Answer:
Step-by-step explanation:
Let the actual width of belt be x
we are given that the ideal width of safety belt strap is 5 cm
So, Difference between actual and ideal = (x-5)cm
We are also given that An actual width can vary by at most 0.35 cm.
So, The difference between actual and ideal width should be less than or equal to 0.35 cm
So, 
We know
and 
and 
and 
So,
Hence the range of acceptable widths is
Answer:
Step-by-step explanation:
The given relation between length and width can be used to write an expression for area. The equation setting that equal to the given area can be solved to find the shed dimensions.
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<h3>Given relation</h3>
Let x represent the width of the shed. Then the length is (2x+3), and the area is ...
A = LW
20 = (2x+3)(x) . . . . . area of the shed
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<h3>Solution</h3>
Completing the square gives ...
2x² +3x +1.125 = 21.125 . . . . . . add 2(9/16) to both sides
2(x +0.75)² = 21.125 . . . . . . . write as a square
x +0.75 = √10.5625 . . . . . divide by 2, take the square root
x = -0.75 +3.25 = 2.50 . . . . . subtract 0.75, keep the positive solution
The width of the shed is 2.5 feet; the length is 2(2.5)+3 = 8 feet.
Answer:
The fourth proportion is 15
2:5 :: 6 : 15
Answer:
1. 2.5 steps per second
2. 1.25 steps
Step-by-step explanation:
Answer:
-13/8
Step-by-step explanation:
u divied by -8 to get n by itself
