Jade solved the following equation:

The width of a rectangle is 6 kilometers less than twice its length. if its area is 108 square kilometers, find the dimensions

Art [367]

Hi there!

Answer:
length = 9 kilometres
Width = 12 kilometres

Let's solve this problem step by step!
To find our answer we need to set up and solve an equation.

Let the length of the rectangle be represented by x.
The width of the rectangle can therefore be expressed by 2x - 6.

The area of a rectangle can be found by using the formula:
A = width × length

Plug in the data from the formula
A = x (2x - 6).

Simplify using rainbow technique.
$x(2x - 6) = 2 {x}^{2} - 6x$

Now we've found the simplified expression that expresses the area of the rectangle. Therefore, we can now set up and start solve the equation.

$2 {x}^{2} - 6x = 108$
Subtract 108

$2 {x}^{2} - 6x - 108 = 0$
Divide by 2.

${x}^{2} - 3x - 54$
$(x - 9)(x + 6) = 0$
Rule AB = 0, gives A is 0 or B is 0.

$x - 9 = 0 \\ x = 9 \\ \\ x + 6 = 0 \\ x = - 6$

The length of the rectangle, which was represented by x, must be 9 (since it cannot be a negative number).

Length
$x = 9$
Width
$2x - 6 = 2 \times 9- 6 = 18 - 6 = 12$

Answer:
length = 9 kilometres
Width = 12 kilometres

~ Hope this helps you!

6 0

3 years ago

Devin has started a lawn care service. He will charge a mowing fee of $7 plus $2 per square yard of lawn. He also offers trimmin

Maru [420]

Answer:7+2xy

Step-by-step explanation:apex

4 0

3 years ago

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Someone help with this

elena-14-01-66 [18.8K]

The quadrants are I, II, III, and IV, (meaning 1, 2, 3, and 4). The first quadrant is in the upper right, which has both positive x and positive y values. The second quadrant is the upper left, which has negative x and positive y values. The third quadrant is the lower left, which has both negative x and y values. The fourth quadrant is the lower right, which has positive x and negative y values. Using this knowledge and our positive and negative signs, the following are the answers to questions 1-6.

1) (-4, -2) - Quadrant III, both x and y are negative

2) (0, -7) - This point is actually on the y-axis. The x value is 0 and the y value is -7, so the graph is 7 units down from the origin on the y-axis.

3) (0,0) - This point is the origin, or where the x and y axes cross (the middle of the graph).

4) (6, -9) - This point is in Quadrant IV, because it has a positive x value and a negative y value

5) (3,5) - This point is in Quadrant I, because both the x and y values are positive.

6) (8,0) - This point is on the x-axis. The y-value is zero, so this point is 8 units to the right of the origin between Quadrant I and Quadrant IV.

Using the knowledge presented above, to graph the points given to you in the second part of the problem, first you can figure out what quadrant or part of the graph the point is on. Then, you can count the number of units (squares on the graph) in the right direction (remember that up is positive on the y-axis and down is negative, and to the right is positive on the x-axis and to the left is negative) in order to plot the points. Then, you must connect the points that correspond to the same figure in order to create the figures.

Please comment if you have any questions!

Hope this helps!

7 0

3 years ago

1. The sum of two consecutive integers is at least 23. What is the smallest value that will make the statement true?

Volgvan

Answer:

1- 11+ 12 = 23

2- -7 + -6 = -13

7 0

3 years ago

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What do you add to 4 7/8 to make 8

nekit [7.7K]

Answer:

3 1/8

Step-by-step explanation:

4+3=7

7/8+1/8= 1

7+1=8

6 0

3 years ago

Read 2 more answers