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STALIN [3.7K]
2 years ago
13

Manuel bought a balloon (that is a perfect sphere) with a radius of 2 cm. He wanted his balloon to be bigger, so he blew 2 big b

reaths of air into the balloon. Each big breath increased the balloon's radius by 1 cm. Please help!!!!
Mathematics
2 answers:
Lilit [14]2 years ago
6 0

Answer:

I guess we want to evaluate how much the volume of the balloon changes in each blow.

The volume of a sphere can be written as:

v = (4/3)*pi*r^3

where pi = 3.1416 and r is the radius of the sphere.

we have that initially, the radius of the sphere is r = 2cm

then v0 = (4/3)*3.1416*(2cm)^3 = 33.5 cm^3

now, after one blow, the radius is 3 cm, so we have:

v1 = (4/3)*3.1416*(3cm)^3 = 113.1cm^3

this means that the change in volume is  

v1 - v0 = 97.6cm^3

and after the other blow, we have that the radius is 4cm:

v2 = (4/3)*3.1416*(4cm)^3 = 268.1cm^3

the change in the volume is:

v2 - v1 = 154.9 cm^3

So we can see that the change of 1cm in the radius has a bigger impact if the initial radius of the sphere is bigger.

horsena [70]2 years ago
5 0
What is the question asking?
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Answer:

The simplyfied version would be 19/4

Show of work:

(1/4)^-2 = 4^2

3 × 8^2/3 × 1 = 12

(9/16)^1/2 = 3/4

4^2 - 12 + 3/4

Convert elements to fractions:

-12 × 4 + 3

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

4 4

Since the denominators are equal combine the fractions:

-12 × 4 + 3

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

4

-12 × 4 + 3 = -45

= -45/4

=4^2 - 45/4

4^2 = 16

16 - 45/4

16 × 4 - 45. 16 × 4 - 45

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

4 4. 4

-> 16 × 4 - 45 = 19

= 19/4

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