer!

yanalaym [24]4 years ago
4 0

Answer:

(D) 70 greater-than t + 55   =  70>t+55

Step-by-step explanation:

70 greater-than 7 minus 55

70 less-than t minus 55

70 less-than-or-equal-to t + 55

70 greater-than t + 55

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The volume of a sphere whose diameter is 18 centimeters is _ cubic centimeters. If it’s diameter we’re reduced by half, it’s vol
kaheart [24]
<h2>Answer:</h2>

<u>First Part</u>

Given that

Volume = \frac{4}{3} \pi r^{3}

We have that

Volume = \frac{4}{3} \pi r^{3} =  \frac{4}{3} \pi (\frac{Diameter}{2})^{3} =  \frac{4}{3} \pi 9^{3} = 972\pi cm^{3} \approx 3053.63 cm^{3}

<u>Second Part</u>

Given that

Volume = \frac{4}{3} \pi r^{3}

If the Diameter were reduced by half we have that

Volume = \frac{4}{3} \pi r^{3} =  \frac{4}{3} \pi (\frac{r}{2}) ^{3} = \frac{\frac{4}{3} \pi r^{3}}{8}

This shows that the volume would be \frac{1}{8} of its original volume

<h2>Step-by-step explanation:</h2>

<u>First Part</u>

Gather Information

Diameter = 18cm

Volume = \frac{4}{3} \pi r^{3}

Calculate Radius from Diameter

Radius = \frac{Diameter}{2} = \frac{18}{2} = 9

Use the Radius on the Volume formula

Volume = \frac{4}{3} \pi r^{3} =  \frac{4}{3} \pi 9^{3}

Before starting any calculation, we try to simplify everything we can by expanding the exponent and then factoring one of the 9s

Volume = \frac{4}{3} \pi 9^{3} = \frac{4}{3} \pi 9 * 9 * 9 = \frac{4}{3} \pi 9 * 9 * 3 * 3

We can see now that one of the 3s can be already divided by the 3 in the denominator

Volume = \frac{4}{3} \pi 9 * 9 * 3 * 3 = 4 \pi 9 * 9 * 3

Finally, since we can't simplify anymore we just calculate it's volume

Volume = 4 \pi 9 * 9 * 3 = 12 \pi * 9 * 9 = 12 * 81 \pi = 972 \pi cm^{3}

Volume \approx 3053.63 cm^{3}

<u>Second Part</u>

Understanding how the Diameter reduced by half would change the Radius

Radius =\frac{Diameter}{2}\\\\If \\\\Diameter = \frac{Diameter}{2}\\\\Then\\\\Radius = \frac{\frac{Diameter}{2} }{2} = \frac{\frac{Diameter}{2}}{\frac{2}{1}} = \frac{Diameter}{2} * \frac{1}{2} = \frac{Diameter}{4}

Understanding how the Radius now changes the Volume

Volume = \frac{4}{3}\pi r^{3}

With the original Diameter, we have that

Volume = \frac{4}{3}\pi (\frac{Diameter}{2}) ^{3} = \frac{4}{3}\pi \frac{Diameter^{3}}{2^{3}}\\\\ = \frac{4}{3}\pi \frac{Diameter^{3}}{2 * 2 * 2} = \frac{4}{3}\pi \frac{Diameter^{3}}{8}\\\\

If the Diameter were reduced by half, we have that

Volume = \frac{4}{3}\pi (\frac{Diameter}{4}) ^{3} = \frac{4}{3}\pi \frac{Diameter^{3}}{4^{3}}\\\\ = \frac{4}{3}\pi \frac{Diameter^{3}}{4 * 4 * 4} = \frac{4}{3}\pi \frac{Diameter^{3}}{4 * 2 * 2 * 4} = \frac{4}{3}\pi \frac{Diameter^{3}}{8 * 8} = \frac{\frac{4}{3}\pi\frac{Diameter^{3}}{8}}{8}

But we can see that the numerator is exactly the original Volume!

This shows us that the Volume would be  \frac{1}{8} of the original Volume if the Diameter were reduced by half.

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2 years ago
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Answer:

Hello,

16\pi

Step-by-step explanation:

I=\dfrac{Area}{4} =\int\limits^4_0 {\sqrt{16-x^2} } \, dx \\\\Let\ say\ x=4*sin(t),\ dx=4*cos(t) dt\\\\\displaystyle I=\int\limits^\frac{\pi }{2} _0 {4*\sqrt{1-sin^2(t)} }*4*cos(t) \, dt \\\\=16*\int\limits^\frac{\pi }{2} _0 {cos^2(t)} \, dt \\\\=16*\int\limits^\frac{\pi }{2} _0 {\frac{1-cos(2t)}{2}} \, dt \\\\=8*[t]^\frac{\pi }{2} _0-[\frac{sin(2t)}{2} ]^\frac{\pi }{2} _0\\\\=4\pi -0\\\\=4\pi\\\\\boxed{Area=4*I=16\pi}\\

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3 years ago
A school is painting its logo in the shape of a triangle in the middle of its sports field. The school wants the height of the t
pochemuha

Answer:

Possible base lengths are 1, 2, 3, 4, 5

Step-by-step explanation:

Area = base x height / 2

Let base be b

b x 8 / 2 < 24 (Area < 24)

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How much would $125 invested at 8% interest compounded continuously be worth after 16 years? Round your answer to the nearest ce
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D.

Step-by-step explanation:

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