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Tutor profile: Kayla H.

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Kayla H.
Graduate Student in Mathematics Education with 6 years of tutoring experience
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Questions

Subject: Pre-Calculus

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Question:

Given that f(x)=x+2 and g(x)=5x-3 find f(g(x)).

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Kayla H.
Answer:

We need to find f(g(x)), so we will begin by substituting g(x) for every x we see in the function f(x). We have f(x)=x+2, so now plugging in g(x)=5x-3 for x we obtain that f(g(x))=(5x-3)+2 f(g(x))=5x -1 (After combining like terms)

Subject: Calculus

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Question:

Find the equation of the tangent line to the function f(x)=x^2+3x+4 at the point x=4

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Kayla H.
Answer:

Recall the equation of the tangent line to a function f(x) at the point x=a is y - f(a) = f'(a)(x-a) In this case, a=4 and f(x)=x^2+3x+4 So the equation of the tangent line in our case is y- f(4) = f'(4) (x-4) We need to find f(4) and f'(4) First let's find f'(x). We know f(x)=x^2+3x+4, then f'(x)=2x+3 (From the power rule of taking derivatives. So, f'(4)=2(4)+3 f'(4)= 8+3 f'(4)=11 Also, f(4)= (4)^2+3(4)+4 f(4)=16+12+4 f(4)=32 Now substituting these into y-f(4)=f'(4) (x-4). we get: y - 32 = 11(x-4) y-32=11x-44 y=11x - 12 (After adding 32 to both sides)

Subject: Algebra

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Question:

Solve the following system of equations: 4x+2y=10. (Equation 1) x+3y=15. (Equation 2)

Inactive
Kayla H.
Answer:

From Equation 2, we can solve for x by first subtracting 3y from both sides of the equation and obtain x=15-3y Then, we use the direct substitution method to substitute this value for x into Equation 1. We have: 4x+2y=10 4(15-3y)+2y=10. (After substituting the value for x) 60-12y+2y=10 (Distribute 4 into the set of parentheses by multiplying each term by 4) 60-10y=10 (Combining like terms) -10y=-50 (Subtracting 60 from both sides) y=5. (Divide each side of the equation by -10). Now, we can use this value of y=5 and plug it into either Equation 1 or Equation 2 to solve for x. Let's plug it into equation 1: 4x+2y=10 4x+2(5)=10. (Substituting 5 for y) 4x+10 = 10 (Simplifying 2 times 5) 4x=0. (Subtracting 10 from both sides) x=0 (Dividing both sides of the equation by 4) So, we have y=5 and x=0. REmember that we can always check our work to make sure by substituting these values into both equations. Equation 1: 4x+2y=10 4(0)+2(5) =? 10 0+10 =? 10 10 =? 10. TRUE Equation 2: x+3y=15 0+3(5) =? 15 0+15 =? 15 15 =? 15 TRUE Since both equations are true, the final answer of x=0 and y=5 is correct and we have solved this system of equations.

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