Answer:
120
Step-by-step explanation:
We are given that the function for the number of students enrolled in a new course is 
.
It is asked to find the average increase in the number of students enrolled per hour between 2 to 4 hours.
We know that the average rate of change is given by,
,
where f(x)-f(a) is the change in the function as the input value (x-a) changes.
Now, the number of students enrolled at 4 = f(4) = 
= 255 and the number of students enrolled at 2 = f(2) = 
= 15
So, the average increase 
= 
= 
= 120.
Hence, the average increase in the number of students enrolled is 120.
Answer: 644,800
Step-by-step explanation:
This can also be solved using the terms of Arithmetic Progressions.
Let the 13 years be number of terms of the sequences (n)
Therefore ;
T₁₃ = a + ( n - 1 )d , where a = 310,000 and d = 9% of 310,000
9% of 310,000 = 9/100 x 310,000
= 27,900
so the common difference (d)
d = 27,900
Now substitute for the values in the formula above and calculate
T₁₃ = 310,000 + ( 13 - 1 ) x 27,900
= 310,000 + 12 x 27,900
= 310,000 + 334,800
= 644,800.
The population after 13 years = 644,800.
Answer:
Difference in the lengths of the polygons is (x + 7) units.
Step-by-step explanation:
Lengths of the given polygon = (7x + 2) units and (6x - 5) units
Therefore, difference in the given lengths = (7x + 2) - (6x - 5)
= (7x - 6x) + (2 + 5)
= x + 7
Therefore, difference in the lengths of the polygons is (x + 7) units.
Answer:
The length of each red rod is 10 cm and the length of each blue rod is 14 cm
Step-by-step explanation:
Let
x ----> the length of each red rod in centimeters
y ----> the length of each blue rod in centimeters
we know that
----> equation A
----> equation B
Solve the system by graphing
Remember that the solution of the system of equations is the intersection point both graphs
using a graphing tool
The solution is the point (10,14)
see the attached figure
therefore
The length of each red rod is 10 cm and the length of each blue rod is 14 cm
Answer:
690 sq in
Step-by-step explanation:
SA = LA + 2B where LA is the lateral area and B is the area of the base
The triangular base has an area of
B = 1/2bh 1/2(5)(12) = 30
LA = ph where p is the perimeter of the base and h is the height of the prism
LA = (12 + 5 + 13)(21) = 30(21) = 630
SA = 630 + 2(30) = 630 + 60 = 690 sq in
