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# Tutor profile: Naaser M.

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Naaser M.
Software programmer with Masters in Engineering
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## Questions

### Subject:Pre-Calculus

TutorMe
Question:

Solve for x ln(x+3) - ln(x-5) = ln(x+2)

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Naaser M.

To solve for x, we need to recall the property of natural log specifically, ln(A/B) = ln(A) - ln(B) Using the above property, we can simplify the Left hand side (LHS) of the equation as follows LHS: ln(x+3) - ln(x-5) =>ln((x+3)/(x-5)) Right hand side (RHS) of the equation is still ln(x+2) Now, we can apply another property of natural log: e^ln(x) = x Applying the same to both sides of the equation =>e^(ln((x+3)/(x-5))) = e^(ln(x+2)) => (x+3)/(x-5) = (x + 2) Sending (x-5) to RHS => (x+3) = (x+2)(x-5) Expanding RHS => x+3 = x^2 -5x + 2x -10 Sending x+3 to the other side => x^2-5x+2x-10-x-3=0 =>x^2-4x-13=0 Solving using the quadratic equation formula (4+sqrt(16-4*1*(-13)))/2 and (4-sqrt(16-4*1*(-13)))/2 =>(4+sqrt(16+52))/2 and (4 - sqrt(16+52))/2 =>(4+sqrt(4*17))/2 and (4-sqrt(4*17))/2 =>2+sqrt(17) and 2-sqrt(17) Therefore, x=2+sqrt(17) and/or x=2-sqrt(17)

### Subject:Calculus

TutorMe
Question:

Differentiate y = Sin(x^3)

Inactive
Naaser M.

To differentiate the above equation we have to use the chain rule. The chain rule states as follows: The differentiation of f(g(x)) = f'(g(x)) * g'(x). Here, g(x) = x^3 = u and f(g(x)) = Sin(x^3). We know that differentiation of Sin(x) = Cos(x) and differentiation of x^3 = 3*x^2. Therefore, differentiation is y' = Cos(x^3) * 3x^2

### Subject:Algebra

TutorMe
Question:

In a grocery store, 5 apples and 6 oranges cost \$28. In the same store 3 apples and 8 oranges cost \$30. How much would 7 apples and 2 oranges cost?

Inactive
Naaser M.

To solve this question, first we need to consider the cost of each apple as x and cost of each orange as y. This will help us in developing linear equations. For the first part of the question, the equation comes out to be 5x + 6y = 28 (equation 1) and for the second part, 3x + 8y = 30 (equation 2). Now that we have two equations and two variables (x & y), we can solve for x and y. We can solve the two linear equations by using either of the following methods. 1) Elimination To solve using Elimination method, we need to eliminate one of the variables. We can eliminate a variable by making the coefficient of one variable equal and subtracting the two equations. To make the coefficient of x equal, we can multiply equation 1 by 3 and equation 2 by 5. 3(5x+6y) = 3*28 5(3x+8y) = 5*30 15x + 18y = 84 15x + 40y = 150 We can now subtract equation 1 from equation 2. 15x + 40y = 150 -15x - 18y = -84 ---------------------- 0x + 22y = 66 ---------------------- =>y = 66/22 =>y = 3 Now using the value of y in equation 1, we can solve for x 5x+6*3 = 28 => 5x + 18 = 28 => 5x = 28 - 18 => 5x = 10 => x =2 Now to get the cost of 7 apples and 2 oranges, we can use the values of x and y to calculate the total cost. 7*2 + 2*3 => 14 + 6 =>\$20 2) Substitution To solve using Substitution method, we get the value for either x or y from equation 1 and substitute in equation 2. Equation 1 5x + 6y = 28 => 6y = 28 - 5x => y = (28 - 5x)/6 Using the above value of y, we can substitute in equation 2 3x + 8 (28 - 5x)/6)) = 30 Multiplying by 6 on both sides => 6*3x + 8 (28-5x) * 6/6 = 6*30 => 18x + 224 - 40x = 180 => -22x = 180 - 224 => -22x = -44 => x = 2 With the value of x we can solve for y y = (28 - 5*2)/6 =>y = (28 - 10)/6 =>y = 18/6 =>y = 3 Now to get the cost of 7 apples and 2 oranges, we can use the values of x and y to calculate the total cost. 7*2 + 2*3 => 14 + 6 =>\$20

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