For an equal number of servings, the bananas will have three times as much sugar.
To put it in perspective:
When we have one serving of each fruit, the carrot will give you 5 grams of sugar whereas banana will give you 15 grams.
Hope this helps!
Answer:
Step-by-step explanation:
Suppose at t = 0 the person is 1m above the ground and going up
Knowing that the wheel completes 1 revolution every 20s and 1 revolution = 2π rad in angle, we can calculate the angular speed
2π / 20 = 0.1π rad/s
The height above ground would be the sum of the vertical distance from the ground to the bottom of the wheel and the vertical distance from the bottom of the wheel to the person, which is the wheel radius subtracted by the vertical distance of the person to the center of the wheel.
(1)
where 
is vertical distance from the ground to the bottom of the wheel, 
is the vertical distance from the bottom of the wheel to the person, R = 10 is the wheel radius, 
is the vertical distance of the person to the center of the wheel.
So solve for in term of t, we just need to find the cosine of angle θ it has swept after time t and multiply it with R
Note that is negative when angle θ gets between π/2 (90 degrees) and 3π/2 (270 degrees) but that is expected since it would mean adding the vertical distance to the wheel radius.
Therefore, if we plug this into equation (1) then
(125 km) x (1mile / 1.609344 km) = <em>77.6714 miles</em> (rounded)
Let t, f, and s be the amount of 25%, 40%, and 60% solutions used.
We are told s=3f. The general equation is:
100(0.25t+0.4f+0.6s)/(t+s+f)=45, and using s=3f we have:
(0.25t+0.4f+1.8f)/(t+4f)=0.45
(0.25t+2.2f)/(t+4f)=0.45
0.25t+2.2f=0.45t+1.8f
0.25t+0.4f=0.45t
0.4f=0.2t
t=2f, remembering that s=3f, will allow us to solve for f
t+s+f=60, using t and s from above
2f+3f+f=60
6f=60
f=10 L then
s=3f=30 L and t=2f=20L
So 20L of 25%, 10L of 40%, and 30L of 60% acid solutions need to be mixed to make 60L of 45% acid solution.
(1+1)x+1 and 4x+10
Explanation
I distributed the 2 to form a new equation and I put together the distributed equation in the second one to form a new one,
