What is the approximate area of the following circle?

Mathematics

2 answers:

balu736 [363]2 years ago

8 0

Answer:

$38.47 {m}^{2}$

Step-by-step explanation:

The formula to find the area of a circle is

$a= \pi {r}^{2}$

We know that

$\pi = 3.14 \: \: \: \: \: \: r = 3.5$

Therefore the area is

$a= \pi {r}^{2} \\ \\ a = (3.14) {(3.5)}^{2} \\ \\ a = 3.14 \times 12.25 \\ \\ a = 38.47 {m}^{2}$

PtichkaEL [24]2 years ago

3 0

Answer: 38.47 m^2

work is shown in picture :)

You might be interested in

What is the answer to 8= −6+x

Aliun [14]

Answer:

x = 14

Step-by-step explanation:

add 6 to both sides:

8+6 = 6-6+x

14 = x

x = 14

6 0

3 years ago

Pat delivers newspapers every morning. On average she can deliver newspapers to 17 homes in 12 minutes. When another driver is s

garik1379 [7]

Answer:

About 220 minutes.

What you have to do is divide 17 by 12 and multiply it by 155. After that, round the decimal up since the decimal part still counts as time.

4 0

3 years ago

PLEASE HELP ASAP: TRIGONOMETRY

ratelena [41]

Recall that sine is the ratio of opposite over hypotenuse, so we say

sin(theta) = opposite/hypotenuse

If we have a triangle labeled as such in the diagram below, then we can use symbols to write sin(theta) = A/C

Similarly, cos(theta) = B/C since cosine is the ratio of adjacent over hypotenuse.

Then from here we use a bit of algebra to compute the left hand side of the given equation. The goal is to see if we can simplify it to 1.

$\sin^2\left(\theta\right)+\cos^2\left(\theta\right)\\\\\left(\sin\left(\theta\right)\right)^2+\left(\cos\left(\theta\right)\right)^2\\\\\left(\frac{A}{C}\right)^2+\left(\frac{B}{C}\right)^2\\\\\frac{A^2}{C^2}+\frac{B^2}{C^2}\\\\\frac{A^2+B^2}{C^2}\\\\\frac{C^2}{C^2}\\\\1$

So this shows that the left hand side is identically the same as the right hand side (both sides are 1), and it doesn't matter what you replace theta with. Theta can be any real number and the equation will always be true.

Note: if you're wondering how A^2+B^2 became C^2, it's because of the pythagorean theorem. So that's why this trig identity is often referred to as the pythagorean trig identity.