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gayaneshka [121]
2 years ago
11

What is the correct answer to this?

Mathematics
2 answers:
ra1l [238]2 years ago
7 0

Answer:

33.4933 ≈ 33.5 mm³

Step-by-step explanation:

Given that,

→ Radius (r) = 2 mm

Formula we use,

→ 4/3 πr³

Let's solve for the volume of sphere,

→ 4/3 πr³

→ 4/3 × 3.14 × 2 × 2 × 2

→ 4/3 × 3.14 × 8

→ 4/3 × 25.12

→ 100.48/3

→ 33.4933

→ [ ≈ 33.5 mm³ ]

Hence, volume of sphere is 33.5 mm³.

Maksim231197 [3]2 years ago
4 0
  • Radius=r=2mm

Now

Volume

  • V=4/3πr³
  • V=4/3π(2)³
  • V=4/3π(8)
  • V=32π/3
  • V=33.49mm³
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Given f(x) = 4x^2 + 19x - 5 and g(x) = 4x^2 - x what is (f/g)(x)?
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Use induction to prove: For every integer n > 1, the number n5 - n is a multiple of 5.
nignag [31]

Answer:

we need to prove : for every integer n>1, the number n^{5}-n is a multiple of 5.

1) check divisibility for n=1, f(1)=(1)^{5}-1=0  (divisible)

2) Assume that f(k) is divisible by 5, f(k)=(k)^{5}-k

3) Induction,

f(k+1)=(k+1)^{5}-(k+1)

=(k^{5}+5k^{4}+10k^{3}+10k^{2}+5k+1)-k-1

=k^{5}+5k^{4}+10k^{3}+10k^{2}+4k

Now, f(k+1)-f(k)

f(k+1)-f(k)=k^{5}+5k^{4}+10k^{3}+10k^{2}+4k-(k^{5}-k)

f(k+1)-f(k)=k^{5}+5k^{4}+10k^{3}+10k^{2}+4k-k^{5}+k

f(k+1)-f(k)=5k^{4}+10k^{3}+10k^{2}+5k

Take out the common factor,

f(k+1)-f(k)=5(k^{4}+2k^{3}+2k^{2}+k)      (divisible by 5)

add both the sides by f(k)

f(k+1)=f(k)+5(k^{4}+2k^{3}+2k^{2}+k)

We have proved that difference between f(k+1) and f(k) is divisible by 5.

so, our assumption in step 2 is correct.

Since f(k) is divisible by 5, then f(k+1) must be divisible by 5 since we are taking the sum of 2 terms that are divisible by 5.

Therefore, for every integer n>1, the number n^{5}-n is a multiple of 5.

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