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4 years ago

Frank borrows$2000 for 2years and pay back $2160 . What simple interest rate was he charged?

Mathematics

1 answer:

Mrac [35]4 years ago

6 0

He had to pay back 1080 per year or 90 per month in total he had to pay $160 more dollars of interest then what he got

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-4 + 8 - 5

pochemuha

Answer:

-1

Step-by-step explanation:

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1 year ago

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Bianca makes a scale drawing of a rectangular garden. The length is 9 in., and the width is 6 in. Bianca changes the scale of th

Artist 52 [7]

Old scale : 1 in. to 3 ft.
length is 9 in.....9 * 3 = 27 ft (old scale length)
width is 6 in......6 * 3 = 18 ft (old scale width)

new scale : 1 in. to 4 ft
length is 9 in.....9 * 4 = 36 ft (new scale length)
width is 6 in......6 * 4 = 24 ft (new scale width)

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3 years ago

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Find the sum of the first four terms of the geometric sequence shown below. 4, 4/ 3, 4/9, ...

user100 [1]

It's evident that the first four terms are 4, 4/3, 4/9, and 4/27. So the fourth partial sum of the series is

$S_4=4+\dfrac43+\dfrac49+\dfrac4{27}$

It's as easy as adding up the fractions, but I bet this is supposed to be an exercise in taking advantage of the fact that the series is geometric and use the well-known formula for computing such a sum.

Multiply the sum by 1/3 and you have

$\dfrac13S_4=\dfrac43+\dfrac49+\dfrac4{27}+\dfrac4{81}$

Now subtracting this from $S_4$ gives

$S_4-\dfrac13S_4=4-\dfrac4{81}$

That is, all the matching terms will cancel. Now solving for $S_4$ , you
have

$\dfrac23S_4=4\left(1-\dfrac1{81}\right)$
$S_4=6\left(1-\dfrac1{81}\right)$
$S_4=\dfrac{480}{81}=\dfrac{160}{27}$

3 0

3 years ago

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

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