Find the area of a circle that has a radius of 14 feet. Approximate Π as 3.14. Round your answer to the nearest hundredth.

Parallelogram ABCD is a rectangle. What are the slopes of the sides that make this quadrilateral a rectangle?

snow_tiger [21]

Answer: -3 and 1/3

Step-by-step explanation:

There’s your answer

6 0

3 years ago

Find an expression which represents the sum of (-X – 7) and (-9x + 6) in simplest terms.

Strike441 [17]

Answer:

sum is -(10x + 1)

Step-by-step explanation:

$( - x - 7) + ( - 9x + 6) \\ = ( - x - 9x) + (6 - 7) \\ = - 10x - 1 \\ = - (10x + 1)$

4 0

3 years ago

Read 2 more answers

1287×6 tens=? Hundreds or ? Ones

Sauron [17]

Boa tarde! 1287 x 6 = 7722 Unidades Ou 77 Centenas e 22 Unidades Espero que eu tenha lhe ajudado!

7 0

3 years ago

HELPPPPP PLZZZZ What real number a and b make the equation true? (a-3)+(b+8)i=3-4i

Radda [10]

Answer:

a= 6

b= -12

Step-by-step explanation:

In complex number algebra, you need to sum the real values with each others and the imaginary values with each others.

In order to determine which number is a real value and which one is an imaginary, we must expand the parenthesis terms:

$(a-3) + (b+8)i =a - 3 + bi + 8i = 3 - 4i$

Now we match real numbers with real numbers and imaginary numbers with imaginary numbers:

$a - 3 = 3\\a= 3 - (-3) = 6\\\\b + 8 = -4 \\b= -4 -8 = -12\\$

6 0

3 years ago

Identify the lines that are perpendicular: A. = −5 and = 2 are perpendicularB. + 5 = 2 and − 5 + = 3 are perpendicularC. = 13 +

scZoUnD [109]

In order to check if the lines are perpendicular, we need to check if their slopes have the following relation (to find the slope we can use the slope-intercept form y = mx + b):

$m_1=-\frac{1}{m_2}$

A.

In this option, y = -5 is an horizontal line and x = 2 is a vertical line, therefore they are perpendicular.

B.

First let's find the slope of each line:

$\begin{gathered} x+\frac{y}{5}=2 \\ 5x+y=10 \\ y=-5x+10\to m=-5 \\ \\ -\frac{x}{5}+y=3 \\ y=\frac{x}{5}+3\to m=\frac{1}{5} \end{gathered}$

These slopes obey the relation stated above, so the lines are perpendicular.

C.

$\begin{gathered} y=\frac{1}{3}x+1\to m=\frac{1}{3} \\ \\ y-1=-3(x-5) \\ y-1=-3x+15 \\ y=-3x+16\to m=-3 \end{gathered}$

These slopes obey the relation stated above, so the lines are perpendicular.

D.

$\begin{gathered} y-5=x+1 \\ y=x+6\to m=1 \\ \\ x+y=3 \\ y=-x+3\to m=-1 \end{gathered}$

These slopes obey the relation stated above, so the lines are perpendicular.

E.

$\begin{gathered} y+2=\frac{1}{3}(x-6) \\ y+2=\frac{1}{3}x-2 \\ y=\frac{1}{3}x-4\to m=\frac{1}{3} \\ \\ y=3x+4\to m=3 \end{gathered}$

These slopes don't obey the relation stated above, so the lines aren't perpendicular.

The correct options are A, B, C and D.

5 0

2 years ago