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A cone has a height of 7ft and a radius of 4ft which equation can find the volume of the cone v=1.3 (7.2)

Margaret [11]

Answer:

The volume of a cone is:

V=(hπr^2)/3, where h=height and r=radius

V=(7π4^2)/3

V=112π/3 ft^3

V≈117.29 ft^3 (to nearest hundredth of a cubic foot)

Step-by-step explanation:

4 0

3 years ago

Sign in if y = 6, what is the value of 5y? a. 1 b. 11 c. 30 d. 56

Daniel [21]

Y = 6
5y = 5 x y
5y = 5 x 6
5y = 30.
It is C. 30.

Hope This Helps You!
Good Luck Studying :)

3 0

3 years ago

Anna said that the product of 78·112=72. How can you tell that her answer is wrong?

Andreyy89

Her answer is wrong because the product of 78.112 = 8736.

5 0

3 years ago

Read 2 more answers

Make a table of ordered pairs for the equation. y=−1/3x+4 Then plot two points to graph the equation.

Zigmanuir [339]

This should help you

8 0

3 years ago

Read 2 more answers

Pls show full working out

sweet-ann [11.9K]

Answer:

Question 1a. i) $x=1.8m$

Question 1a. ii) $Volume=27.6m^3$

Question 1b) $Volume=65.9m^3$

Explanation:

Question 1 a. i) Find the value of x.

$tan(\theta )=\dfrac{opposite\text{ }leg}{adjacent\text{ }leg}$

For the smalll triangle you can write:

$tan(\theta )=\dfrac{x}{1m}$

For tthe big triangle:

$tan(\theta )=\dfrac{x+2.7m}{2.5m}$

Substitute:

$\dfrac{x}{1m}=\dfrac{x+2.7m}{2.5m}$

Solve for x:

$2.5x=x+2.7m\\\\2.5x-x=2.7m\\\\1.5x=2.7m\\\\x=2.7m/1.5\\\\x=1.8m$

Question 1a ii) Find the volume of the frustrum

Find the volume of a cone with height = 2.7m + 1.8m = 4.5m, and radius = 2.5m

Formula:

$Volume=(1/3)\pi \times radius^2\times height$

Substitute:

$Volume=(1/3)\pi \times (2.5m)^2\times 4.5m=9.375\pi m^3$

Find the volume of a cone with heigth = 1.8m and radius = 1m

$Volume=(1/3)\pi \times (1m)^2\times 1.8m=0.6\pi m^3$

Subtract the volume of the small cone from the volume of the big cone

$Volume\text{ }of\text{ }frustrum=9.375\pi m^3-0.6\pi m^3=8.775\pi m^3\approx 27.6m^3$

Question 1b. Calculate the volume of the bin

i) Upper frustrum

This is the same frustrum from the equation of above, thus ist volume is 27.6m³.

ii) Lower frustrum

$\dfrac{x}{2.0m}=\dfrac{x+2.4m}{2.5m}$

$2.5x=2(x+2.4m)\\\\2.5x=2x+4.8m\\\\0.5x=4.8m\\\\x=9.6m$

$Volume=(1/3)\pi \times (2.5m)^2\times (9.6m+2.4m)-(2.0m)^2\times (9.6m)$

$Volume=38.3m^3$

iii) Add the volume of the two frustrums

$Volume=27.6m^3+38.3m^3=65.9m^3$

6 0

3 years ago