Answer:
Class 8 has 27 and class 7 has 40
Step-by-step explanation:
8th (r) class has 13 fewer than 7th (x).
x -13 = r
Total for both classes = 67
r + x - 13 = 67
r + x = 80
80 divided by two = 40 (per class)
40 - 13 = 27
Answer:
The whole number dimension that would allow the student to maximize the volume while keeping the surface area at most 160 square is 6 ft
Step-by-step explanation:
Here we are required find the size of the sides of a dunk tank (cube with open top) such that the surface area is ≤ 160 ft²
For maximum volume, the side length, s of the cube must all be equal ;
Therefore area of one side = s²
Number of sides in a cube with top open = 5 sides
Area of surface = 5 × s² = 180
Therefore s² = 180/5 = 36
s² = 36
s = √36 = 6 ft
Therefore, the whole number dimension that would allow the student to maximize the volume while keeping the surface area at most 160 square = 6 ft.
The sine of any acute angle is equal to the cosine of its complement. The cosine of any acute angle is equal to the sine of its complement. of any acute angle equals its cofunction of the angle's complement.
A
C
B
Since m∠A = 22º is given, we know m∠B = 68º since there are 180º in the triangle. Since the measures of these acute angles of a right triangle add to 90º, we know these acute angles are complementary. ∠A is the complement of ∠B, and ∠B is the complement of ∠A.
If we write, m∠B = 90º - m∠A (or m∠A = 90º - m∠B ), and we substitute into the original observation, we have:
Answer: She would have to score 91 or more points to have an average of 60.
Step-by-step explanation: