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2 years ago

The radius of a circle is 8 feet. What is the circle's area? Use 3.14 for .

Mathematics

2 answers:

Tasya [4]2 years ago

4 0

Answer:

50.27

Step-by-step explanation:

remember that the circumference of a circle is 2πr

so 3.15x8 and then multiply by two

Ainat [17]2 years ago

3 0

Answer:

200.96 ft²

Step-by-step explanation:

Step 1: Define/formula

To find the area of a circle, multiply the radius by itself twice, then multiply the product by pi.

$a=\pi r^2(3.14*radius*radius)$

Step 2: Solve

According to the word problem, the radius is 8.

From here, we plug the numbers into the formula.

$\pi 8^2(3.14*8*8)=200.96ft^2.$

Step 3: Conclude

Therefore, the area of this circle is 200.96 ft.²

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One of the legs of a right triangle measures 8 cm and its hypotenuse measures 14 cm.

Ugo [173]

Answer:

about 11.5 cm.

Step-by-step explanation:

I know the measurement of the other leg is about 11.5 cm. I know because I used the Pythagorean theorem.

a^2+b^2=c^2.

"a" and "b" are the values of the legs of the triangle, while "c" is the measure of the hypotenuse. We know that 8cm is the measure of one of the legs, and 14 cm is a measure of the hypotenuse.

8^2+b^2=14^2 simplified: 64+b^2=196

Then, I subtracted 64 on both sides, so I would have "b" by itself.

b^2=132

Next, I found the square root of both b^2 and 132, so I would find the true value of "b."

b=11.4891252931

So, the measure of the other leg rounded to the nearest tenth is about 11.5.

3 0

2 years ago

In the diagram of circle A, what is m angle LMN? О 75° O90° О 120° 135° ОО

Jet001 [13]

Answer:

Step-by-step explanation:

Angle Formed Outside = 1/2(DIFFERENCE of Intercepted Arcs)

ARC LN = 360 - 270 =90

LMN = 1/2(270 -90)

= 1/2 (180)

= 90

3 0

3 years ago

Read 2 more answers

Explain why the graph of a linear equation in the form of y=c is a horizontal line

xeze [42]

Let's say "c" is a constant, hmmmm any constant, for any value whatsoever of "x", "y" is always that constant, for example, say c = 3, thus y = 3, so a table for it will look like

$\bf \begin{array}{ccll}
x&y\\
\text{\textemdash\textemdash\textemdash}&\text{\textemdash\textemdash\textemdash}\\
-\infty&3\\
-6&3\\
-5&3\\
-4&3\\
-3&3\\
-2&3\\
-1&3\\
0&3\\
1&3\\
2&3\\
3&3\\
4&3\\
5&3\\
6&3\\
+\infty&3
\end{array}$

now, if you plot those points, it'd looks like the picture below.