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marusya05 [52]
2 years ago
12

the length of a rectangle is 3 cm more than twice its width. Find the largest possible width if the perimeter is at most 66cm

Mathematics
2 answers:
MA_775_DIABLO [31]2 years ago
7 0
Let the length be L

Let the width be W.

L is 3cm more than twice its width.

L = 2W + 3


P = 2*(L + W)

P = 2*(2W + 3 + W)

P = 2*(3W + 3)

P = 6W + 6

Since Perimeter is at most 66, so P is less than or equal to 66.

P ≤ 66

6W + 6 ≤ 66

6W ≤ 66 - 6

6W ≤ 60

W ≤ 60/6

W ≤ 10

So the largest possible width is 10cm.

Largest possible width is 10cm.

I hope this helped.
Vanyuwa [196]2 years ago
6 0
If you let the width be x, then the length will be 2x + 3. The perimeter P is twice the width plus twice the length, i.e., P = 2*x + 2*(2x + 3) = 2x + 4x + 6 = 6x + 6. So if the perimeter is 66 cm then 66 = 6x + 6 ---> 6x = 60 ---> x = 10 cm., and so if the perimeter is at most 66 cm., the width can be at most 10 cm..
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Step-by-step explanation:

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-6

Step-by-step explanation:  6  6  15  40

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Write the equation of the line shown in point-slope form. (2,1) (3,8)
Ulleksa [173]

<u>Answer:</u>

The line equation that passes through the given points is 7x – y = 13

<u>Explanation:</u>

Given:

Two points are A(2, 1) and B(3, 8).

To find:

The line equation that passes through the given two points.

Solution:

We know that, general equation of a line passing through two points (x1, y1), (x2, y2) in point slope form is given by

\frac{(y- y1)}{(x-x_1)}= \frac{((y_2- y_1)}{(x_2- x_1 )}

{(y- y1)= \frac{((y_2- y_1)}{(x_2- x_1 )}\times(x-x_1)..........(1)

here, in our problem x1 = 3, y1 = 8, x2 = 2 and y2 = 1.

Now substitute the values in (1)

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7x – y = 13  

Hence, the line equation that passes through the given points is 7x – y = 13

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