Answer:
Kendra needs 3 coins (1 coin of 25 cent & 2 coins of 1 cent) to add 27 cents in 73 cents to make 1 dollar
Step-by-step explanation:
Given :
Kendra has 73 cents
She need 1 dollar to by a ball
∴ She requires additional money = 
Standard Coins available are 50 cents , 25 cents, 10 cents, 5 cents and 1 cent.
To make 27 cents =
= 1 coin of 25 cent & 2 coins of 1 cent
Remark
You know the most about the trip on the way back and the total of the two trips.
Way Back
r = 30 mph
d = distance traveled
t = t - 1 where t = the time to go there which we know nothing about
Total Distance
2d = 25 mph * (t + t - 1)
2d = 25 * (2t - 1)
2d = 50t - 25 Divide by 2
2d = 50t/2 - 25/2
d = 25t - 12.5
Equation coming back
d = 30*(t - 1)
d = 30t - 30
Comment
Since the distances are the same, equate them.
Solution
25t - 12.5 = 30t - 30 Subtract 25t from both sides
25t - 25t -12.5 = 30t - 25t - 30 Combine like terms
-12.5 = 5t - 30 Add 30 to both sides
-12.5 + 30 = 5t - 30 + 30 Combine
17.5 = 5t Divide by 5
17.5/5 = t Divide and Switch
t = 3.5
t is the time going there
t - 1 is the time coming back
Total time = 2*t - 1 = 2*3.5 - 1 = 6
The total time is 6 hours. Answer
Answer:
x²+10xy+25y²
Step-by-step explanation:
(x+5y)²=(x+5y)(x+5y)=x×x+5y×x+x×5y+5y×5y=x²+5xy+5xy+25y²=x²+10xy+25y²
But it can be much easier
(a+b)²=a²+2ab+b²
The coefficient of the third term in this expression is +2 from the term (+2x).
Answer:
Below in bold.
Step-by-step explanation:
The surface area of the box
= x^2 + 4hx where x = a side of the square base and h is the height.
So x^2 + 4hx = 8
The volume of the box
V = x^2h
From the first equation we solve for h
4hx = 8 - x^2
h = (8 - x^2) / 4x
Now we substitute for h in the formula for the volume:
V = x^2 * (8 - x^2) / 4x
V = 8x^2 - x^4 / 4x
V = 2x - 0.25x^3
Finding the derivative:
V' = 2 - 0.75x^2 = 0 for max/mimn values
x^2 = 2/ 0.75 = 2.667
x = 1.633.
So the length and width of the base is 1.633 m and the height
= ( 8 - 2.667) / (4*1.633)
= 0.816 m
The maximum volume = 0.816 * 2.667 = 2.177 m^2.
The answers are correct to the nearest thousandth.