Something you own but u don't use a lot what is it

A sector of a circle has a central angle of 100 degrees. If the area of the sector is 50pi, what is the radius of the circle

MrMuchimi

The radius of the circle having the area of the sector 50π, and the central angle of the radius as 100° is <u>6√5 units</u>.

An area of a circle with two radii and an arc is referred to as a sector. The minor sector, which is the smaller section of the circle, and the major sector, which is the bigger component of the circle, are the two sectors that make up a circle.

Area of a Sector of a Circle = (θ/360°) πr², where r is the radius of the circle and θ is the sector angle, in degrees, that the arc at the center subtends.

In the question, we are asked to find the radius of the circle in which a sector has a central angle of 100° and the area of the sector is 50π.

From the given information, the area of the sector = 50π, the central angle, θ = 100°, and the radius r is unknown.

Substituting the known values in the formula Area of a Sector of a Circle = (θ/360°) πr², we get:

50π = (100°/360°) πr²,

or, r² = 50*360°/100° = 180,

or, r = √180 = 6√5.

Thus, the radius of the circle having the area of the sector 50π, and the central angle of the radius as 100° is <u>6√5 units</u>.

Learn more about the area of a sector at

brainly.com/question/22972014

#SPJ4

8 0

1 year ago

What is the expansion of (3+x)^4

Vlad1618 [11]

Answer:

$\left(3+x\right)^4:\quad x^4+12x^3+54x^2+108x+81$

Step-by-step explanation:

Considering the expression

$\left(3+x\right)^4$

Lets determine the expansion of the expression

$\left(3+x\right)^4$

$\mathrm{Apply\:binomial\:theorem}:\quad \left(a+b\right)^n=\sum _{i=0}^n\binom{n}{i}a^{\left(n-i\right)}b^i$

$a=3,\:\:b=x$

$=\sum _{i=0}^4\binom{4}{i}\cdot \:3^{\left(4-i\right)}x^i$

Expanding summation

$\binom{n}{i}=\frac{n!}{i!\left(n-i\right)!}$

$i=0\quad :\quad \frac{4!}{0!\left(4-0\right)!}3^4x^0$

$i=1\quad :\quad \frac{4!}{1!\left(4-1\right)!}3^3x^1$

$i=2\quad :\quad \frac{4!}{2!\left(4-2\right)!}3^2x^2$

$i=3\quad :\quad \frac{4!}{3!\left(4-3\right)!}3^1x^3$

$i=4\quad :\quad \frac{4!}{4!\left(4-4\right)!}3^0x^4$

$=\frac{4!}{0!\left(4-0\right)!}\cdot \:3^4x^0+\frac{4!}{1!\left(4-1\right)!}\cdot \:3^3x^1+\frac{4!}{2!\left(4-2\right)!}\cdot \:3^2x^2+\frac{4!}{3!\left(4-3\right)!}\cdot \:3^1x^3+\frac{4!}{4!\left(4-4\right)!}\cdot \:3^0x^4$