Answer:
3 1/2
Step-by-step explanation:
Quarter bag plus three quarter bag=1 bag
2 half bags is one bag
2 three quarters full is 1 1/2 bag :)
Answer:
2. B
3. B
4. 2(7k+5h)
5. $325
Step-by-step explanation:
2) "X decreased by the sum of 2x + 5" Decreased means to be lowered, so that means this problem has subtraction in it. It also says x is being decreased BY the sum of 2x + 5 so that means 2x + 5 needs to be solved, which calls for parentheses. B best fits the description.
3) It is B because if you distribute the 8 to 2a and 3b, you'll get 16a and 24b, which is the given expression.
4) The given expression is 10h + 14k and in the equivalent expression, you are given 5h. To get from 10 to 5, you must divide by 2. This tells you that 2 is being distributed, so you put 2 before the parentheses. Since you divided 10 by 2, you want to do the same to 14, which is 7. Therefore, it is 2(7k+5h).
5) You are given the hourly charge, which is $25. This means that they charge $25 per hour/for every hour. You want to find the cost for 13 hours, so you multiply the hourly charge by 13 hours. 25 x 13 = 325.
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; 
You have to multiply 8 and 2 1/2 which is 20. Then divide 20 and 200. Then you get 10 and that is your answer.