All categories

2 years ago

Anyone help me please. A circle has a circumference

Mathematics

2 answers:

Elis [28]2 years ago

8 0

Answer:

$R=5.7$

Step-by-step explanation:

$Radius=\frac{Circumference}{2\pi }$

$R=\frac{36}{2(3.142)}$

$R=\frac{36}{6.284}$

$R=5.728835137$

$R=5.7$

Hope this helped :)

Ket [755]2 years ago

6 0

The circumference of a circle is pi x 2 x radius.

To find the radius, divide the circumference by pi, then divide that by 2:

36/3.142 = 11.4576703

Radius = 11.4576703/2 = 5.7288

Rounded to one decimal place the radius is 5.7 units.

You might be interested in

How can you get coefficients that are equal for X? 3x + 2y = 12 6x + 3y = 21

Verizon [17]

Step-by-step explanation:

To make the coefficient of x equal we can multiply both sides of (1) by 2. the answer is B

5 0

3 years ago

What is 4004 times 72

scoundrel [369]

4004×72=288288

~Hope this helps

5 0

2 years ago

Read 2 more answers

Please help me find 'x' your help would be much appreciated -Thanks

Vladimir79 [104]

Answer:

x=25.

Step-by-step explanation:

We know because of corresponding angles, the sum of the angle adjacent to (2x-5) and (2x-5) is 180 degrees

2x-5+5x+10=180

7x+5=180

7x=175

x=25

4 0

3 years ago

Name the orperation you should perform first. i need help pls and thank you!!

pishuonlain [190]

Answer:

For the first one on the left, you need to evaluate the expressions inside the parenthesis. For Question number 6, evaluate the exponent first.

Step-by-step explanation:

Think of PEMDAS, order of operations.

Parenthesis

Exponent

Multiplication

Division

Addition

Subtraction

6 0

2 years ago

1) Use power series to find the series solution to the differential equation y'+2y = 0 PLEASE SHOW ALL YOUR WORK, OR RISK LOSING

iogann1982 [59]

$y=\displaystyle\sum_{n=0}^\infty a_nx^n$

then

$y'=\displaystyle\sum_{n=1}^\infty na_nx^{n-1}=\sum_{n=0}^\infty(n+1)a_{n+1}x^n$

The ODE in terms of these series is

$\displaystyle\sum_{n=0}^\infty(n+1)a_{n+1}x^n+2\sum_{n=0}^\infty a_nx^n=0$

$\displaystyle\sum_{n=0}^\infty\bigg(a_{n+1}+2a_n\bigg)x^n=0$

$\implies\begin{cases}a_0=y(0)\\(n+1)a_{n+1}=-2a_n&\text{for }n\ge0\end{cases}$

We can solve the recurrence exactly by substitution:

$a_{n+1}=-\dfrac2{n+1}a_n=\dfrac{2^2}{(n+1)n}a_{n-1}=-\dfrac{2^3}{(n+1)n(n-1)}a_{n-2}=\cdots=\dfrac{(-2)^{n+1}}{(n+1)!}a_0$

$\implies a_n=\dfrac{(-2)^n}{n!}a_0$

So the ODE has solution

$y(x)=\displaystyle a_0\sum_{n=0}^\infty\frac{(-2x)^n}{n!}$

which you may recognize as the power series of the exponential function. Then

$\boxed{y(x)=a_0e^{-2x}}$

7 0

3 years ago