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Butoxors [25]
3 years ago
12

What is the volume of a sphere with a radius of 6.6 cm, rounded to the nearest tenth of a cubic centimeter?

Mathematics
1 answer:
Dvinal [7]3 years ago
6 0

Answer:

Volume=1204.26cm3

Step-by-step explanation:

V=

4

3

π

r

3

=

4

3

π

6.6

3

≈

1204.26043cm³

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Find the cosine of angle B.
Alexandra [31]

\boxed { \text {Cosine Rule : } a^2 = b^2 + c^2 - 2bc \cos(A) }

-

9² = 12² + 15² - 2 (12) (15) cos (B)

81 = 144 + 225 - 360 cos(B)

81 = 369 - 360 cos (B)

360 cos (B) = 369 - 81

360 cos (B) = 288

cos (B) = 0.8

-

Answer: Cosine B = 0.8

-

\boxed { \text {Cosine Rule : } a^2 = b^2 + c^2 - 2bc \cos(A) }

-

12² = 15² + 9² - 2 (15)(9) cos (A)

144 = 225 + 81 - 270 Cos A

144 = 306 - 270 Cos A

270 Cos A = 162

Cos A = 3/5 or 0.6

-

Answer: Cosine Angle A = 3/5

4 0
3 years ago
Set up a system of equations needed to solve each probler
goblinko [34]

Answer:

This is not a solvable question, since the least amount of money you can have with this combination is $4.20.

Step-by-step explanation:

Let q = quarters

Let d = dimes

Equation: 0.1d + 0.25q = 3.00

Maximum amount of dimes: 0.1d *42 = 4.2d

8 0
3 years ago
Each side of a rhombus is 6 cm. The length of the diagonals are 10 cm and 12 cm. What is the
ladessa [460]

Answer:

60cm²

Step-by-step explanation:

The area of a rhombus is: A=d1×d2/2, where d1 and d2 are its diagonals.

A=10×12/2=60cm²

3 0
3 years ago
Fifteen more than twice a number 9 is
Leokris [45]
The question is asking 15+2*9=? so to do this first multiply 2*9 which = 18 and then add 15, 18+15= 33
6 0
3 years ago
Read 2 more answers
Show that 1n^ 3 + 2n + 3n ^2 is divisible by 2 and 3 for all positive integers n.
Dmitry_Shevchenko [17]

Prove:

Using mathemetical induction:

P(n) = n^{3}+2n+3n^{2}

for n=1

P(n)  = 1^{3}+2(1)+3(1)^{2} = 6

It is divisible by 2 and 3

Now, for n=k, n > 0

P(k) = k^{3}+2k+3k^{2}

Assuming P(k) is divisible by 2 and 3:

Now, for n=k+1:

P(k+1) = (k+1)^{3}+2(k+1)+3(k+1)^{2}

P(k+1) = k^{3}+3k^{2}+3k+1+2k+2+3k^{2}+6k+3

P(k+1) = P(k)+3(k^{2}+3k+2)

Since, we assumed that P(k) is divisible by 2 and 3, therefore, P(k+1) is also

divisible by 2 and 3.

Hence, by mathematical induction, P(n) = n^{3}+2n+3n^{2} is divisible by 2 and 3 for all positive integer n.

3 0
3 years ago
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