So if 4 times the width would be $4w$ and 3 more than would be $+ 3$ , the equation would be $l = 4w + 3$ . If the perimeter is 56, then the new equation, $2(4w + 3) + 2(w) = 56$ can be formed. This simplifies to $8w + 6 + 2w = 56$ , then to $10w + 6 = 56$ , then to $10w = 50$ , and finally, $w = 5$ . So the width is 5, and now plug it into the first equation.

$l = 4(5) + 3$ , becomes $l = 20 + 3$ , becomes $l = 23$ .

So the dimensions are: length = 23 cm, width = 5 cm.

*And just to be sure, $2(23 + 5) = 2(28) = 56$ which is correct.

4 0

2 years ago

In a different plan for area codes, the first digit could be any number from 4 through 8, the second digit was either 3, 4, 5,

Snezhnost [94]

For the first digit, we have 5 options that are 4,5,6,7,8 . For the second digit, we have 4 options which are 3,4,5 or 6 and for the third digit, we have the options of all numbers except 2 or 5 that is 1,3,4,6,7,8,9,0 . SO we have 8 options for third digit . So to find the total number of options, we need to multiply all the possible options for each digit that is 5 times 4 times 8 = 160 . So the number of possible options are 160 .

7 0

3 years ago

2. What step did Mikayla do wrong when she solved this

Lunna [17]

Mikayla subtracted 4 from both sides in the beginning when she should've added it to both sides to cancel it out on the right. the correct answer should actually be x=1

5 0

3 years ago

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