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4 years ago

What’s the area of a circle with a diameter of 4

Mathematics

2 answers:

umka21 [38]4 years ago

7 0

Answer:

$4\pi$ ≈ 12.57 units squared

Step-by-step explanation:

The area of a circle is denoted by: $A=\pi r^{2}$ , where r is the radius.

Remember that diameter is twice the radius, so since the diameter equals 4, we know that the radius r = 4/2 = 2.

Now, plug this in: $A=\pi *2^2=4\pi$

We can round 4pi to 12.57 units squared.

Hope this helps!

hoa [83]4 years ago

3 0

Answer:

12.56 units²

Step-by-step explanation:

Radius = 4/2 = 2

Area = pi × r²

3.14 × 2²

12.56

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A statistics professor plans classes so carefully that the lengths of her classes are uniformly distributed between 47.047.0 and

alisha [4.7K]

Answer:

0.025 = 2.5% probability that a given class period runs between 51.5 and 51.75 minutes.

Step-by-step explanation:

An uniform probability is a case of probability in which each outcome is equally as likely.

For this situation, we have a lower limit of the distribution that we call a and an upper limit that we call b.

The probability that we find a value X lower than x is between c and d is given by the following formula

$P(c \leq X \leq d) = \frac{d - c}{b-a}$

For this problem, we have that:

$a = 47, b = 57, c = 51.5, d = 51.75$ . So

$P(c \leq X \leq d) = \frac{d - c}{b-a}$

$P(51.5 \leq X \leq 51.75) = \frac{51.75 - 51.5}{57 - 47} = 0.025$

0.025 = 2.5% probability that a given class period runs between 51.5 and 51.75 minutes.

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3 years ago

In the middle school debate club, 30% of the members are in sixth grade. If there are 12 sixth graders in the debate club, how m

cupoosta [38]

The answer to this would be 40 because you multiply 12x0.30 and get 40 which mean 12 is 30% of 40 total

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3 years ago

PLEASE HELP!!! WILL MARK BRAINLIEST!! THX! A golfer hits a ball with an initial velocity of 32.7 m/s from the ground. Find the f

OLEGan [10]

Answer:

See below

Step-by-step explanation:

First Problem

The ball hits the ground when $h(t)=0$ , therefore:

$h(t)=-4.9t^2+v_0t+h_0$

$0=-4.9t^2+32.7t$

$0=t(-4.9t+32.7)$

$t=0$ and $t=\frac{32.7}{4.9}\approx6.67$

Since the ball is in the air before it hits the ground, $t=6.67$ (seconds) is the more appropriate choice.

Second Problem

The maximum height of the ball is determined when $t=-\frac{b}{2a}$ , therefore:

$t=-\frac{b}{2a}$

$t=-\frac{32.7}{2(-4.9)}$

$t=-\frac{32.7}{-9.8}$

$t\approx3.34$

This means that the height of the ball is at its maximum after 3.34 seconds:

$h(t)=-4.9t^2+32.7t$

$h(3.34)=-4.9(3.34)^2+32.7(3.34)$

$h(3.34)\approx54.55$

Thus, the answer is 54.55 (meters).

Third Problem

Refer to the second problem

Fourth Problem

$h(t)=-4.9t^2+32.7t$

$h(4.3)=-4.9(4.3)^2+32.7(4.3)$

$h(4.3)\approx50.01$

Therefore, the height of the ball after 4.3 seconds is 50.01 (meters).

Fifth Problem

The ball will be 24 meters off the ground when $h(t)=24$ , therefore:

$h(t)=-4.9t^2+32.7t$

$24=-4.9t^2+32.7t$

$0=-4.9t^2+32.7t-24$

$t=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

$t=\frac{-32.7\pm\sqrt{(32.7)^2-4(-4.9)(-24)}}{2(-4.9)}$

$t_1\approx0.84$

$t_2\approx5.83$

Therefore, the ball will be 24 meters off the ground after 0.84 (seconds) and 5.83 (seconds)

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2 years ago