6 phone numbers are possible for one area code if the first four numbers are 202-1
<u>Solution:</u>
Given that, the first four numbers are 202-1, in that order, and the last three numbers are 1-7-8 in any order
We have to find how many phone numbers are possible for one area code.
The number of way “n” objects can be arranged is given as n!
Then, we have three places which changes, so we can change these 3 places in 3! ways
Hence 3! is found as follows:
So, we have 6 phone numbers possible for one area code.
Answer:
$287.50
Step-by-step explanation:
31.25 x 9.20 = $287.50
Answer:
Point F is at -3
Step-by-step explanation:
In general, the average rate of change of f (x) on the interval a, b is given by f(b) – f(a) / b – a. The average rate of alteration of a function, f (x) on an interval is well-defined to be the variance of the function values at the endpoints of the interim divided by the difference in the x values at the endpoints of the interval. this is also known as the difference quotient that tells how on average, the y values of a function are changing in connection to variations in the x values. A positive or negative rate of change is applicable which match up to an increase or decrease in the y value among the two data points. It is called zero rate of change when a quantity does not change over time.
Answer:
t≈8.0927
Step-by-step explanation:
h(t) = -16t^2 + 128t +12
We want to find when h(t) is zero ( or when it hits the ground)
0 = -16t^2 + 128t +12
Completing the square
Subtract 12 from each side
-12 = -16t^2 + 128t
Divide each side by -16
-12/-16 = -16/-16t^2 + 128/-16t
3/4 = t^2 -8t
Take the coefficient of t and divide it by 8
-8/2 = -4
Then square it
(-4) ^2 = 16
Add 16 to each side
16+3/4 = t^2 -8t+16
64/4 + 3/4= (t-4)^2
67/4 = (t-4)^2
Take the square root of each side
±sqrt(67/4) =sqrt( (t-4)^2)
±1/2sqrt(67) = (t-4)
Add 4 to each side
4 ±1/2sqrt(67) = t
The approximate values for t are
t≈-0.092676
t≈8.0927
The first is before the rocket is launched so the only valid answer is the second one
