Can somebody help me with these to questions how do you find the area of the circle

Mathematics

1 answer:

Grace [21]3 years ago

6 0

13)

Circumference (C) = 2π·r

11π = 2π·r

11/2 = r

******************************

Area (A) = π·r²

= π·(11/2)²

= 121π/4

≈ 94.985 in²

14)

diameter = 2r

42 = 2r

21 = r

*******************

Area (A) = π·r²

= π·(21)²

= 441π

≈ 1384.74 in²

***************************

$280/1384.74 in²

x = $0.20

Answer: about 20 cents per square inch

You might be interested in

Evaluate the function represented by the graph for f(-2).

Alex_Xolod [135]

Answer:

f(-2) = 2

Step-by-step explanation:

The question is asking you to find the y value when the x value is -2. When observing the graph, we can see when the x value is -2, the y value is 2. Therefore, f(-2) = 2.

4 0

2 years ago

Use Part 1 of the Fundamental Theorem of Calculus to find the derivative of the function.

seropon [69]

Answer:

$h'(x)=\frac{3r^{2}}{2\sqrt{r^3+5}}$

Step-by-step explanation:

1) The Fundamental Theorem of Calculus in its first part, shows us a reciprocal relationship between Derivatives and Integration

$g(x)=\int_{a}^{x}f(t)dt \:\:a\leqslant x\leqslant b$

2) In this case, we'll need to find the derivative applying the chain rule. As it follows:

$h(x)=\int_{a}^{x^{2}}\sqrt{5+r^{3}}\therefore h'(x)=\frac{\mathrm{d} }{\mathrm{d} x}\left (\int_{a}^{x^{2}}\sqrt{5+r^{3}}\right )\\h'(x)=\sqrt{5+r^{3}}\\Chain\:Rule:\\F'(x)=f'(g(x))*g'(x)\\h'=\sqrt{5+r^{3}}\Rightarrow h'(x)=\frac{1}{2}*(r^{3}+5)^{-\frac{1}{2}}*(3r^{2}+0)\Rightarrow h'(x)=\frac{3r^{2}}{2\sqrt{r^3+5}}$