Create a number line. Plot the following numbers on the number line: 5, -10, 20, -5, 0, 10, -20

Naddika [18.5K]

I think the answer is 0.8 or 0.78.
You divide 140/180 which gives you 0.77777778
You can either round up to the nearest hundredth or tenth.

7 0

3 years ago

A statement of an employee's biweekly earnings is given below. Earnings Deductions Week Ended Regular FED. SOC. MED STATE WITH.

DiKsa [7]

Answer: d. $852.96

Step-by-step explanation:

Given: The net pay = $667.17

Since net pay is the amount of money your employees take home after all deductions have been taken out.

We know that gross pay is the amount of money that employees receive before any taxes and deductions are taken out.

Thus to find the gross pay, we need to add all of the deductions to the net pay as:

Gross pay $=667.17+98+52.88+12.37+22.54 =\$852.96$

Hence, D is the right option.

8 0

4 years ago

Read 2 more answers

Imagine you are given the diameter of a circle and asked to find its volume. Select the statement that describes what you should

ivanzaharov [21]

Answer:

We need to take the diameter and divide by 2 to find the radius

Step-by-step explanation:

The volume of a sphere is found by

V = 4/3 pi r^3

So we need the radius

We know the diameter

r = d/2

We need to take the diameter and divide by 2

3 0

4 years ago

Maggie graphed the image of a 90 counterclockwise rotation about vertex A of . Coordinates B and C of are (2, 6) and (4, 3) and

Lemur [1.5K]

Answer:

A(2,2)

Step-by-step explanation:

Let the vertex A has coordinates $(x_A,y_A)$

Vectors AB and AB' are perpendicular, then

$\overrightarrow {AB}=(2-x_A,6-y_A)\\ \\\overrightarrow {AB'}=(-2-x_A,2-y_A)\\ \\\overrightarrow {AB}\perp\overrightarrow {AB'}\Rightarrow \overrightarrow {AB}\cdot \overrightarrow {AB'}=0\Rightarrow (2-x_A)(-2-x_A)+(6-y_A)(2-y_A)=0$

Vectors AC and AC' are perpendicular, then

$\overrightarrow {AC}=(4-x_A,3-y_A)\\ \\\overrightarrow {AC'}=(1-x_A,4-y_A)\\ \\\overrightarrow {AC}\perp\overrightarrow {AC'}\Rightarrow \overrightarrow {AC}\cdot \overrightarrow {AC'}=0\Rightarrow (4-x_A)(1-x_A)+(3-y_A)(4-y_A)=0$

Now, solve the system of two equations:

$\left\{\begin{array}{l}(2-x_A)(-2-x_A)+(6-y_A)(2-y_A)=0\\ \\(4-x_A)(1-x_A)+(3-y_A)(4-y_A)=0\end{array}\right.\\ \\\left\{\begin{array}{l}-4-2x_A+2x_A+x_A^2+12-6y_A-2y_A+y^2_A=0\\ \\4-4x_A-x_A+x_A^2+12-3y_A-4y_A+y_A^2=0\end{array}\right.\\ \\\left\{\begin{array}{l}x_A^2+y_A^2-8y_A+8=0\\ \\x_A^2+y_A^2-5x_A-7y_A+16=0\end{array}\right.$

Subtract these two equations:

$5x_A-y_A-8=0\Rightarrow y_A=5x_A-8$

Substitute it into the first equation:

$x_A^2+(5x_A-8)^2-8(5x_A-8)+8=0\\ \\x_A^2+25x_A^2-80x_A+64-40x_A+64+8=0\\ \\26x_A^2-120x_A+136=0\\ \\13x_A^2-60x_A+68=0\\ \\D=(-60)^2-4\cdot 13\cdot 68=3600-3536=64\\ \\x_{A_{1,2}}=\dfrac{60\pm8}{2\cdot 13}=\dfrac{34}{13},2$