1 gallon = 4 quarts
2570 x 4 = 10,280 quarts of gas
10280/7 = 1468.571 quarts per day
rounded to nearest hundredth is 1468.57 quarts
Answer:
- h(x) = -2x^2 -x +3
- Reservoir A releases the same amount of water as Reservoir B over 1 week
Step-by-step explanation:
<u>Given</u>:
f(x) = x^2 -7x +5
g(x) = 3x^2 -6x +2
h(x) = f(x) -g(x)
<u>Find</u>:
h(x)
h(1)
<u>Solution</u>:
The expression for h(x) is found by evaluating its definition:
h(x) = (x^2 -7x +5) -(3x^2 -6x +2)
h(x) = x^2(1 -3) +x(-7 -(-6)) +(5 -2)
h(x) = -2x^2 -x +3
Then h(1) is found by substituting 1 for x:
h(1) = -2(1^2) -(1) +3 = -2 -1 +3
h(1) = 0 . . . . difference in release amounts after 1 week is 0
The true statements are ...
- h(x) = -2x^2 -x +3
- Reservoir A releases the same amount of water as Reservoir B over 1 week
Kepler's third law described the relation between semi-major axis (or average distance to the star) and
the orbital period (how long it takes to complete one lap) as follows:
a^3 / p^2 = constant
In the case of our Solar system the constant is 1
This means that, for this problem:
a^3 / p^2 = 1
p^2 = a^3
p = a^(3/2)
The semi major axis is given as 101 million km. We need to convert this into AU where 1 AU is approximately 150 million Km
101 million Km = (101x1) / 150 = 0.67 AU
Now, we substitute in the equation to get the orbital period as follows:
p = (0.67)^(3/2) = 0.548 earth years
