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2 years ago

Find the missing lengths of the sides.

Mathematics

1 answer:

lbvjy [14]2 years ago

6 0

Answer:

Step-by-step explanation:

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What is the volume of a right circular cylinder with a radius of 5cm and a height of 12cm?

Tom [10]

Answer: $V=942.47cm^{3}$

Step-by-step explanation:

1. To solve this problem you must use the following formula for calculate the volume of a right circular cylinder, which is shown below:

$V=\pi*r^{2}*h$

Where r is the radius and h is the height.

2. Now, you must substitute the values given in the problem. Therefore, you obtain the following result:

$V=\pi*(5cm)^{2}*(12cm)$

$V=942.47cm^{3}$

3 0

3 years ago

Read 2 more answers

There are 27 performances in a dance recital. The ratio of hip hop to ballet is 5:4.

Aloiza [94]

Answer:

15 hip hop performances

12 ballet performances

Step-by-step explanation:

5:4 (multiply by 3)

15:12

6 0

3 years ago

How many axes of symmetry has a rhombus

Kitty [74]

Rhombus has 2 axes of symetry (they contain diagonals).

If rhombus is a square then has 4 axes of symetry.

3 0

3 years ago

It’s a yes or no question<br> !!!!!!!!!!!!<br> For math

vesna_86 [32]

I’m confused on what the question is exactly

5 0

3 years ago

At noon, ship A is 130 km west of ship B. Ship A is sailing east at 25 km/h and ship B is sailing north at 20 km/h. How fast is

Hitman42 [59]

Answer:

answer = 12.87 km/h

Step-by-step explanation:

Given

Ship A is sailing east at 25 km/h = $\frac{dx}{dt}$

ship B is sailing north at 20 km/h = $\frac{dy}{dt}$

here x and y are the sailing at t = 4 : 00 pm for ship A and B respectively

so we get x = 4 ×25 =100 km/h

y = 4× 20 = 80 km/h

let z is the distance between the ships, we need to find $\frac{dz}{dt}$ at t = 4 hr

At noon, ship A is 130 km west of ship B (12:00 pm)

so equation will be

$z^2 = (130-x)^2 + y^2......................(i)\\put x = 100 and y = 80 \\\\we | | get \\$

$z^2 = 30^2 + 80^2\\z =\sqrt{7300} km/h$

derivative first equation w . r. to t we get

$2z\frac{dz}{dt} =$ $-2(130-x)\frac{dx}{dt}$ $+2y\frac{dy}{dt}$

$\frac{dz}{dt} =\frac{1}{z}[(x -130)\frac{dx}{dt} +y\frac{dy}{dt}]$

$\frac{dz}{dt} = \frac{( -20\times25 + 80\times20)}{\sqrt{7300} }$

$= \frac{1100}{85.44}\\ = 12.87km/h$

8 0

3 years ago