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slava [35]
3 years ago
10

What is the volume of a hemisphere with a diameter of 45.7 m, rounded to the nearest tenth of a cubic meter?

Mathematics
1 answer:
worty [1.4K]3 years ago
6 0

24987.2 meters cubed.

Step-by-step explanation:

A Hemisphere is 1/2 of a sphere, so let's find the volume of a sphere and then cut it in half.

The volume of a sphere is V= (4/3)πr3

The diameter is double the size of the radius, so we can find r, the radius, by dividing the diameter,

45.7m, by 2. So r = 22.85 meters

so the volume of the sphere is (4/3)π(22.85 meters)3, which is 49974.35787 meters cubed.

Since we're actually looking for the Hemisphere, we can divide this volume in half to get the volume of the hemisphere as 24987.17894 meters cubed.

And because the answer must be to 1/10th of a cubic meter, that means we only want one decimal point. So we round to 24987.2 meters cubed.

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How do I solve this?
Snowcat [4.5K]
M>B = 65 so arc AC = 2(65) = 130
arc ABC = 360 - 130 = 230

m<D  = 1/2(230 - 130)
m<D = 1/2(100) = 50

answer
m<D = 50 first choice
6 0
3 years ago
I need help with #61 to #64 ASAP please and thank you …. Could someone please help me with it… I need to get it done ASAP
aivan3 [116]

Problem 61

The nth triangular number is

T(n) = n(n+1)/2

I'll rewrite this into

T(n) = 0.5n(n+1)

The triangular number right after this is

T(n+1) = 0.5(n+1)(n+2)

I replaced every n with n+1 and simplified

Let's see what we get when we add up the two expressions

T(n) + T(n+1)

0.5n(n+1) + 0.5(n+1)(n+2)

0.5n^2+0.5n + 0.5(n^2+3n+2)

0.5n^2 + 0.5n + 0.5n^2 + 1.5n + 1

n^2+2n+1

(n+1)^2

This shows that the sum of any two consecutive triangular numbers results in a square number

Here's a few examples

  • 0+1 = 1
  • 1+3 = 4
  • 3+6 = 9
  • 6+10 = 16
  • 10+15 = 25

Note each sum is a perfect square, which visually would plot out a square figure.

For quick reference, the set of the first few triangular numbers is {0, 1, 3, 6, 10, 15, 21, 28, 36, 45, 55,...}

<h3>Answer: Square number</h3>

==========================================================

Problem 64

Let's say we go with n = 5.

This means,

T(n) = 0.5n(n+1)

T(n-1) = 0.5(n-1)(n-1+1)

T(n-1) = 0.5n(n-1)

T(5-1) = 0.5*5(5-1)

T(4) = 10

This says that when n = 5, the 4th triangular number is 10

Triple that result and add on n = 5

3*T(4) + n = 3*10+5 = 35

This result is beyond obvious which category of figurate number it belongs to. It's not a triangular number since it's not in the form n(n+1)/2. It's not a square number either.

Through a bit of trial and error, you should find it's a pentagonal number

Pentagonal numbers are of the form n(3n-1)/2

If you plugged n = 5 into that, it leads to 35

n(3n-1)/2 = 5*(3*5-1)/2 = 5*14/2 = 70/2 = 35

The diagram shown below represents the first few pentagonal numbers. The number of blue dots corresponds to the pentagonal number itself. Note the equal spacing when dealing with dots on each segment (eg: some interior blue dots are midpoints, others are quarter points, etc.)

<h3>Answer: Pentagonal number</h3>

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3 9/100 would be written as a decimal like this, 3.09. You say three and nine one hundredths. The 9 has to be in the hundredths place, so you put a 0 in front of it. And since the 3 is hole, you put it in front of the decimal point.
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