I have no clue what to do on these questions.
1.Two resistors a and b are connected in parallel. The total resistance R in such a circuit is found by formula R= 1/a + 1/b. If R=5/6 ohms and b is one more ohm than a, find the number of ohms in a and b.
2. The profit function for producing blank CD's was projected to be P(x)=x^2 + 5x-6 where x is the number of CD's produced in hundreds. The average profit for a cd is found by P(x)/x. At what level of production must the company produce the CD's for the average profit to be greater than zero?
Where is P(x) zero?
(x+6)(x-1) = 0
You can not make a negative number of blank CDs
so where x>1 the profit is positive, so is P(x)/x
1. aren't you told that
b is one more ohm than a ?
so -----> b = a+1
sub in the equation
5/6 = 1/a + 1/(a+1)
multiply by 6a(a+1) , the LCD
5a(a+1) = 6(a+1) + 6a
5a^2 + 5a = 6a + 6 + 6a
5a^2 - 7a - 6 = 0
(a-2)(5a + 3) = 0
a = 2 or a = -3/5, but I believe that a > 0
so a = 2 ohms and b = 3 ohms
check:
1/2 + 1/3 = 5/6 , all is good!
5/6 = 1/a + 1/(a+1)
5 a(a+1) = 6(a+1) + 6 a
5 a^2 + 5 a = 6 a + 6 + 6 a
5 a^2 - 7a - 6 = 0
(5a+3)(a-2) = 0
a = 2
b = 3
1. Ohm my, these resistors are causing quite a "parallel"lem! Don't worry, I'll help you untangle this circuitry confusion.
Let's solve it step by step. We know that R = 5/6 ohms, and b is one more ohm than a. So, we can rewrite the formula as R = 1/a + 1/(a+1).
Now, let's substitute R in the equation:
5/6 = 1/a + 1/(a+1)
To make things easier, let's find a common denominator:
(5/6) = [(1*(a+1) + a)/(a*(a+1))]
Simplify further:
5*(a*(a+1)) = 2*(2a+1)
Expand:
5a^2 + 5a = 4a + 2
Rearrange:
5a^2 + a - 2 = 0
Solve for a using your favorite quadratic formula:
a = (-1 ± √(1^2 - 4*5*(-2))) / (2*5)
Simplifying that gives us two solutions, but we can discard the negative one since we're dealing with resistances.
So, a = (-1 + √41) / 10
Now, to find b, we add 1 to the value of a:
b = a + 1
Calculating that gives us the number of ohms for both a and b. Keep in mind that the quadratic formula can sometimes be a bit irrational. Thankfully, resistors don't judge!
2. Ah, the magical world of CD profits! Let's find the level of production where the average profit is greater than zero.
We're given the profit function P(x) = x^2 + 5x - 6 (where x is in hundreds).
The average profit for a CD is found by dividing the profit function by x: P(x)/x.
So, to find where the average profit is greater than zero, we need to determine when P(x)/x > 0.
Let's set up the inequality:
(x^2 + 5x - 6)/x > 0
Now, let's break it down and find the x values that satisfy this condition:
Case 1: x > 0
If x > 0, we can cancel out the x from both sides of the inequality, resulting in:
x + 5 - 6/x > 0
x^2 + 5x - 6 > 0
We solve this quadratic inequality by factoring:
(x - 1)(x + 6) > 0
We have two factors, and for the inequality to hold true, either both factors are greater than 0 or both factors are less than 0.
Case 1.1: (x - 1) > 0 and (x + 6) > 0
This gives us:
x > 1 and x > -6
Since x must be greater than both 1 and -6, the minimum value that satisfies this condition is x = 1.
Case 1.2: (x - 1) < 0 and (x + 6) < 0
This gives us:
x < 1 and x < -6
Since x must be less than both 1 and -6, there is no x value that satisfies this condition.
Therefore, in Case 1, the values of x that make P(x)/x > 0 are x > 1.
Finally, let's double-check our answer by substituting values into the equation. At x = 1, we have P(1) = 1^2 + 5(1) - 6 = 0. Since it equals 0, it does not satisfy the condition of being greater than zero.
However, at x > 1, let's say x = 2, we have P(2) = 2^2 + 5(2) - 6 = 10. This satisfies the condition of being greater than zero!
So, the company must produce at least 200 CDs (x > 1) for the average profit to be greater than zero. It's time to crank up the CD production and make some music to our ears, cha-ching!
For question 1, we are given the formula for calculating the total resistance of resistors connected in parallel: R = 1/a + 1/b. We are also given that the total resistance, R, is 5/6 ohms and that b is one more ohm than a. We need to find the number of ohms in a and b.
To solve this, we can substitute the given values into the formula and then solve for a and b:
1/a + 1/b = 5/6
Since b is one more ohm than a, we can rewrite b as a + 1:
1/a + 1/(a + 1) = 5/6
To eliminate the fractions, we can find a common denominator:
(a + 1)/[a(a + 1)] + a/[a(a + 1)] = 5/6
Now, combine the fractions:
(a + 1 + a)/(a(a + 1)) = 5/6
Simplify:
(2a + 1)/(a^2 + a) = 5/6
Cross-multiply:
6(2a + 1) = 5(a^2 + a)
Expand:
12a + 6 = 5a^2 + 5a
Rearrange to form a quadratic equation:
5a^2 + 5a - 12a - 6 = 0
Combine like terms:
5a^2 - 7a - 6 = 0
Now, we can solve this quadratic equation using factoring, completing the square, or the quadratic formula. In this case, factoring is the most convenient method:
(5a + 2)(a - 3) = 0
From here, we can set each factor equal to zero and solve for a:
5a + 2 = 0
a = -2/5
a - 3 = 0
a = 3
Since resistance cannot be negative, we discard the negative value. Therefore, a = 3 ohms.
To find b, we use the fact that b is one more ohm than a:
b = a + 1
b = 3 + 1
b = 4 ohms
So, the number of ohms in resistor a is 3, and the number of ohms in resistor b is 4.
For question 2, we are given the profit function for producing blank CDs in terms of the number of CDs produced, P(x) = x^2 + 5x - 6, where x is the number of CDs produced in hundreds. We need to find the level of production at which the average profit is greater than zero.
The average profit is given by the formula P(x)/x. To find the level of production where the average profit is greater than zero, we set up the following inequality:
P(x)/x > 0
Substituting the profit function into the inequality:
(x^2 + 5x - 6)/x > 0
Now, we need to determine the values of x for which the expression is greater than zero. To do this, we look at the sign of the numerator:
(x + 6)(x - 1)/x > 0
We have three critical points: x = -6, x = 0, and x = 1. We create a sign chart to determine the intervals where the expression is positive:
| (-∞) | -6 | 0 | 1 | (∞) |
|________|________|_______|_______|________|
| | - | + | - | + |
From the sign chart, we see that the expression is positive on the intervals (-∞, -6) and (0, 1). This means that the average profit is greater than zero when the level of production is between -6 and 0, or between 1 and infinity.
Therefore, the company must produce the CDs for the average profit to be greater than zero at any level of production between -6 and 0, or between 1 and infinity (excluding 0 itself).