The best answer would be D) 67 because the other numbers are not prime.
31 degrees, 31 degrees, 118 degrees
Step-by-step explanation:
Step 1 :
Let x be the measure of 2 angles of the given isosceles triangle with same measure
Let y be the measure of 3rd angle
So we have x + x + y = 180
Step 2 :
Given that the measure of 3rd angle of triangle is 25° more than three times the measure of either of the other two angles
So we have , y = 3 x + 25
Step 3:
Substituting for y in the first equation we have,
x + x + 3 x + 25 = 180
=> 5 x + 25 = 180
=> 5 x = 180-25 = 155
=> x = 155/5 = 31
Hence the 2 angles of the triangle are 31 degrees.
Step 4:
we have y = 3 x + 25
=> y = 3 * 31 + 25 = 118
Hence the 3rd angle of given triangle is 118 degrees
Answer:
the rate of change of the water depth when the water depth is 10 ft is; 
Step-by-step explanation:
Given that:
the inverted conical water tank with a height of 20 ft and a radius of 8 ft is drained through a hole in the vertex (bottom) at a rate of 4 ft^3/sec.
We are meant to find the rate of change of the water depth when the water depth is 10 ft.
The diagrammatic expression below clearly interprets the question.
From the image below, assuming h = the depth of the tank at a time t and r = radius of the cone shaped at a time t
Then the similar triangles ΔOCD and ΔOAB is as follows:
( similar triangle property)


h = 2.5r

The volume of the water in the tank is represented by the equation:



The rate of change of the water depth is :

Since the water is drained through a hole in the vertex (bottom) at a rate of 4 ft^3/sec
Then,

Therefore,

the rate of change of the water at depth h = 10 ft is:




Thus, the rate of change of the water depth when the water depth is 10 ft is; 