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2 years ago

What is the circumference of area=64pi mi^2

Mathematics

1 answer:

azamat2 years ago

4 0

Answer:

16pi mi

Step-by-step explanation:

The smallest circumference for a given area is that of a circle with that area. The area is given by ...

A = πr²

so we can find r as ...

64π = πr²

r = √64 = 8 . . . . miles

The circumference of this circle is ...

C = 2πr = 2π·(8 mi) = 16π mi

The circumference of the area is at least 16π miles.

_____

If the shape is not constrained to a circle, the circumference can be anything you like.

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The solution to -3x + 5x - 7 = -11 is _____.

Lemur [1.5K]

Answer:

x = -2

Step-by-step explanation:

Isolate the variable by dividing each side by factors that don't contain the variable.

6 0

2 years ago

How many cups are in 7 1/2 gallons?

saul85 [17]

Answer:

120 cups

Step-by-step explanation:

there are 16 cups in one gallon. 7x16=112 and half of a gallon is 8 cups. So 112+8=120

4 0

2 years ago

Read 2 more answers

What is the slop of 2x+3y-9=0

MAVERICK [17]

Answer:

Step-by-step explanation:

We need to write this equation in y=mx+b where m is the slope.

3y= -2x +9

y= -2/3 +9

So now that it is in the form of y=mx+b, we can find the slope. -2/3 is the slope of the equation. Hope that helps!

3 0

2 years ago

What's the domaing and range of this <br><br> (sorry for the quality, its the best I could get)

Schach [20]

Answer:

domain: [-3,2]

range: [-3,2]

Step-by-step explanation:

domain: all possible x values *maximum values of x

range: all possible y values *maximum values of x

brackets means the max while parenthesis mean infinite

8 0

2 years ago

Solve the system by using a matrix equation.<br> --4x - 5y = -5<br> -6x - 8y = -2

evablogger [386]

Answer:

Solution : (15, - 11)

Step-by-step explanation:

We want to solve this problem using a matrix, so it would be wise to apply Gaussian elimination. Doing so we can start by writing out the matrix of the coefficients, and the solutions ( - 5 and - 2 ) --- ( 1 )

$\begin{bmatrix}-4&-5&|&-5\\ -6&-8&|&-2\end{bmatrix}$

Now let's begin by canceling the leading coefficient in each row, reaching row echelon form, as we desire --- ( 2 )

Row Echelon Form :

$\begin{pmatrix}1\:&\:\cdots \:&\:b\:\\ 0\:&\ddots \:&\:\vdots \\ 0\:&\:0\:&\:1\end{pmatrix}$

Step # 1 : Swap the first and second matrix rows,

$\begin{pmatrix}-6&-8&-2\\ -4&-5&-5\end{pmatrix}$

Step # 2 : Cancel leading coefficient in row 2 through $R_2\:\leftarrow \:R_2-\frac{2}{3}\cdot \:R_1$ ,

$\begin{pmatrix}-6&-8&-2\\ 0&\frac{1}{3}&-\frac{11}{3}\end{pmatrix}$

Now we can continue canceling the leading coefficient in each row, and finally reach the following matrix.

$\begin{bmatrix}1&0&|&15\\ 0&1&|&-11\end{bmatrix}$

As you can see our solution is x = 15, y = - 11 or (15, - 11).

4 0

2 years ago