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# Tutor profile: Jakob H.

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Jakob H.
Physics Student at University of South Florida
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## Questions

### Subject:Trigonometry

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Question:

You want find the height of a flagpole. You notice that at a particular moment of the day the flagpole casts a shadow 20 feet long, and the tip of the shadow lined up with the top of the flagpole creates an angle of 60 degrees. What is the height of the flagpole?

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Jakob H.

This situation creates a right triangle where the right angle is at the point where the shadow and flagpole meet, and at the tip of the shadow we have an angle of 60 degrees. Let x be the hypotenuse of this triangle (the length from the tip of the flagpole to the tip of its shadow). Thus the length of the shadow should be 20 =xcos(60). From the unit circle we know that cos(60)=1/2, thus we can solve that x=40. The height of the flagpole should then be 40sin(60)=40*sqrt(3)/2=20*sqrt(3)=34.64 feet.

### Subject:Calculus

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Question:

Find F'(x) where F(x)=[(e^(3x))(ln(x^2))]/(x^3).

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Jakob H.

While this is a normal derivative problem, it requires a good understanding of the product,quotient, and chain rules of derivation and keeping an eye on the algebra within them. Via the quotient rule: F'(x) = ([(e^(3x))(ln(x^2))]'(x^3)-[(e^(3x))(ln(x^2)](x^3)')/(x^6) Notice that we will need to use the product rule AND chain rule on the first term. F'(x)= [(3e^(3x))(ln(x^2))+(e^3x)(2/x)](x^3)-(3x^2)[(e^(3x))(ln(x^2)]/(x^6) = [e^(3x)[3x^3(ln(x^2)+2x^2-3x^2(ln(x^2))]]/x^6 = [e^(3x)[3(ln(x^2))(x-1)+2]]/(x^4)

### Subject:Algebra

TutorMe
Question:

John is 5 years older than Brian. John's age times 4 is 10 more than Brian's age times 5. How old are John and Brian?

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Jakob H.

Let x be John's age and y be Brian's age. The first sentence can be written into the equation x-y=5, whereas the second equation can be written as 4x-5y=10. Thus we have a system of two linear equations so we can solve this in one of two ways, for our purposes let's do both. First we use substitution: In equation 1 add y to both sides, giving us x=5+y, substitute this into the second equation we get 4(5+y)-5y=20+4y-5y=20-y=10 which is now an equation only of y's. Thus we get y=10, which we can plug into either the first or second equation and solve to get x=15. Secondly we could use a linear combination of both equations to get rid of one of the variables: Multiplying the first equation by -4 gives us -4x+4y=-20. We can add this to the second equation which gives us -4x+4x+4y-5y=-20+10 which reduces to -y=-10, or y=10. Again we can plug this into either equation and we get x=15. Thus John is 15 and Brian is 10.

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