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OleMash [197]
3 years ago
9

5 A swimming pool is being filled using a pípe A at x gallons per hour. After two hours, pipe B is used

Mathematics
1 answer:
Tpy6a [65]3 years ago
6 0

Answer:

Example 1:

A tank can be filled by pipe A in 3 hours and by pipe B in 5 hours. When the tank is full, it can be drained by pipe C in 4 hours. if the tank is initially empty and all three pipes are open, how many hours will it take to fill up the tank?

Solution:

Step 1: Assign variables:

Let x = time taken to fill up the tank

Step 2: Use the formula:

Since pipe C drains the water it is subtracted.

1/3+1/5-1/4=1/x

Step 3: Solve the equation

The LCM of 3, 4 and 5 is 60

Multiply both sides with 60

solve the eqn

Answer: The time taken to fill the tank is 3 9/17 hours.

 

 

Work Problem: Pumps draining a tank

Example:

A swimming pool can be emptied in 6 hours using a 10-horsepower pump along with a 6-horsepower pump. The 6-horsepower pump requires 5 hours more than the 10-horsepower pump to empty the pool when working by itself.

How long will it take to empty the pool using just the 10-horsepower pump?

Show Step-by-step Solutions

Cooperative Work Word Problems (Time to Finish)

Examples:

1. Pump A can empty a pool in 20 hours and pump B can empty it in 24 hours. Working together, how long will it take to empty the pool?

2. A painter can paint a building in 15 days and a coworker can do the same job in 10 days. If the first painter starts and 3 days later the coworker joins in to help finish the job, how many days doe it take to paint the building?

Show Step-by-step Solutions

Rates of Performing Work Problems

Example:

It takes 12 hours to fill a water tank. It takes 16 hours to drain the same water tank. How long will it take to fill the tank if the drain is left open?

Show Step-by-step Solutions

Step-by-step explanation:

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Explanation:

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Of these, behavior 2 will ultimately look like one of the others.

For polynomials, the function will always approach ±∞ as the independent variable approaches ±∞. Whether the signs of the infinities agree or not depends on the even/odd degree of the polynomial, and the sign of its leading coefficient.

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For trig functions sine and cosine, the end behavior is the same as the "middle" behavior: the function oscillates between two extreme values.

For rational functions (ratios of polynomials), the end behavior will depend on the difference in degree between numerator and denominator. If the degree of the denominator is greater than or equal to that of the numerator, the function will have a horizontal asymptote. If the degree of the numerator is greater, then the end behavior will asymptotically approach the quotient of the two functions—often a "slant asymptote".

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2. A polynomial inequality written in the form f(x) ≥ 0, or f(x) > 0, will be solved by first identifying the real zeros of the function f(x), including the multiplicity of each. For positive values of x greater than the largest zero, the sign of the function will match the sign of the leading coefficient. The sign will change at each zero that has odd multiplicity, so one can work right to left to identify the sign of the function in each interval between odd-multiplicity zeros.

The value of the function will be zero at each even-multiplicity zero, but will not change sign there. Obviously, the zero at that point will not be included in the solution interval if the inequality is f(x) > 0, but will be if it is f(x) ≥ 0. Once the sign of the function is identified in each interval, the solution to the inequality becomes evident.

As a check on your work, you will notice that the sign of the function for x > max(zeros) will be the same as the sign of the function for x < min(zeros) if the function is of even degree; otherwise, the signs will be different.

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3. A composite function is a function of a function, or a function of a composite function. For example f(g(x)) is a composite function. The composition can be written using either of the equivalent forms ...

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It can be easy to confuse an improperly written composition operator with a multiplication symbol, so the form f(g(x)) is preferred when the appropriate typography is not available.

When simplifying the form of a composition, the Order of Operations applies. That is, inner parenthetical expressions are evaluated (or simplified) first. As with any function, the argument of the function is substituted wherever the independent variable appears.

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Note that the expression f(g(x)) is written as the composition shown above. The expression g(f(x)) would be written using the composition operator with g on the left of it, and f on the right of it:

  (g\circ f)(x)\ \Leftrightarrow\ g(f(x))

That is, with respect to the argument of the composition, the functions in a composition expression are right-associative. For example, ...

  for h(x)=2x+3, g(x)=x^2, f(x)=x-2 we can evaluate f(g(h(x)) as follows:

  f(g(h(x)) = f(g(2x+3) = f((2x+3)^2) = (2x+3)^2 -2

It should be obvious that g(h(f(x)) will have a different result.

  g(h(f(x)) = g(h(x-2)) = g(2(x-2)+3) = (2(x-2)+3)^2

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