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3 years ago

A cone has a height of 11 meters and a radius of 7 meters . What is the volume?

Mathematics

2 answers:

Levart [38]3 years ago

5 0

Answer:

564.512 m³

Step-by-step explanation:

Formula to find the volume of a cone is :

Volume = $\frac{1}{3}$ π r² h

Here,

r = radius = 7 m

h = height = 11 m

Let us solve now.

Volume = $\frac{1}{3}$ π r² h

Volume = $\frac{1}{3}$ π ×7² × 11

Volume = $\frac{1}{3}$ π × 49 × 11

Volume = $\frac{1}{3}$ π × 539

Volume = $\frac{1}{3}$ × 1693.538

Volume = 564.512 m³

Hope this helps you :-)

Let me know if you have any other questions :-)

baherus [9]3 years ago

3 0

Answer:

V = 564.2 m^3

Step-by-step explanation:

The volume of a cone is given by

V = 1/3 pi r^2 h where r is the radius and h is the height

V = 1/3 pi (7)^2 (11)

V =539 /3 * pi

Letting pi be approximated by 3.14

V = 564.1533333 m^3

Rounding to the nearest tenth

V = 564.2 m^3

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3044 in a base 5 to a base 10

nikklg [1K]

3,059 I don’t really care

4 0

3 years ago

In a naval engagement, one-third of the fleet was captured, one-sixth was sunk, and two ships were destroyed by fire. One-sevent

Otrada [13]

Answer:

60 ships.

Step-by-step explanation:

Let the total number of ships in the naval fleet be represented by x

One-third of the fleet was captured = 1/3x

One-sixth was sunk = 1/6x

Two ships were destroyed by fire = 2

Let surviving ships be represented by y

One-seventh of the surviving ships were lost in a storm after the battle = 1/7y

Finally, the twenty-four remaining ships sailed home

The 24 remaining ships that sailed home =

y - 1/7y = 6/7y of the surviving fleet sailed home.

Hence

24 = 6/7y

24 = 6y/7

24 × 7/ 6

y = 168/6

y = 28

Therefore, total number of ships that survived is 28.

Surviving ships lost in the storm = 1/7y = 1/7 × 28 = 4

Total number of ships in the fleet(x) =

x = 1/3x + 1/6x + 2 + 28

Collect like terms

x - (1/3x + 1/6x) = 30

x - (1/2x) = 30

1/2x = 30

x = 30 ÷ 1/2

x = 30 × 2

x = 60

Therefore, ships that were in the fleet before the engagement were 60 ships.

3 0

3 years ago

If sin tetha=root3/2, what is cos tetha?

Minchanka [31]

Answer:

The value of cos Ф is ± $\frac{1}{2}$

Step-by-step explanation:

There are important rules for sin Ф and cos Ф

sin²Ф + cos²Ф = 1

sin²Ф = 1 - cos²Ф

cos²Ф = 1 - sin²Ф

∵ sin Ф = $\frac{\sqrt{3}}{2}$

∴ sin²Ф = $(\frac{\sqrt{3}}{2})^{2}$

∴ sin²Ф = $\frac{3}{4}$

→ By using the third rule above

∵ cos²Ф = 1 - sin²Ф

∴ cos²Ф = 1 - $\frac{3}{4}$

∴ cos²Ф = $\frac{1}{4}$

→ Take square root for both sides

∴ cos Ф = ± $\frac{1}{2}$

∴ The value of cos Ф is ± $\frac{1}{2}$

6 0

3 years ago

Write a polynomial function, p(x) with degree 3 that has p(7)=0

MArishka [77]

Answer:

$p (x) = x^{3} - 21x^{2}+ 147x - 343$

is the required polynomial with degree 3 and p ( 7 ) = 0

Step-by-step explanation:

Given:

p ( 7 ) = 0

To Find:

p ( x ) = ?

Solution:

Given p ( 7 ) = 0 that means substituting 7 in the polynomial function will get the value of the polynomial as 0.

Therefore zero's of the polynomial is seven i.e 7

Degree : Highest raise to power in the polynomial is the degree of the polynomial

We have the identity,

$(a -b)^{3} = a^{3}-3a^{2}b +3ab^{2} - b^{3}$

Take a = x

b = 7

Substitute in the identity we get

$(x -7)^{3} = x^{3}-3x^{2}(7) +3x(7)^{2} - 7^{3}\\(x -7)^{3} = x^{3}-21x^{2} +147x - 343$

Which is the required Polynomial function in degree 3 and if we substitute 7 in the polynomial function will get the value of the polynomial function zero.

p ( 7 ) = 7³ - 21×7² + 147×7 - 7³

p ( 7 ) = 0

$p (x) = x^{3} - 21x^{2}+ 147x - 343$

4 0

3 years ago

The volume of the cone shown is 5 cubic inches. What is the height of a cone with the same base diameter but a volume of 10 cubi

Alisiya [41]

Answer:

The height of the big cone is double the one in the small cone

h2 = 2h1

Step-by-step explanation:

Given that:

The volume of the small cone: 5 cubic inches
The volume of the big cone: 10 cubic inches

As we know, the volume of a cone is as following:

V = (1/3)*area of the base*height

If the base diameter are unchanged => area of the base of the two cones are unchanged and from the given information, the volume of the big cone is double the volume of the small cone. So the height of the big cone is double the one in the small cone

<=> h2 = 2h1

3 0

3 years ago