The perimeter of the flag is 2080 ft what is the flags width and length of the length is 420 ft greater than the width
1 answer:
Answer:
The length of the flag is 730 ft and the width is 310 ft
Step-by-step explanation:
<u><em>The correct question is</em></u>
The perimeter of the flag is 2080 ft what is the flags width and length <u>if</u> the length is 420 ft greater than the width
Let
L ----> the length of the flag
W ---> the width of the flag
we know that
The perimeter of the flag (rectangle) is equal to
we have
so
----> equation A
---> equation B
Solve the system by substitution
substitute equation B in equation A
solve for W
<em>Find the value of L</em>
---->
therefore
The length of the flag is 730 ft and the width is 310 ft
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1a. 10. 24
1b. 125/243
2a. 0. 0048
2b. 32/3125
3a. 0. 64
3b. 0. 0031
4a. 2/5
4b. 48. 735
5a. 2. 36
5b. 1. 39
<h3>How to determine the values</h3>
1a. Given the values
(2/5)^2/(1/2)^6
Multiply both the numerator and denominator by the powers
⇒
To find the common ration, multiply thus;
⇒ <em> × </em>
⇒
= 10. 24
1b. (5/7)^2 × (5/7)^1
= ×
=
= 125/243
2a. 0. 6^1 × 0. 2^3
= 0. 6 × 0. 008
= 0. 0048
2b. (2/5)^3 × (2/5)^2
= ×
<em>= </em>
= 32/ 3125
3a. 1^99 - 0. 6^2
= 1 - 0. 36
= 0. 64
3b. (0. 2 ) ^1 × (1/8)^2
= 0. 2 × 1/64
= 0. 2 × 0. 016
= 0. 0031
4a. (1/2)^2/ (5/8)^1
=
Take the inverse of the denominator
= ×
= 2/5
4b. 7^2 - 0. 5^3
= 49 - 0. 125
= 48. 875
5a. 3^1 - 0. 8 ^2
= 3 - 0. 64
= 2. 36
5b. 0. 7^2 + 0. 9^1
= 0. 49 + 0. 9
= 1. 39
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