All categories

3 years ago

Height=20diameter=14cmfind tsa

Mathematics

1 answer:

dusya [7]3 years ago

3 0

Answer:

Let the radius of the cylinder be r and the height be h.

Curved surface area of a cylinder = 2 π r h

Curved surface area of a cylinder = 2 π r h= 2 × 22 / 7 × 7 × 20 = 880 cm^2.

Or of an:

Acute Isosceles Triangle

Side a = 20

Side b = 20

Side c = 14

Angle ∠A = 69.513° = 69°30'46" = 1.21323 rad

Angle ∠B = 69.513° = 69°30'46" = 1.21323 rad

Angle ∠C = 40.975° = 40°58'29" = 0.71514 rad

Area = 131.14496

Perimeter p = 54

Semiperimeter s = 27

Hope this helps! If so please Mark brainliest and rate/heart if it did!

You might be interested in

For his long distance phone service, Juan pays a $5 monthly fee plus 9 cents per minute. Last month, Juan's long distance bill w

Lera25 [3.4K]

Answer:

80 or 81 minutes

Step-by-step explanation:

12.28-5 =7.28 7.28/.09= 80.88

4 0

3 years ago

I just need help solving this problem. Also please don't put the answer into a file because I can't open them. It's also really

dem82 [27]

Answer:

Step-by-step explanation:

5 0

3 years ago

Given MN= PQ, NO=QR,.and OM=RP

shtirl [24]

Answer:

Step-by-step explanation:

4 0

2 years ago

An airplane travels 6111 kilometers against the wind in 9 hours and 7911 kilometers with the wind in the same amount of time. Wh

natima [27]

Answer:

Speed of the plane in still air: $779\; {\rm km \cdot h^{-1}}$ .

Windspeed: $100\; {\rm km \cdot h^{-1}}$ .

Step-by-step explanation:

Assume that $x\; {\rm km \cdot h^{-1}}$ is the speed of the plane in still air, and that $y\; {\rm km \cdot h^{-1}}$ is the speed of the wind.

When the plane is travelling against wind, the ground speed of this plane (speed of the plane relative to the ground) would be $(x - y)\; {\rm km \cdot h^{-1}}$ .
When this plane is travelling in the same direction as the wind, the ground speed of this plane would be $(x + y)\; {\rm km \cdot h^{-1}}$ .

The question states that when going against the wind ( $v = (x - y)\; {\rm km \cdot h^{-1}}$ ,) the plane travels $6111\; {\rm km}$ in $9\; {\rm h}$ . Hence, $9\, (x - y) = 6111$ .

Similarly, since the plane travels $7911\; {\rm km}$ in $9\; {\rm h}$ when travelling in the same direction as the wind ( $v = (x + y)\; {\rm km \cdot h^{-1}}$ ,) $9\, (x + y) = 7911$ .

Add the two equations to eliminate $y$ . Subtract the second equation from the first to eliminate $x$ . Solve this system of equations for $x$ and $y$ : $x = 779$ and $y = 100$ .

Hence, the speed of this plane in still air would be $779\; {\rm km \cdot h^{-1}}$ , whereas the speed of the wind would be $100\; {\rm km \cdot h^{-1}}$ .

3 0

1 year ago

On a number line, the coordinates of X, Y, Z, and W are negative 7, negative 4, 3, and 7, respectively. Find the lengths

Vinil7 [7]

Djdndisinx83 Jdjeijb

6 0

3 years ago

Height=20diameter=14cmfind tsa​

Height=20diameter=14cmfind tsa