A team math contest has $11$ teams of $6$ students each. Before the contest begins, every student high-fives every other student
1 answer:
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I'm going to assume that the room is a rectangle.
The area of a rectangle is A = lw, where l=length of the rectangle and w=width of the rectangle.
You're given that the length, l = (x+5)ft and the width, w = (x+4)ft. You're also told that the area, A = 600 sq. ft. Plug these values into the equation for the area of a rectangle and FOIL to multiply the two factors:
Now subtract 600 from both sides to get a quadratic equation that's equal to zero. That way you can factor the quadratic to find the roots/solutions of your equation. One of the solutions is the value of x that you would use to find the dimensions of the room:
Now you know that x could be -29 or 20. For dimensions, the value of x must give you a positive value for length and width. That means x can only be 20. Plugging x=20 into your equations for the length and width, you get:
Length = x + 5 = 20 + 5 = 25 ft.
Width = x + 4 = 20 + 4 = 24 ft.
The dimensions of your room are 25ft (length) by 24ft (width).
The area of a rectangle is A = lw, where l=length of the rectangle and w=width of the rectangle.
You're given that the length, l = (x+5)ft and the width, w = (x+4)ft. You're also told that the area, A = 600 sq. ft. Plug these values into the equation for the area of a rectangle and FOIL to multiply the two factors:
Now subtract 600 from both sides to get a quadratic equation that's equal to zero. That way you can factor the quadratic to find the roots/solutions of your equation. One of the solutions is the value of x that you would use to find the dimensions of the room:
Now you know that x could be -29 or 20. For dimensions, the value of x must give you a positive value for length and width. That means x can only be 20. Plugging x=20 into your equations for the length and width, you get:
Length = x + 5 = 20 + 5 = 25 ft.
Width = x + 4 = 20 + 4 = 24 ft.
The dimensions of your room are 25ft (length) by 24ft (width).
Answer:
a)
The combined resistance of a circuit consisting of two resistors in parallel is given by:
where
R is the combined resistance
are the two resistors
We can re-write the expression as follows:
Or
In order to see if the function is increasing in r1, we calculate the derivative with respect to r1: if the derivative if > 0, then the function is increasing.
The derivative of R with respect to r1 is:
We notice that the derivative is a fraction of two squared terms: therefore, both factors are positive, so the derivative is always positive, and this means that R is an increasing function of r1.
b)
To solve this part, we use again the expression for R written in part a:
We start by noticing that there is a limit on the allowed values for r1: in fact, r1 must be strictly positive,
So the interval of allowed values for r1 is
From part a), we also said that the function is increasing versus r1 over the whole domain. This means that if we consider a certain interval
a ≤ r1 ≤ b
The maximum of the function (R) will occur at the maximum value of r1 in this interval: so, at