The answer for this question depends on the model being used to model the growth of the bacteria. I'm going to assume it's exponential. So number of bacteria is modeled as A = Sekt. A is the amount of bacteria, S is the starting amount of bacteria, k is a growth constant and t...

There are 23C20 ways to pick questions on the test (read as 23 choose 20). This represents the various ways that the 20 can be picked without respect to order. There are 23! total ways and 20! ways to arrange 20 items and there's 3! ways to arrange the remaining 3 questions.
This...

Looking at the equation we can tell it will be an ellipse because of the coefficients of x2 and y2 are not the same but they are the same sign (so not a hyperbola). It's easiest to find the critical points like the center foci and vertices if it's written in the form: (x-h)2/(a2)+(y-k)2/(b2)...

Starting with 56, the multiples of 56 (the product if 8&7) are:
56,112,168,224,280,336, and 392. Of those 7 only 168 and 336 are divisible by 6. Thus 169 and 336 are the only numbers we need to exclude from our count of multiples of 6.
Now for the multiples of 6. We use the...

Starting with 105, the multiples of 35 (the product if 5&7) are:
105,140, and 175. Of those three only 105 is divisible by three. Thus 105 is the only number we need to exclude from our count of multiples of three.
Now for the multiples of three. We use the formula for...

I think you mean √(3b2)/(27b4)
in this case, the 3 and 27 reduce to a 9 in the denominator. The b2/b4 reduces down to b2 in the denominator. so you now have √(1/(9b2). This is a perfect square so you get ± 1/(3b). The plus or minus is necessary from...

For question 1, we translate english directly into math: Cost = $31 + $0.16*miles or C = .16m + 31. We are given the total cost C, so we can directly solve for m.
For question 2, the most direct way is to differentiate the equation with respect to x and set it...

For this parametric equation we have to be careful because of the squares on the t variable. We will start with x = -t2. Multiplying by negative 1, we get -x = t2. We substitute this into the other equation (for the t2) and get y = 2(-x) - 2 or y = -2x-2. Now we have...

To eliminate the parameter T, from our equations we have to substitute one into the other. The equation y = t allows us to substitute into x=1/4t2. We get the relation: x = 1/4y2. If you want it in function form, you'll end up with two equations, one positive and one negative due...

This is problem similar to continuously compounded interest. It's a simple function where t is the independent variable and B the dependent variable. It's just a matter of plugging in two years.

The domain of a function can be restricted pretty much anyway you want. However, in this case, they are looking for the set of x values that can be put into the function without causing a division by zero or a point discontinuity. In the case of f(x) = 3x+4, there are no values of...

Translating the problem directly into an equation:
y (number of quizzes) = 2*(the number of quizzes per week) x (the number of weeks) =>
y = 2x

If the equation is written correctly above, then the solution looks like this:
5x-6x=-9
-x=-9 (Combining like terms)
x=9 (mult both sides by -1)
Therefore it has one real solution.