All categories

3 years ago

A circle has circumference of 20.7cm. What is the area of the circle? Using 3.14

Mathematics

2 answers:

Alex787 [66]3 years ago

4 0

Circumference=2pi*r
20.7=2pi*r
r=3.296 or 3.3
A=pir^2
a=34.1946 or about 34.19

krek1111 [17]3 years ago

4 0

Hey!

Circumference = 20.7 cm
Circumference of circle= 2πr

20.7= 2×3.14×r
20.7=6.28×r
20.7/6.28= r
r= 3.29 cm

Area of circle = πr^2
Area= 3.14×3.29×3.29
Area= 33.98 cm^2

Hope it helps...!!!

You might be interested in

Simple mathhh help please!!

Neporo4naja [7]

Answer:

Step-by-step explanation:

85/100 = 85%

8 0

3 years ago

If f(1) = 10, what is f(3)?

Alexus [3.1K]

Answer:

Step-by-step explanation:

multiply each f by the same root numeral

4 0

3 years ago

A certain number was divided by 3 and 3 was added to the result. The final answer was 8 what was the number

nevsk [136]

Answer:

Step-by-step explanation:

x/3 + 3 = 8

x/3 + 9/3 = 8

x + 9 = 24

x = 15

5 0

2 years ago

Find the differential coefficient of <br><img src="https://tex.z-dn.net/?f=e%5E%7B2x%7D%281%2BLnx%29" id="TexFormula1" title="e^

Gemiola [76]

Answer:

$\rm \displaystyle y' = 2 {e}^{2x} + \frac{1}{x} {e}^{2x} + 2 \ln(x) {e}^{2x}$

Step-by-step explanation:

we would like to figure out the differential coefficient of $e^{2x}(1+\ln(x))$

remember that,

the differential coefficient of a function y is what is now called its derivative y', therefore let,

$\displaystyle y = {e}^{2x} \cdot (1 + \ln(x) )$

to do so distribute:

$\displaystyle y = {e}^{2x} + \ln(x) \cdot {e}^{2x}$

take derivative in both sides which yields:

$\displaystyle y' = \frac{d}{dx} ( {e}^{2x} + \ln(x) \cdot {e}^{2x} )$

by sum derivation rule we acquire:

$\rm \displaystyle y' = \frac{d}{dx} {e}^{2x} + \frac{d}{dx} \ln(x) \cdot {e}^{2x}$

Part-A: differentiating $e^{2x}$

$\displaystyle \frac{d}{dx} {e}^{2x}$

the rule of composite function derivation is given by:

$\rm\displaystyle \frac{d}{dx} f(g(x)) = \frac{d}{dg} f(g(x)) \times \frac{d}{dx} g(x)$

so let g(x) [2x] be u and transform it:

$\displaystyle \frac{d}{du} {e}^{u} \cdot \frac{d}{dx} 2x$

differentiate:

$\displaystyle {e}^{u} \cdot 2$

substitute back:

$\displaystyle \boxed{2{e}^{2x} }$

Part-B: differentiating ln(x)•e^2x

Product rule of differentiating is given by:

$\displaystyle \frac{d}{dx} f(x) \cdot g(x) = f'(x)g(x) + f(x)g'(x)$

let

$f(x) \implies \ln(x)$
$g(x) \implies {e}^{2x}$

substitute

$\rm\displaystyle \frac{d}{dx} \ln(x) \cdot {e}^{2x} = \frac{d}{dx}( \ln(x) ) {e}^{2x} + \ln(x) \frac{d}{dx} {e}^{2x}$

differentiate:

$\rm\displaystyle \frac{d}{dx} \ln(x) \cdot {e}^{2x} = \boxed{\frac{1}{x} {e}^{2x} + 2\ln(x) {e}^{2x} }$

Final part:

substitute what we got:

$\rm \displaystyle y' = \boxed{2 {e}^{2x} + \frac{1}{x} {e}^{2x} + 2 \ln(x) {e}^{2x} }$

and we're done!

6 0

3 years ago

If their food and beverages cost 25.30 and there is an 8% meals tax, how much is the bill?

pentagon [3]

25.3 × 1.08
27.324

The total bill is $27.32

7 0

2 years ago