All categories

2 years ago

A circle with circumference 18 has an arc with a 120° central angle. What is the length of the arc?

Mathematics

1 answer:

ArbitrLikvidat [17]2 years ago

3 0

Answer:

Step-by-step explanation:

A circle is 360 degrees

The central angle is 120 degrees

120/360 = 1/3

The arc is 1/3 of the circle or 1/3 of the circumference

1/3 * 18 = 6

You might be interested in

Rectangle ABCD is translated and then reflected to create rectangle A'B'C'D'. Do rectangle ABCD and rectangle A'B'C'D' have the

lions [1.4K]

Answer:

is there a graph for this that I can see?

5 0

2 years ago

Can someone please explain this

pickupchik [31]

Answer:

Step-by-step explanation:

Transformation #1 (blue grafic): rotation in y axis

Transformation #2 (green grafic): translation in y axis.

I hope I've helped you.

I hope you've understand me.

3 0

3 years ago

Peter is making an "X marks the spot" flag for a treasure hunt. The flag is made of a square white flag with sides of 12 centime

zheka24 [161]

Answer:

34 cm

Step-by-step explanation:

The actual question is How many inches of ribbon Peter will need to make the "X".

Using the Pythagorean theorem, the length of the diagonal of a square with 12 cm sides is:

$d^2=12^2+12^2\\d=\sqrt{288}\\ d=16.97\ cm$

Since the "X" requires two diagonals, the length of ribbon required is:

$L=2*16.97\\L=34\ cm$

The length required, rounded to the nearest centimeter, is 34 cm.

6 0

3 years ago

Please solve the problem with steps

Debora [2.8K]

Answer:

Infinite series equals 4/5

Step-by-step explanation:

Notice that the series can be written as a combination of two geometric series, that can be found independently:

$\frac{3^{n-1}-1}{6^{n-1}} =\frac{3^{n-1}}{6^{n-1}} -\frac{1}{6^{n-1}} =(\frac{1}{2})^{n-1} -\frac{1}{6^{n-1}}$

The first one: $(\frac{1}{2})^{n-1}$ is a geometric sequence of first term ( $a_1$ ) "1" and common ratio (r) " $\frac{1}{2}$ ", so since the common ratio is smaller than one, we can find an answer for the infinite addition of its terms, given by: $Infinite\,Sum=\frac{a_1}{1-r} = \frac{1}{1-\frac{1}{2} } =\frac{1}{\frac{1}{2} } =2$

The second one: $\frac{1}{6^{n-1}}$ is a geometric sequence of first term "1", and common ratio (r) " $\frac{1}{6}$ ". Again, since the common ratio is smaller than one, we can find its infinite sum: