All categories

3 years ago

What is the area with circumference 36pi cm

1 answer:

5 0

$\bf \textit{circumference of a circle}\\\\
C=2\pi r\qquad 
\begin{cases}
r=radius\\
-----\\
C=36\pi 
\end{cases}\implies 36\pi =2\pi r
\\\\\\
\cfrac{36\pi }{2\pi }=r\implies \boxed{18=r}\\\\
-------------------------------\\\\
\textit{area of a circle}\\\\
A=\pi r^2\qquad A=\pi \cdot \boxed{18}^2$

You might be interested in

If i can type 420 words in 8 minutes, what would be the unit rate for time per word

Tasya [4]

420÷8=52.5
the answer is 52.5 words per minute

7 0

3 years ago

Read 2 more answers

Find the x intercepts of the parabola with vertex (4,75) and y intercept (0,27)

user100 [1]

The equation for a parabola with vertex $(h, k)$ is
$y=a(x-h)^{2}+k$

We plug in:
$y=a(x-4)^{2}+75$

We find $a$ by plugging in the point $(0, 27)$ :
$27=a(0-4)^{2}+75$
$\Rightarrow 27=a(-4)^{2}+75$
$\Rightarrow 27=16a+75$
$\Rightarrow 16a=-48$
$\Rightarrow a=-3$

So our equation is
$y=-3(x-4)^{2}+75$

To find x-intercepts, we set $y=0$ :
$-3(x-4)^{2}+75=0$
$\Rightarrow -3(x-4)^{2}=-75$
$\Rightarrow (x-4)^{2}=25$
$\Rightarrow x-4=5, -5$
$\Rightarrow x=9, -1$

So the x-intercepts are 9 and -1.

8 0

3 years ago

I need something equivalent to 35x + 14 + 14

DerKrebs [107]

The awnser to your equasion is 63

I hope this helps you

7 0

3 years ago

Use the table of integrals, or a computer or calculator with symbolic integration capabilities, to find the indefinite integral.

andriy [413]

Answer:

$\frac{2}{3}(\frac{1}{5}ln|3x-5|-\frac{1}{5}ln|x|)+C$

Step-by-step explanation:

We have been given a indefinite integral $\int \frac{2}{3x\left(3x-5\right)}dx$ . We are asked to find the indefinite integral.

We will use partial fraction formula to solve our given problem.

$\frac{2}{3x\left(3x-5\right)}=\frac{3}{5(3x-5)}-\frac{1}{5x}$

$\int \frac{2}{3x\left(3x-5\right)}dx=\frac{2}{3}\int \frac{1}{x\left(3x-5\right)}dx$

$\frac{2}{3}\int \frac{1}{x\left(3x-5\right)}dx=\frac{2}{3}\int \frac{3}{5(3x-5)}-\frac{1}{5x}dx$

Using difference rule of integrals, we will get:

$\frac{2}{3}(\int \frac{3}{5(3x-5)}dx-\int \frac{1}{5x}dx)$

Now, we need to use u-substitution as:

Let $u=3x-5$ .

$\frac{du}{dx}=3$

$dx=\frac{1}{3}du$

$\int \frac{3}{5(3x-5)}dx= \frac{3}{5}\int \frac{1}{(u)}*\frac{1}{3}du=\frac{3}{5}*\frac{1}{3}\int \frac{1}{(u)}du=\frac{1}{5}ln|u|=\frac{1}{5}ln|3x-5|$