The approximate amount of water that remains in the tub after the 6 spherical balls are placed in the tub are 3479.12 in³.
<h3>What is the approximate amount of water that remains in the tub?</h3>
The first step is to determine the volume of the cylinder.
Volume of the tub = πr²h
Where:
- r = radius = diameter / 2 = 18/2 = 9 inches
- h = height
- π = 3.14
3.14 x 9² x 20 = 5086.8 in³
The second step is to determine the volume of the 6 balls.
Volume of a sphere= 4/3πr³
r = diameter / 2 = 8/2 = 4 inches
6 x (3.14 x 4/3 x 4³) = 1607.68 in³
Volume that remains in the tub = 5086.8 in³ - 1607.68 in³ = 3479.12 in³
To learn more about the volume of a sphere, please check: brainly.com/question/13705125
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The square is 130 in squared and the other shape is 54 in squared. If you're looking for the sum of the whole thing, its 185 in squared.
square: 13*10=130
triangle: 7*4= 28/2=14
rectangle: 10*4=40
Answer:
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Step-by-step explanation:
Given
See attachment for complete question
Required
Match equivalent expressions
Solving (a):
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The expression can be written as:
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--- 2
---- 3
---- 4
For the nth term, the expression is:
---- n
So, the summation is:
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Solving (b):
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The expression can be written as:
--- 0
---- 1
--- 2
---- 3
---- 4
For the nth term, the expression is:
---- n
So, the summation is:
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Solving (c):
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The expression can be written as:
--- 0
---- 1
--- 2
---- 3
---- 4
For the nth term, the expression is:
---- n
So, the summation is:
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Solving (d):
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The expression can be written as:
--- 0
---- 1
--- 2
---- 3
---- 4
For the nth term, the expression is:
---- n
So, the summation is:
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Answer:
About 8.6
Step-by-step explanation:
d=(x2-x1)^2+(y2-y1)^2
8=x1 8=y1 3=x2 1=y2
(3-8)^2+(1-8)^2
(-5)^2+(-7)^2
25+49
74
take the square root and you get about 8.6
Answer:
The dimensions on paper are 1.992ft by 5.976ft with a scale factor of 7.53
Step-by-step explanation:
The first step will be to find the diagonal of the rea life mural.
We can use Pythagoras' Theorem to do this.
Diagonal =
=47.43 feet.
Now we have the real-life diagonal, we will now relate the diagonal of the painting outside with the one on paper. We can do this by dividing the two diagonals.
This will be 47.43 / 6.3 units = 7.53.
The scale factor is 7.53
To get the dimensions of the length and the breadth on paper, we divide the outside dimensions by the scale factor.
This will be
Length = 15/ 7.53 = 1.992
Breadth = 45/7.53 = 5.976
Therefore, the dimensions on paper are 1.992ft by 5.976ft with a scale factor of 7.53