Answer:
the answer is <em><u>1</u></em><em><u>1</u></em><em><u>7</u></em><em><u>8</u></em><em><u>8</u></em> :) your welcome
Answer:
5.40 is not the same as 7.00
Step-by-step explanation:
Answer:
The rate is 1 ÷ 2 yards per hour
Step-by-step explanation:
The computation of the rate does the amount of yarn changes is given below:
= 3 3 ÷ 4 - 1 1 ÷8
= 15 ÷ 4 - 9 ÷ 8
= 21 ÷ 8 yards
Now divide the above fraction by the number of hours
= 21 ÷ 8 yards ÷ 5 1 ÷ 4
= 21 ÷ 8 × 4 ÷ 21
= 1 ÷ 2 yards per hour
Hence, the rate is 1 ÷ 2 yards per hour
The same would be relevant
Answer:
y is 1 and x is 4
Step-by-step explanation:
becasue the y intercepts at 1 and the x intercepts at 4
Answer:
% change in stopping distance = 7.34 %
Step-by-step explanation:
The stooping distance is given by
![T = 2.5 x + 0.5 x^{2}](https://tex.z-dn.net/?f=T%20%3D%202.5%20x%20%2B%200.5%20x%5E%7B2%7D)
We will approximate this distance using the relation
![f (x + dx) = f (x)+ f' (x)dx](https://tex.z-dn.net/?f=f%20%28x%20%2B%20dx%29%20%3D%20f%20%28x%29%2B%20f%27%20%28x%29dx)
dx = 26 - 25 = 1
T' = 2.5 + x
Therefore
![f(x)+f'(x)dx = 2.5x+ 0.5x^{2} + 2.5 +x](https://tex.z-dn.net/?f=f%28x%29%2Bf%27%28x%29dx%20%3D%202.5x%2B%200.5x%5E%7B2%7D%20%2B%202.5%20%2Bx)
This is the stopping distance at x = 25
Put x = 25 in above equation
2.5 × (25) + 0.5×
+ 2.5 + 25 = 402.5 ft
Stopping distance at x = 25
T(25) = 2.5 × (25) + 0.5 × ![25^{2}](https://tex.z-dn.net/?f=25%5E%7B2%7D)
T(25) = 375 ft
Therefore approximate change in stopping distance = 402.5 - 375 = 27.5 ft
% change in stopping distance =
× 100
% change in stopping distance = 7.34 %