Distance between the two cities:
453 - 333 = 120 miles.
Rest area is 2/3 of the way:
120 x 2/3 = 240/3 = 80 miles.
Divide the miles to the rest stop by his speed:
80 miles/ 60 miles per hour = 1 and 1/3 hours as a fraction. 1.3333 as a decimal( round as needed.
( 1 hour and 20 minutes)
Answer:
The pairs are (13,15) and (-15,-13).
Step-by-step explanation:
If n is an odd integer, the very next odd integer will be n+2.
n+1 is even (so we aren't using this number)
The sum of the squares of (n) and (n+2) is 394.
This means
(n)^2+(n+2)^2=394
n^2+(n+2)(n+2)=394
n^2+n^2+4n+4=394 since (a+b)(a+b)=a^2+2ab+b^2
Combine like terms:
2n^2+4n+4=394
Subtract 394 on both sides:
2n^2+4n-390=0
Divide both sides by 2:
n^2+2n-195=0
Now we need to find two numbers that multiply to be -195 and add up to be 2.
15 and -13 since 15(-13)=-195 and 15+(-13)=2
So the factored form is
(n+15)(n-13)=0
This means we have n+15=0 and n-13=0 to solve.
n+15=0
Subtract 15 on both sides:
n=-15
n-13=0
Add 13 on both sides:
n=13
So if n=13 , then n+2=15.
If n=-15, then n+2=-13.
Let's check both results
(n,n+2)=(13,15)
13^2+15^2=169+225=394. So (13,15) looks good!
(n,n+2)=(-15,-13)
(-15)^2+(-13)^2=225+169=394. So (-15,-13) looks good!
The ball will be at height 19 feet first at 0.7 sec and then at 1.42 sec.
<u>Explanation:</u>
We need to find the time at which the ball will be at height 19 feet.
Equation:
h = 3 + 34t - 16t²
19 = 3 + 34t - 16t²
16 = -16t² + 34t
-16t² + 34t - 16 = 0
On solving the equation, we get
t1 = 0.7 s and t2 = 1.42s
Therefore, the ball will be at height 19 feet first at 0.7 sec and then at 1.42 sec.
Answer:
y-intercept = 7
Step-by-step explanation:
The y-intercept is also the starting point
