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Probblem #6
solve by multiplying through with the common denominator

(11)/(3)x + (1)/(6) = (8)/(3)x + (19)/(6)

Problem #20

the perimeter of a rectangle is to be no greater than 300in. and the length must be 125in. write an inequality representing the maximum perimeter.

i know that the perimeter is length times width...

#6. First, find the common denominator.

(22/6)X + 1/6 = (16/6)X + 19/6

(22/6)X - (16/6)X = 19/6 - 1/6

You should be able to take it from here.

#20. Area — NOT perimeter — is length times width. Perimeter is obtained by adding the length of all the sides.

300 > 2(125) + 2X, where X = width

I hope this helps. Thanks for asking.

To solve problem #6, you need to multiply through with the common denominator to eliminate fractions. Let's go step by step:

The common denominator for the terms in the equation is 6. Multiply every term by 6:
6 * (11/3)x + 6 * (1/6) = 6 * (8/3)x + 6 * (19/6)

Simplifying each term, you get:
(66/3)x + 1 = (48/3)x + 19

Now, let's simplify the fractions further:
(22/1)x + 1 = (16/1)x + 19

Next, subtract (16/1)x from both sides:
(22/1)x - (16/1)x + 1 = 19
(22 - 16)/1)x + 1 = 19
(6/1)x + 1 = 19
6x + 1 = 19

Finally, subtract 1 from both sides:
6x + 1 - 1 = 19 - 1
6x = 18

Divide both sides by 6 to solve for x:
(6x)/6 = 18/6
x = 3

So, the solution to problem #6 is x = 3.

Now, let's move on to problem #20.

The perimeter of a rectangle is obtained by adding the length of all the sides. In this case, you have a rectangle with a length of 125 inches, and you want to find the maximum perimeter that is no greater than 300 inches.

The perimeter of a rectangle is given by the formula: P = 2L + 2W, where P is the perimeter, L is the length, and W is the width.

In this case, the length is fixed at 125 inches. So, the inequality to represent the maximum perimeter can be written as:
P <= 300

Substituting the formula for perimeter, we get:
2L + 2W <= 300

Since you want to find the maximum perimeter, the inequality is less than or equal to (<=).

I hope this helps! Let me know if you have any further questions.