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Sloan [31]
3 years ago
7

The length of a rectangle is 3 times the width. If the perimeter is to be greater than 56 meter. What are the lossible values fo

r the width?( use w as the width)
Mathematics
2 answers:
tatyana61 [14]3 years ago
8 0

Answer:

w > 7m

Step-by-step explanation:

let l is the length of rectangle and w is the width of rectangle,

Perimeter of rectangle is 2(l+w).

From the question statement, we observe that

l=3w

2(l+w)>56

Put l=3w in above inequality,

2(3w+w)>56

2(4w)>56

8w>56

Multiplying by 1/8 on both sides of equation we get

(1/8)8w>(1/8)56

w>7m

We get the conclusion that width must be greater than 7m.


lidiya [134]3 years ago
7 0

Answer:

w > 7

Step-by-step explanation:

L = 3w

2L + 2w > 56

sub for L

2(3w) + 2w > 56

6w + 2w > 56

8w > 56

w > 7


Check: L = 3(8)

L = 24

2(24) + 2(8) > 56

48 + 16 > 56

64 > 56

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Answer:

G. 7

Explanation:

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The ratio of the area of ∆ABC to the area of ∆DEF is; 1:100.

<h3>What is the ratio of the area of ∆ABC to the area of ∆DEF?</h3>

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