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balandron [24]
3 years ago
5

Find the volume of each sphere. Round to the nearest tenth.. volume V=4/3

Mathematics
1 answer:
FrozenT [24]3 years ago
3 0
The complete question is
Find the volume of each sphere for the given radius. <span>Round to the nearest tenth

we know that
[volume of a sphere]=(4/3)*pi*r</span>³

case 1) r=40 mm
[volume of a sphere]=(4/3)*pi*40³------> 267946.66 mm³-----> 267946.7 mm³

case 2) r=22 in
[volume of a sphere]=(4/3)*pi*22³------> 44579.63 in³---->  44579.6 in³

case 3) r=7 cm
[volume of a sphere]=(4/3)*pi*7³------> 1436.03 cm³---->  1436 cm³

case 4) r=34 mm
[volume of a sphere]=(4/3)*pi*34³------> 164552.74 mm³----> 164552.7 mm³

case 5) r=48 mm
[volume of a sphere]=(4/3)*pi*48³------> 463011.83 mm³----> 463011.8 mm³

case 6) r=9 in
[volume of a sphere]=(4/3)*pi*9³------> 3052.08 in³----> 3052 in³

case 7) r=6.7 ft
[volume of a sphere]=(4/3)*pi*6.7³------> 1259.19 ft³-----> 1259.2 ft³

case 8) r=12 mm
[volume of a sphere]=(4/3)*pi*12³------>7234.56 mm³-----> 7234.6 mm³
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xeze [42]
<h3>Answer: 1</h3>

where x is nonzero

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

Explanation:

We'll use two rules here

  • (a^b)^c = a^(b*c) ... multiply exponents
  • a^b*a^c = a^(b+c) ... add exponents

------------------------------

The portion [ x^(a-b) ]^(a+b) would turn into x^[ (a-b)(a+b) ] after using the first rule shown above. That turns into x^(a^2 - b^2) after using the difference of squares rule.

Similarly, the second portion turns into x^(b^2-c^2) and the third part becomes x^(c^2-a^2)

-------------------------------

After applying rule 1 to each of the three pieces, we will have 3 bases of x with the exponents of (a^2-b^2),  (b^2-c^2) and (c^2-a^2)

Add up those exponents (using rule 2 above) and we get

(a^2-b^2)+(b^2-c^2)+(c^2-a^2)

a^2-b^2+b^2-c^2+c^2-a^2

(a^2-a^2) + (-b^2+b^2) + (-c^2+c^2)

0a^2 + 0b^2 + 0c^2

0+0+0

0

All three exponents add to 0. As long as x is nonzero, then x^0 = 1

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