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3 years ago

Labor time = 8 hours

2 answers:

6 0

8(1.65)R + 98.70 = 242.58 13.20R = 143.88 [subtract 98.70 from both sides] R = $10.90 / hr. [divide both sides by 13.20]

Masteriza [31]3 years ago

5 0

Answer:

Hourly Rate for labor = 10.9%

Step-by-step explanation:

Labor time = 8 hours

Overhead rate = 65%

Retail price of parts = $98.70

Total cost of job = $242.58

Let hourly rate of the labor be r

⇒ Labor time × (1 + overhead rate) × r + Retail price = Total cost

⇒ 8 × (1 + 0.65) × r + 98.70 = 242.58

⇒ 8 × 1.65 × r = 242.58 - 98.70

⇒ 13.2 × r = 143.88

⇒ r = 10.9

Thus, Hourly Rate for labor = 10.9%

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Lily has a stamp collection .she has 54 more U.S stamps than foreign stamps . she has 652 stamps in all.

NISA [10]

Answer:

foreign stamps = 299

US stamps = 353

Step-by-step explanation:

Let number of foreign stamps be x.

Given she has 54 more U.S stamps than foreign , US stamps = x +54

Total stamps = US + Foreign

652 = x + 54 + x

652 - 54 = 2x

598 = 2x

x = 299

7 0

3 years ago

Solve the system of equations. −2x+5y =−35 7x+2y =25

Otrada [13]

Answer:

The equations have one solution at (5, -5).

Step-by-step explanation:

We are given a system of equations:

$\displaystyle{\left \{ {{-2x+5y=-35} \atop {7x+2y=25}} \right.}$

This system of equations can be solved in three different ways:

Graphing the equations (method used)
Substituting values into the equations
Eliminating variables from the equations

Graphing the Equations

We need to solve each equation and place it in slope-intercept form first. Slope-intercept form is $\text{y = mx + b}$ .

Equation 1 is $-2x+5y = -35$ . We need to isolate y.

$\displaystyle{-2x + 5y = -35}\\\\5y = 2x - 35\\\\\frac{5y}{5} = \frac{2x - 35}{5}\\\\y = \frac{2}{5}x - 7$

Equation 1 is now $y=\frac{2}{5}x-7$ .

Equation 2 also needs y to be isolated.

$\displaystyle{7x+2y=25}\\\\2y=-7x+25\\\\\frac{2y}{2}=\frac{-7x+25}{2}\\\\y = -\frac{7}{2}x + \frac{25}{2}$

Equation 2 is now $y=-\frac{7}{2}x+\frac{25}{2}$ .

Now, we can graph both of these using a data table and plotting points on the graph. If the two lines intersect at a point, this is a solution for the system of equations.

The table below has unsolved y-values - we need to insert the value of x and solve for y and input these values in the table.

$\begin{array}{|c|c|} \cline{1-2} \textbf{x} & \textbf{y} \\ \cline{1-2} 0 & a \\ \cline{1-2} 1 & b \\ \cline{1-2} 2 & c \\ \cline{1-2} 3 & d \\ \cline{1-2} 4 & e \\ \cline{1-2} 5 & f \\ \cline{1-2} \end{array}$

$\bullet \ \text{For x = 0,}$

$\displaystyle{y = \frac{2}{5}(0) - 7}\\\\y = 0 - 7\\\\y = -7$

$\bullet \ \text{For x = 1,}$

$\displaystyle{y=\frac{2}{5}(1)-7}\\\\y=\frac{2}{5}-7\\\\y = -\frac{33}{5}$

$\bullet \ \text{For x = 2,}$

$\displaystyle{y=\frac{2}{5}(2)-7}\\\\y = \frac{4}{5}-7\\\\y = -\frac{31}{5}$

$\bullet \ \text{For x = 3,}$

$\displaystyle{y=\frac{2}{5}(3)-7}\\\\y= \frac{6}{5}-7\\\\y=-\frac{29}{5}$

$\bullet \ \text{For x = 4,}$

$\displaystyle{y=\frac{2}{5}(4)-7}\\\\y = \frac{8}{5}-7\\\\y=-\frac{27}{5}$

$\bullet \ \text{For x = 5,}$

$\displaystyle{y=\frac{2}{5}(5)-7}\\\\y=2-7\\\\y=-5$

Now, we can place these values in our table.

As we can see in our table, the rate of decrease is $-\frac{2}{5}$ . In case we need to determine more values, we can easily either replace x with a new value in the equation or just subtract $-\frac{2}{5}$ from the previous value.

For Equation 2, we need to use the same process. Equation 2 has been resolved to be $y=-\frac{7}{2}x+\frac{25}{2}$ . Therefore, we just use the same process as before to solve for the values.

$\bullet \ \text{For x = 0,}$

$\displaystyle{y=-\frac{7}{2}(0)+\frac{25}{2}}\\\\y = 0 + \frac{25}{2}\\\\y = \frac{25}{2}$

$\bullet \ \text{For x = 1,}$

$\displaystyle{y=-\frac{7}{2}(1)+\frac{25}{2}}\\\\y = -\frac{7}{2} + \frac{25}{2}\\\\y = 9$

$\bullet \ \text{For x = 2,}$

$\displaystyle{y=-\frac{7}{2}(2)+\frac{25}{2}}\\\\y = -7+\frac{25}{2}\\\\y = \frac{11}{2}$

$\bullet \ \text{For x = 3,}$

$\displaystyle{y=-\frac{7}{2}(3)+\frac{25}{2}}\\\\y = -\frac{21}{2}+\frac{25}{2}\\\\y = 2$

$\bullet \ \text{For x = 4,}$

$\displaystyle{y=-\frac{7}{2}(4)+\frac{25}{2}}\\\\y=-14+\frac{25}{2}\\\\y = -\frac{3}{2}$

$\bullet \ \text{For x = 5,}$

$\displaystyle{y=-\frac{7}{2}(5)+\frac{25}{2}}\\\\y = -\frac{35}{2}+\frac{25}{2}\\\\y = -5$

And now, we place these values into the table.

$\begin{array}{|c|c|} \cline{1-2} \textbf{x} & \textbf{y} \\ \cline{1-2} 0 & 25/2 \\ \cline{1-2} 1 & 9 \\ \cline{1-2} 2 & 11/2 \\ \cline{1-2} 3 & 2 \\ \cline{1-2} 4 & -3/2 \\ \cline{1-2} 5 & -5 \\ \cline{1-2} \end{array}$

When we compare our two tables, we can see that we have one similarity - the points are the same at x = 5.

Equation 1 Equation 2

$\begin{array}{|c|c|} \cline{1-2} \textbf{x} & \textbf{y} \\ \cline{1-2} 0 & -7 \\ \cline{1-2} 1 & -33/5 \\ \cline{1-2} 2 & -31/5 \\ \cline{1-2} 3 & -29/5 \\ \cline{1-2} 4 & -27/5 \\ \cline{1-2} 5 & -5 \\ \cline{1-2} \end{array}$ $\begin{array}{|c|c|} \cline{1-2} \textbf{x} & \textbf{y} \\ \cline{1-2} 0 & 25/2 \\ \cline{1-2} 1 & 9 \\ \cline{1-2} 2 & 11/2 \\ \cline{1-2} 3 & 2 \\ \cline{1-2} 4 & -3/2 \\ \cline{1-2} 5 & -5 \\ \cline{1-2} \end{array}$

Therefore, using this data, we have one solution at (5, -5).

4 0

3 years ago

If you have 6 lessons a week in mathematics, each lesson being 38 minutes long, what is the total time spent in mathematics less

Lynna [10]

Answer:

2280 minutes.

Step-by-step explanation:

Time spent each week in lessons = (Lessons per week)(Time per lesson)

= (6)(38)

= 228 minutes/week

Time spent in a 10-week period:

10 weeks * 228 minutes/week

= 2280 minutes

Use conversion factors to determine the time in the desired units (i.e. Hours, seconds, etc.).

Hope this helps!

5 0

1 year ago

Helppp 8th grade algebra C. Six units left D.six units right

yanalaym [24]

D, six units to the right. This is the answer because the pre-image is six to the left of the newest image. Just count the spaces away!

5 0

3 years ago

Which of the following represents the distance

Lunna [17]

Answer:

D. d = sqrt( ( x2-x1)^2 + (y2-y1)^2)

C. d = sqrt( ( x1-x2)^2 + (y1-y2)^2)

B. d = sqrt( (|x2-x1|^2 + |y2-y1|^2)

Step-by-step explanation:

Given two points (x1, y1) and (x2,y2) we can find the distance using

d = sqrt( ( x2-x1)^2 + (y2-y1)^2)

The order of the terms inside the square doesn't matter

d = sqrt( ( x1-x2)^2 + (y1-y2)^2)

When we are squaring are term, we can take the absolute value before we square and it does not change the value

d = sqrt( (|x2-x1|^2 + |y2-y1|^2)

3 0

3 years ago