PLEASE HELP!!! Find the area of the shaded polygon:

The distance around the middle of a sphere (the circumference) equals 37.699 in. what is the surface area of the sphere?

podryga [215]

To solve this problem you must apply the proccedure shown below:

1. Clear the radius from the formula for the circumference of a circle and substitute values, as following:

$C=2\pi r\\ r=C/2\pi \\ r=37.699in/2\pi \\ r=5.99in$

2. Substitute the radius calculated into the formula for calculate the surface area of the sphere:

$SA=4\pi r^{2} \\ SA=4\pi(5.99in^{2})\\ SA=450.88in^{2}$

The answer is: $450.88in^{2}$

5 0

3 years ago

In the equation 9^2 x 27^3 = 3^x, what is the value of x? <br> Show your work on scratchpaper

IceJOKER [234]

Answer:

x=13

Step-by-step explanation:

9^2 * 27^3 = 3^x

We need to get each term with a base of 3

9^2 = (3^2) ^2

We know that a^b^c = a^(b*c)

(3^2) ^2 = 3^(2+2) = 3^4

27^3 = (3^3) ^3 = 3^(3*3) = 3^9

Replacing these in the original equation

3^4 * 3^9 = 3^x

We know that a^b *a^c = a^(b+c)

3^4 * 3^9 =3^(4+9) = 3^13 = 3^x

The bases are the same, so the exponents must be the same

x=13

7 0

3 years ago

What would you do first to solve 5x−24=6

Alenkasestr [34]

answer: x=6

Step-by-step explanation:

move all terms that don't contain x to the right side and solve.

8 0

4 years ago

Can anybody help me?

Sloan [31]

Answer:

7⅓

Step-by-step explanation:

2 ÷ 3⁄11 → 2 × 3⅔ >> 7⅓

Dividing by a fraction is the exact same as multiplying by its multiplicative inverse.

* 3⅔ = 11⁄3

* 7⅓ = 22⁄3

I am joyous to assist you anytime.

6 0

3 years ago

The system of equations may have a unique solution, an infinite number of solutions, or no solution. Use matrices to find the ge

Leno4ka [110]

Answer:

Infinite number of solutions.

Step-by-step explanation:

We are given system of equations

$5x+4y+5z=-1$

$x+y+2z=1$

$2x+y-z=-3$

Firs we find determinant of system of equations

Let a matrix A= $\left[\begin{array}{ccc}5&4&5\\1&1&2\\2&1&-1\end{array}\right]$ and B= $\left[\begin{array}{ccc}-1\\1\\-3\end{array}\right]$

$\mid A\mid=\begin{vmatrix}5&4&5\\1&1&2\\2&1&-1\end{vmatrix}$

$\mid A\mid=5(-1-2)-4(-1-4)+5(1-2)=-15+20-5=0$

Determinant of given system of equation is zero therefore, the general solution of system of equation is many solution or no solution.

We are finding rank of matrix

Apply $R_1\rightarrow R_1-4R_2$ and $R_3\rightarrow R_3-2R_2$

$\left[\begin{array}{ccc}1&0&1\\1&1&2\\0&-1&-3\end{array}\right]$ : $\left[\begin{array}{ccc}-5\\1\\-5\end{array}\right]$

Apply $R_2\rightarrow R_2-R_1$

$\left[\begin{array}{ccc}1&0&1\\0&1&1\\0&-1&-3\end{array}\right]$ : $\left[\begin{array}{ccc}-5\\6\\-5\end{array}\right]$

Apply $R_3\rightarrow R_3+R_2$

$\left[\begin{array}{ccc}1&0&1\\0&1&1\\0&0&-2\end{array}\right]$ : $\left[\begin{array}{ccc}-5\\6\\1\end{array}\right]$

Apply $R_3\rightarrow- \frac{1}{2}$ and $R_2\rightarrow R_2-R_3$

$\left[\begin{array}{ccc}1&0&1\\0&1&0\\0&0&1\end{array}\right]$ : $\left[\begin{array}{ccc}-5\\\frac{13}{2}\\-\frac{1}{2}\end{array}\right]$

Apply $R_1\rightarrow R_1-R_3$

$\left[\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right]$ : $\left[\begin{array}{ccc}-\frac{9}{2}\\\frac{13}{2}\\-\frac{1}{2}\end{array}\right]$

Rank of matrix A and B are equal.Therefore, matrix A has infinite number of solutions.

Therefore, rank of matrix is equal to rank of B.

4 0

4 years ago