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svet-max [94.6K]
2 years ago
6

A ball as a radius of 18cm. What is the approximate volume of the ball?

Mathematics
1 answer:
KatRina [158]2 years ago
7 0

Answer:

what I got was 24438.9cm³

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Step-by-step explanation:

The error carolyn has made is that she hasn't subtracted properly.

2x + 3y - (2x +6y) = 2x - 2x + 3y - 6y = -3y.

Carolyn probably didn't use parentheses...

To do the thing in substitution, we single out x or y in any equation. In this case, i'm gonna single out the x in the 2nd eqn. In other words, i'll make x the subject of the 2nd eqn.

2x + 6y = 10

2x         = 10 - 6y

x           = 5 - 3y

Now we simply plug in 5 - 3y wherever x apperas in the 1st equation.

2 (5-3y) + 3y = 8

10 - 6y + 3y   = 8

10 - 3y           = 8

-3y                 =  -2

y                    = 2/3

Then x = 5 - 3*2/3 = 3

x=3

y=2/3

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Built around 2600 BCE, the Great Pyramid of Giza in Egypt is 485 ft high (due to erosion, its current height is slightly less) a
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Answer:1.69*10^12 J

Step-by-step explanation:

From figure above, using triangle ratio

485/755.5=y/l. Cross multiplying 485l=755.5y Divide via 485) hence l= 755.5y/485

Consider a slice volume Vslice= (755.5y/485)^2∆y; recall density =150lb/ft^3

Force slice = 150*755.5^2.y^2.∆y/485^2

From figure 2 in the attachment work done for elementary sclice

Wslice= 150.755.5^2.y^2.∆y.(485-y)/485^2

= (150*755.5^2*y^2)(485-y)∆y/485

To calculate the total work we integrate from y=0 to y= 485

Ie W=[ integral of 150*755.5^2 *y^2(485-y)dy/485] at y=0 and y= 485

Integrating the above

W= 150*755.5^2/485[485*y^3/3-y^4/4] at y= 0 and y=485

W= 150*755.5^2/485(485*485^3/3-484^4/4)-(485.0^3/3-0^4/4)

Work done 1.69*10^12joules

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