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3 years ago

5

A rectangular prism has a length of 12 inches, a height of 16 inches, and a width of 20 inches. What is its volume, in cubic inc

hes?
Please explain

1 answer:

drek231 [11]3 years ago

3 0

add 12 with 16 by 20 it gives you 48 cubic inches

12 +16+20=48 cubic inches

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there are 25 students in Hansels class and 17 of the students went on the class trip to the zoo. what percentage of students wen

drek231 [11]

It is given in the problem that out of 25 students in Hansels class 17 students went to the class trip to the zoo. From the above information it is easy to find the percentage of students that went for the class trip to the zoo.
Total number of students in the class = 25
Number of students that went for the class trip to the zoo = 17
Then
Percentage of students that went to the zoo = (17/25) * 100
= (17 * 4) percent
= 68 percent
So 68% of the students in Hansels class went for the class trip to the zoo.

6 0

3 years ago

Will give brainliest answer

Alik [6]

Answer:

Below.

Step-by-step explanation:

log 10 ( 100 ) = 2

5 0

3 years ago

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5.) What is the quotient of 15.26 and 3.5?

elena55 [62]

Answer:

I got 18.26

Step-by-step explanation:

Added them up

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3 years ago

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Find the surface area of each figure.

timurjin [86]

Answer:

For a cylinder of radius R and height H, the surface area is given by:

A = 4*pi*R^2 + H*(2*pi*R)

Where pi = 3.14

We can just use that formula for each one of the given cylinders.

1) We can see that the diameter is 10 yd, and the radius is half of the diameter, then the radius is:

R = 10yd/2 = 5yd

And the height is 3yd, then H = 3yd

Replacing these in the area equation, we get:

A = 4*3.14*(5 yd)^2 + 3yd*(2*3.14*5yd) = 408.2 yd^2

2) Here we can see that the diameter is 24 cm, then the radius is:

R = 24cm/2 = 12cm

And the height is H = 10cm

Then the area is:

A = 4*3.14*(12 cm)^2 + 10cm*(2*3.14*12cm) = 2,562.24 cm^2

3) In this case we have a radius equal to 12 cm, and a height equal to 7 cm, then the area is:

A = 4*3.14*(12 cm)^2 + 7cm*(2*3.14*12cm) = 2,336.16cm^2

4) Here is hard to see the measures, I think that here we have:

diameter = 8m

Then R = 8m/2 = 4m

And the height is also 8m, H = 8m

Then the area is:

A = 4*3.14*(4 m)^2 + 8 m*(2*3.14*4m) =401.92 m^2

6 0

3 years ago

Which of the following is not one of the 8th roots of unity?

Anika [276]

Answer:

1+i

Step-by-step explanation:

To find the 8th roots of unity, you have to find the trigonometric form of unity.

1. Since $z=1=1+0\cdot i,$ then

$Rez=1,\\ \\Im z=0$

and

$|z|=\sqrt{1^2+0^2}=1,\\ \\\\\cos\varphi =\dfrac{Rez}{|z|}=\dfrac{1}{1}=1,\\ \\\sin\varphi =\dfrac{Imz}{|z|}=\dfrac{0}{1}=0.$

This gives you $\varphi=0.$

Thus,

$z=1\cdot(\cos 0+i\sin 0).$

2. The 8th roots can be calculated using following formula:

$\sqrt[8]{z}=\{\sqrt[8]{|z|} (\cos\dfrac{\varphi+2\pi k}{8}+i\sin \dfrac{\varphi+2\pi k}{8}), k=0,\ 1,\dots,7\}.$

Now

at k=0, $z_0=\sqrt[8]{1} (\cos\dfrac{0+2\pi \cdot 0}{8}+i\sin \dfrac{0+2\pi \cdot 0}{8})=1\cdot (1+0\cdot i)=1;$

at k=1, $z_1=\sqrt[8]{1} (\cos\dfrac{0+2\pi \cdot 1}{8}+i\sin \dfrac{0+2\pi \cdot 1}{8})=1\cdot (\dfrac{\sqrt{2}}{2}+i\dfrac{\sqrt{2}}{2})=\dfrac{\sqrt{2}}{2}+i\dfrac{\sqrt{2}}{2};$

at k=2, $z_2=\sqrt[8]{1} (\cos\dfrac{0+2\pi \cdot 2}{8}+i\sin \dfrac{0+2\pi \cdot 2}{8})=1\cdot (0+1\cdot i)=i;$

at k=3, $z_3=\sqrt[8]{1} (\cos\dfrac{0+2\pi \cdot 3}{8}+i\sin \dfrac{0+2\pi \cdot 3}{8})=1\cdot (-\dfrac{\sqrt{2}}{2}+i\dfrac{\sqrt{2}}{2})=-\dfrac{\sqrt{2}}{2}+i\dfrac{\sqrt{2}}{2};$

at k=4, $z_4=\sqrt[8]{1} (\cos\dfrac{0+2\pi \cdot 4}{8}+i\sin \dfrac{0+2\pi \cdot 4}{8})=1\cdot (-1+0\cdot i)=-1;$

at k=5, $z_5=\sqrt[8]{1} (\cos\dfrac{0+2\pi \cdot 5}{8}+i\sin \dfrac{0+2\pi \cdot 5}{8})=1\cdot (-\dfrac{\sqrt{2}}{2}-i\dfrac{\sqrt{2}}{2})=-\dfrac{\sqrt{2}}{2}-i\dfrac{\sqrt{2}}{2};$

at k=6, $z_6=\sqrt[8]{1} (\cos\dfrac{0+2\pi \cdot 6}{8}+i\sin \dfrac{0+2\pi \cdot 6}{8})=1\cdot (0-1\cdot i)=-i;$

at k=7, $z_7=\sqrt[8]{1} (\cos\dfrac{0+2\pi \cdot 7}{8}+i\sin \dfrac{0+2\pi \cdot 7}{8})=1\cdot (\dfrac{\sqrt{2}}{2}-i\dfrac{\sqrt{2}}{2})=\dfrac{\sqrt{2}}{2}-i\dfrac{\sqrt{2}}{2};$

The 8th roots are

$\{1,\ \dfrac{\sqrt{2}}{2}+i\dfrac{\sqrt{2}}{2},\ i, -\dfrac{\sqrt{2}}{2}+i\dfrac{\sqrt{2}}{2},\ -1, -\dfrac{\sqrt{2}}{2}-i\dfrac{\sqrt{2}}{2},\ -i,\ \dfrac{\sqrt{2}}{2}-i\dfrac{\sqrt{2}}{2}\}.$

Option C is icncorrect.

5 0

3 years ago