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Mike D.

WVU Math Ed Masters Student

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Pre-Calculus

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Question:

Solve for x: e^(2x) +2e^(x)-15=0

Mike D.

Answer:

Here we can substitute another variable in for e^x. Call this variable y, so y=e^x. Therefore we have y^2+2y-15=0. This factors to (y+5)(y-3)=0. So, y=-5 and y=3. Substituting back in we have e^x=-5 and e^x=3. e^x has a range of 0 to infinity so it will never equal -5 and we can reject that answer. As for e^x=3, we can solve this by taking the natural log of both sides, giving us x=ln(3) since the natural log of e^x equals x.

Calculus

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Question:

A farmer has 60 feet of fencing and wants to fence in rectangular pen up against a barn. (the barn side does not need fencing, so we are only fencing three sides) What is the maximum amount of area that the farmer can fence in?

Mike D.

Answer:

This is an optimization equation. First we need to figure out what equation we can use as a constraint and what equation we are optimizing. The constraint is the perimeter equation which here is 60= 2x+y (only 1 y because y here is going to represent the side opposite the barn that we don't need to fence). Our optimizing equation is going to be the area equation, A=x*y. Using the constraint equation we can get this in terms of one variable. Take 60=2x+y and solve for y by subtracting 2x from both sides, leaving us with y=60-2x. Substituting in for y in the area equation now we have A=x*(60-2x) which we can simplify to A=60x-2x^2. Now we can derive this, giving us A'=60-4x. Set this equal to zero to find our critical value, giving us 0=60-4x. Algebra then solves x=15. Normally in optimization problems we also check the end points, but here the end points would realistically be ridiculous to consider, being x=0 and x=30 which both clearly would give us an area of 0. So we know x=15 gives us our maximum. Plugging this into the area equation of A=60x-2x^2 gives us A=450 so the maximum area that can be enclosed here is 450 feet.

Algebra

TutorMe

Question:

Find x: 24x-20=-4(1-5x)

Mike D.

Answer:

x=4 First we look at the right side of the equation and distribute the -4 to the (1-5x), giving us 24x-20=-4+20x, then we can add 20 to both sides, giving us 24x=16+20x. Next, we can subtract 20x from both sides, giving us 4x=16. Finally we reach our goal of x being alone by dividing both sides by 4, giving us x=4.

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