I NEED A LOT OF HELP PLEASE! I WILL FAIL IF IT DOESNT GET DONE!

Mathematics

1 answer:

Anna35 [415]3 years ago

5 0

In order to solve for this, we need to find the volumes of both the cylinder and cone described.

In order to find the area of the cylinder:
We use the equation V= $\pi r^{2} h$
So, V= $\pi 6^{2}16$
The volume of the cylinder is therefore approximately 1809.56 $m^{3}$

For the cone:
V= $\pi r^{2} \frac{h}{3}$
So, V= $\pi 6^{2} \frac{3}{3}$
Thus, the volume of the cone is 113.1 $m^{3}$

We add these volumes together to derive the solution, which is 1922.66 $m^{3}$

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Kite EFGH is inscribed in a rectangle where F and H are midpoints of parallel sides. The area of EFGH is 35 square units. What i

sasho [114]

*see attachment for the figure described

Answer:

5 units

Step-by-step explanation:

==>Given the figure attached below, let where FH and EG intercepted be K.

Since FH are midpoints of parallel lines, KE = KG = x.

Given that the area of the kite EFGH = 35 square units, and we know the length of one of the diagonals = HF = KF + KH = 2 + 5 = 7, we can solve for x using the formula for the area of a kite.

Area of kite = ½ × d1 × d2

Where d1 = KH = 7

d2 = EG = KE + KG = x + x = 2x

Area of kite EFGH = 35

THUS:

35 = ½ × 7 × 2x

35 = 1 × 7 × x

35 = 7x

Divide both sides by 7

35/7 = x

x = 5