Answer:
t₁ = 2,75 sec
Step-by-step explanation:
The ball will get h maximum when dh/dt ( its vertical component of the speed is equal to 0. At this moment the ball has flown half of the total flight time
Then
h(t) = - 4t * ( 4*t - 11 )
h(t) = - 16*t² + 44*t
Taking derivatives on both sides of the equation we get:
dh/dt = - 32*t + 44
dh/dt = 0 ⇒ -32*t + 44 = 0
t = 44/32 ⇒ t = 1,375 s
So twice this time
2*t = 2 * 1.375
Total time t₁ = 2,75 sec
Perimeter is just the addition of the side lengths so add them together:
5c+6 + 2c+5 = 7c+11
Option D.
The table does not represent a function.
<h3>
What is a function?</h3>
A function is a relationship that maps elements of the domain into elements of the range. Such that each element of the domain can be mapped into only one element of the range.
Here we have a table.
If you look at the table, you can see that the element of the domain:
x = -18
Is being mapped into two different values in the range.
Into y = -10 and y = 21.
From that, we conclude that this table does not represent a function.
If you want to learn more about functions:
brainly.com/question/2328150
#SPJ1
1)Rewrite the table:
70, 49, 34.3, 24.01, 11.807
2) write the quotient of each number by the number before & notice the value:
49/70= 0.7
34.3/49 = 0.7
24.01/34.3 =0.7
16.0807/24.01 = 0.67 ≈0.7
You notice this is a geometric progression with r 0.7
The last term in a GP =ar^(n-1)
a=70; r= 0.7 n-1= number of terms (days -1)
For last term = 1 ==> then 1=70(0.7)^(n-1)
1/70 = (0.7)^(n-1)
log(1/70) =log[(0.7)^(n-1) ==> log1 - log 70 = (n-1) log(0.7) & you will find that it need 11.92 days to equal on & to be less than it will necessitate 12 Days
3) Domain and Range of this function:
Last term = a₁.rⁿ⁻¹. let last term be y==> f(n) = y =70(0.7)ⁿ⁻¹
or f(n) = y = 70(0.7)ⁿ / 0.7==> f(n) = [(0.7)ⁿ ]/ 100.
This is a decreasing exponential function where the coefficient
raised to n is < 1.
The domain is for all n>= 0.
When n→∞, f(n)→0; For n=0==>f(n) =70. So the range of f(n) is:<=70
