Answer:
Function H h(t) = -6.1 + 19t - 55
Step-by-step explanation:
Let t = 1
<u>Function F</u>
h(t) = 6.1t^2 + 19t - 54.4
h(1) = 6.1(1)^2 + 19(1) - 54.4
h(1) = 6.1 + 19 - 54.4
h(1) = -29.3
<u>Function G</u>
h(t) = -24.1t^2 + 2.8
h(1) = -24.1(1)^2 + 2.8
h(1) = -24.1 + 2.8
h(1) = -21.3
<u>Function H</u>
h(t) = -6.1t^2 + 19t - 55
h(1) = -6.1(1)^2 + 19(1) - 55
h(1) = -6.1 + 19 - 55
h(1) = -42.1
<u>Function J</u>
h(t) = -6.1t^2 + 19t + 0.99
h(t) = -6.1(1)^2 + 19(1) + 0.99
h(t) = -6.1 + 19 + 0.99
h(t) = 13.89
Answer:
Step-by-step explanation:
We are given the first derivative of and the value of
.
To go from the first derivative to the original function, we integrate.
Therefore:
To integrate, we add 1 to the power and divide by the new power:
Because we have an indefinite integral, we have to add the constant, , to the end.
So:
We know so we can find the constant
.
Therefore
Answer:
The number of tables were used that day at lunch is 23 tables.
Step-by-step explanation:
Given as :
The total number of students in the Polk school = 203
On one day the number of students absent = 19
So, The number of student present on that day = Total student - absent students
I.e The number of student present on that day = 203 - 19 = 184
The number of students sit on each table = 8
Let The number of tables were used that day at lunch = n
<u>So, According to question</u>
The number of tables were used that day at lunch =
I.e n =
∴ n = 23 tables
So, The number of tables were used that day at lunch = n = 23 tables
Hence, The number of tables were used that day at lunch is 23 tables. Answer