We know, 1 ounce = 29.57 ml
so, 64 ounces = 1892.71 ml
so, remaining drink would be: 1892.71-700 = 1192.71 ml
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; 
Answer:
All real numbers are solutions.
Step-by-step explanation:
Step 1: Simplify both sides of the equation.
−3(m+8)=−3m−24
(−3)(m)+(−3)(8)=−3m+−24(Distribute)
−3m+−24=−3m+−24
−3m−24=−3m−24
Step 2: Add 3m to both sides.
−3m−24+3m=−3m−24+3m
−24=−24
Step 3: Add 24 to both sides.
−24+24=−24+24
0=0
Answer:
All real numbers are solutions.
All three of those angels are 98 degrees. F is supplementary, as the given angle and it must add up to 180. D is parallel to F. E is the same as D.