What's the surface area of a sphere with radius 5 centimeters

Mathematics

1 answer:

torisob [31]3 years ago

3 0

Surface area: 4*pi*r^2 = 4*pi*5^2= 4*25*pi = 100pi or 314.16

You might be interested in

Two american states are not part of the continental united sates.What percent of states are included in the continental united s

Harrizon [31]

So you know that there are 51 states in the US. 49/51 are apart of the continental US.
To find that percentage, all we need to do is divide 49 by 51 and convert the decimal we get into a percentage.

49/51= 0.9607...
To convert this into a percentage, we multiply it by 100. This is the equivalent of moving the decimal point over two places to the right.

49/51= 96.07% of the states are part of the continental US

Does this make sense? Hope it helped!

4 0

3 years ago

What is −3/5−(−2/5)

EleoNora [17]

I hope it can help you

7 0

3 years ago

(–2) + (–20) step by step

aalyn [17]

(-2) +(-20) =-22
2+20=22 make result negative

6 0

2 years ago

Read 2 more answers

Solve the following system of equations algebraically 3x-y=0 5x+2y=22

Lyrx [107]

3x - y = 0 ⇒ 2(3x - y = 0 ) ⇒ 6x - 2y = 0

5x + 2y = 22 ⇒ 1(5x + 2y = 22) ⇒ 5x + 2y = 22

11x = 22

x = 2

3x - y = 0 ⇒ 3(2) - y = 0 ⇒ 6 - y = 0 ⇒ 6 = y

Answer: x=2, y=6

3 0

3 years ago

A particle is projected with a velocity of <img src="https://tex.z-dn.net/?f=40ms%5E-%5E1" id="TexFormula1" title="40ms^-^1" alt

Katena32 [7]

Answer:

$2\sqrt{55}\text{ m/s or }\approx 14.8\text{m/s}$

Step-by-step explanation:

The vertical component of the initial launch can be found using basic trigonometry. In any right triangle, the sine of an angle is equal to its opposite side divided by the hypotenuse. Let the vertical component at launch be $y$ . The magnitude of $40\text{ m/s}$ will be the hypotenuse, and the relevant angle is the angle to the horizontal at launch. Since we're given that the angle of elevation is $60^{\circ}$ , we have:

$\sin 60^{\circ}=\frac{y}{40},\\y=40\sin 60^{\circ},\\y=20\sqrt{3}$ (Recall that $\sin 60^{\circ}=\frac{\sqrt{3}}{2}$ )

Now that we've found the vertical component of the velocity and launch, we can use kinematics equation $v_f^2=v_i^2+2a\Delta y$ to solve this problem, where $v_f/v_i$ is final and initial velocity, respectively, $a$ is acceleration, and $\Delta y$ is distance travelled. The only acceleration is acceleration due to gravity, which is approximately $9.8\:\mathrm{m/s^2}$ . However, since the projectile is moving up and gravity is pulling down, acceleration should be negative, represent the direction of the acceleration.