All categories

3 years ago

What is the approximate circumference of a semicircle with a radius of 30 centimeters

Mathematics

1 answer:

Dmitry [639]3 years ago

8 0

Approximate circumference of a semicircle with a radius of 30 centimeters is 94.2 centimeter

Solution:

Given semicircle with a radius of 30 centimeters

To find: approximate circumference of a semicircle

To find circumference of semicircle we can divide the circumference of circle by 2

circumference of semicircle = circumference of circle $\div$ 2

$\text{ circumference of semicircle} = \frac{2 \pi r}{2} = \pi r$

Substituting the given radius = 30 cm,

$\text{ circumference of semicircle} = 3.14 \times 30 = 94.2$

Thus approximate circumference of a semicircle with a radius of 30 centimeters is 94.2 centimeter

You might be interested in

Help plzzzzzzzzzzzzzzzzzzzz

bulgar [2K]

Answer:

y = -4

Step-by-step explanation:

First, let's find the slope. Since the line is horizontal, it's slope has to be 0 (this is because slope = rise/run, there is no "rise" in a horizontal line therefore the slope is 0)

Now, we need to find the y-intercept. This is where the line intercepts the y-axis. Looking at the graph, the line intercepts the axis at (0,-4), therefore the y-intercept is -4.

The equation would be:

y=0x-4

y=-4

5 0

2 years ago

How much water consumed by Aguilar family as shown in the meter reading

Sati [7]

Answer:

?????????????????

Step-by-step explanation:

????????

6 0

3 years ago

What is the solution to this equation? 3x-x+8+5x-2=10

Kaylis [27]

3x - x + 8 + 5x - 2 = 10

combine like terms on the left, then add or subtract as asked
(3x - x + 5x) + (8 - 2) = 10

7x + 6 = 10
subtract 6 from both sides

7x = 4
divide both sides by 7

x = 4/7

Hope this helps! :)

7 0

3 years ago

Find the radius of convergence, r, of the series. ∞ xn 2n − 1 n = 1 r = 1 find the interval, i, of convergence of the series. (e

Bingel [31]

Assuming the series is

$\displaystyle\sum_{n\ge1}\frac{x^n}{2n-1}$

The series will converge if

$\displaystyle\lim_{n\to\infty}\left|\frac{\frac{x^{n+1}}{2(n+1)-1}}{\frac{x^n}{2n-1}}\right|$

We have

$\displaystyle\lim_{n\to\infty}\left|\frac{\frac{x^{n+1}}{2(n+1)-1}}{\frac{x^n}{2n-1}}\right|=|x|\lim_{n\to\infty}\frac{\frac1{2n+1}}{\frac1{2n-1}}=|x|-\lim_{n\to\infty}\frac{2n-1}{2n+1}=|x|$

So the series will certainly converge if $-1$ , but we also need to check the endpoints of the interval.

If $x=1$ , then the series is a scaled harmonic series, which we know diverges.

On the other hand, if $x=-1$ , by the alternating series test we can show that the series converges, since

$\left|\dfrac{(-1)^n}{2n-1}\right|=\dfrac1{2n-1}\to0$

and is strictly decreasing.

So, the interval of convergence for the series is $-1\le x$ .

6 0

3 years ago

How do you rationalise this surd?

Ksivusya [100]

$\frac{1 + \sqrt{3} }{ \sqrt{2} } \times \frac{ \sqrt{2} }{ \sqrt{2} } = \frac{ \sqrt{2} + \sqrt{6} }{2}$
here you go

5 0

3 years ago

Read 2 more answers