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3 years ago

What is the volume of a regular pyramid having a base area of 24 in.² and a height of 6 inches?

Mathematics

1 answer:

Assoli18 [71]3 years ago

3 0

Answer:

48 in³

Step-by-step explanation:

The volume (V) of a pyramid is calculated as

V = $\frac{1}{3}$ × area of base × height

here area of base = 24 and height = 6, so

V = $\frac{1}{3}$ × 24 × 6 = 8 × 6 = 48 in³

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Answer all 3 for brainliest

fiasKO [112]

Answer:

7.) 53.38

8.) 18.84

9.) 87.92

Step-by-step explanation:

7.)

For the first question, we can see that the diameter is given. So, we will want to divide that diameter by 2 to get the radius. 17 ÷ 2 = 8.5.

The formula for finding the circumference is 2 × π × 8.5

2 × 3.14(π) = 6.28.

6.28 × 8.5 = 53.38

8.)

For the second question, we can see that the radius is now given to us which is 3ft.

The formula for finding the circumference is 2 × π × 3

2 × 3.14(π) = 6.28.

6.28 × 3 = 18.84

9.)

For the third question the raduis is also given to us, which is 14 in.

The formula for finding the circumference is 2 × π × 14.

2 × 3.14(π) = 6.28.

6.28 × 14 = 87.92

<em>-kiniwih426</em>

7 0

3 years ago

F(5) if f(x) = 3x - 2

iragen [17]

Answer:

f(5)=12

Step-by-step explanation:

Replace all x by 5

f(5)=3(5)-2

multiply 3 and 5 together

f(5)=15-2

Simplify:

f(5)=13

Hope this helps!

7 0

3 years ago

Read 2 more answers

Scientists have discovered a vaccine for a debilitating disease. This year there were 570,000 reported new cases of the disease

Ket [755]

Answer: There are approximately 853827 new cases in 6 years.

Step-by-step explanation:

Since we have given that

Initial population = 570000

Rate at which population decreases is given by

$\frac{2}{3}$

Now,

First year =570000

Second year is given by

$570000\times (\frac{1}{3})$

Third year is given by

$570000(\frac{1}{3})^2$

so, there is common ratio ,

it becomes geometric progression, as there is exponential decline.

so,

$570000,570000\times \frac{1}{3},570000\times( \frac{1}{3})^2,......,570000\times (\frac{1}{3})^6$

a=570000

common ratio is given by

$r=\frac{a_2}{a_1}=\frac{1}{3}$

number of terms = 6

Sum of terms will be given by

$S_n=\frac{a(1-r^n)}{(1-r)}$

We'll put this value in this formula,

$S_6=\frac{570000(1-(\frac{1}{3})^6}{(1-\frac{1}{3})}\\\\=853827.16$

So, there are approximately 853827 new cases in 6 years.

6 0

3 years ago

Pls help me for brainliest answer

Ket [755]

Answer with Step-by-step explanation:

$a)3n + 4 \\ \\n = 1 \\ 3 \times 1 + 4 \\ 3 + 4 \\ =7 \\ \\ n = 2 \\ 3 \times 2 + 4 \\ 6 + 4 \\ = 10 \\ \\ n = 3 \\ 3 \times 3 + 4 \\ 9 + 4 \\ = 13 \\ \\ n = 4 \\ 3 \times 4 + 4 \\ 12 + 4 \\ = 16 \\ \\ n = 10 \\ 3 \times 10 + 4 \\ 30 + 4 \\ = 34 \\$

So in this sequence,

1st term = 7

2nd term = 10

3rd term = 13

4th term = 16

10th term = 34

$b)4n - 5 \\ \\ n = 1 \\ 4 \times 1 - 5 \\ 4 - 5 \\ = - 1 \\ \\ n = 2 \\ 4 \times 2 - 5 \\ 8 - 5 \\ = 3 \\ \\ n = 3 \\ 4 \times 3 - 5 \\ 12 - 5 \\ = 7 \\ \\ n = 4 \\ 4 \times 4 - 5 \\ 16- 5 \\ = 11 \\ \\ n = 10 \\ 4 \times 10 - 5 \\ 40 - 5 \\ = 35$

In this sequence,

1st term = -1

2nd term = 3

3rd term = 7

4th term = 11

10th term =35

5 0

3 years ago

James folds a piece of paper in half several times,each time unfolding the paper to count how many equal parts he sees. After fo

Snezhnost [94]

Answer:

There will be total 2048 parts of the given paper if James if able to fold the paper eleven times.

The needed function is $y = 2 ^n$

Step-by-step explanation:

Let us assume the piece of paper James decides to fold is a SQUARE.

Now, let us assume:

n : the number of times the paper is folded.

y : The number of parts obtained after folds.

Now, if the paper if folded ONCE ⇒ n = 1

Also, when the pap er is folded once, the parts obtained are TWO equal parts.

⇒ for n = 1 , y = 2 ..... (1)

Similarly, if the paper if folded TWICE ⇒ n = 2

Also, when the paper is folded twice, the parts obtained are FOUR equal parts.

⇒ for n = 2 , y = 4 ..... (2)

⇒ $y = 2^2 = 2^n$

Continuing the same way, if the paper is folded SEVEN times ⇒ n = 7

So, $y = 2^ n = 2^7 = 128$

⇒ There are total 128 equal parts.

Lastly, if the paper is folded ELEVEN times ⇒ n = 11

So, $y = 2^ n = 2^{11} = 2048$

⇒ There are total 2048 equal parts.

Hence, there will be total 2048 parts of the given paper if James if able to fold the paper eleven times.

And the needed function is $y = 2 ^n$

8 0

4 years ago