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Alborosie
2 years ago
10

Need a little help!

Mathematics
1 answer:
zhenek [66]2 years ago
5 0
Answer = 7234.6 cm cu.
Reason
Volume of sphere
V= 4/3 x TT x r^3
V = 4/3 x 3.14 x 12^3
V= 7234.56
Rounded to 7234.6
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Javier has a model of a cone with a radius of 3 inches and a height of 10 inches. Javier increased his model’s volume by doublin
statuscvo [17]

Answer:

240 \pi\ in^{3}

Step-by-step explanation:

we know that

If two figures are similar, then the ratio of its volumes is equal to the scale factor elevated to the cube

Let

z------> the scale factor

x-----> the volume of the larger cone

y-----> the volume of the original cone

z^{3}=\frac{x}{y}

In this problem we have

z=2 -----> scale factor

substitute

2^{3}=\frac{x}{y}

8=\frac{x}{y}

x=8y

That means-----> The volume of the larger cone is  8 times the volume of the original cone

Find the volume of the original cone

V=\frac{1}{3}\pi r^{2} h

we have

r=3\ in

h=10\ in

substitute

V=\frac{1}{3}\pi (3^{2})(10)=30 \pi\ in^{3}

therefore

The volume of the larger cone is equal to

8*30 \pi\ in^{3}=240 \pi\ in^{3}

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3 years ago
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Suppose five people are in a room. What is the probability that there is at least one shared birthday among these five people?
iris [78.8K]

I think it would be 73

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Jaden makes 63% of the free throws he shoots. About how many shots do you predict he will have to take to make 15 free throws?
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Answer:

Step-by-step explanation:

49%

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QUESTION IN THE ATTACHMENT
eimsori [14]

Answer:

A. The sum of the first 10th term is 100.

B. The sum of the nth term is n²

Step-by-step explanation:

Data obtained from the question include:

Sum of 20th term (S20) = 400

Sum of 40th term (S40) = 1600

Sum of 10th term (S10) =..?

Sum of nth term (Sn) =..?

Recall:

Sn = n/2[2a + (n – 1)d]

Sn is the sum of the nth term.

n is the number of term.

a is the first term.

d is the common difference

We'll begin by calculating the first term and the common difference. This is illustrated below:

Sn = n/2 [2a + (n – 1)d]

S20 = 20/2 [2a + (20 – 1)d]

S20= 10 [2a + 19d]

S20 = 20a + 190d

But:

S20 = 400

400 = 20a + 190d .......(1)

S40 = 40/2 [2a + (40 – 1)d]

S40 = 20 [2a + 39d]

S40 = 40a + 780d

But

S40 = 1600

1600 = 40a + 780d....... (2)

400 = 20a + 190d .......(1)

1600 = 40a + 780d....... (2)

Solve by elimination method

Multiply equation 1 by 40 and multiply equation 2 by 20 as shown below:

40 x equation 1:

40 x (400 = 20a + 190d)

16000 = 800a + 7600. ........ (3)

20 x equation 2:

20 x (1600 = 40a + 780d)

32000 = 800a + 15600d......... (4)

Subtract equation 3 from equation 4

Equation 4 – Equation 3

32000 = 800a + 15600d

– 16000 = 800a + 7600d

16000 = 8000d

Divide both side by 8000

d = 16000/8000

d = 2

Substituting the value of d into equation 1

400 = 20a + 190d

d = 2

400 = 20a + (190 x 2)

400 = 20a + 380

Collect like terms

400 – 380 = 20a

20 = 20a

Divide both side by 20

a = 20/20

a = 1

Therefore,

First term (a) = 1.

Common difference (d) = 2.

A. Determination of the sum of the 10th term.

First term (a) = 1.

Common difference (d) = 2

Number of term (n) = 10

Sum of 10th term (S10) =..?

Sn = n/2 [2a + (n – 1)d]

S10 = 10/2 [2x1 + (10 – 1)2]

S10 = 5 [2 + 9x2]

S10 = 5 [2 + 18]

S10 = 5 x 20

S10 = 100

Therefore, the sum of the first 10th term is 100.

B. Determination of the sum of the nth term.

First term (a) = 1.

Common difference (d) = 2

Sum of nth term (Sn) =..?

Sn = n/2 [2a + (n – 1)d]

Sn = n/2 [2x1 + (n – 1)2]

Sn = n/2 [2 + 2n – 2]

Sn = n/2 [2 – 2 + 2n ]

Sn = n/2 [ 2n ]

Sn = n²

Therefore, the sum of the nth term is n²

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3 years ago
How many solutions?
Sunny_sXe [5.5K]

Answer: C

Step-by-step explanation: all real numbers can solve this problem

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3 years ago
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