# Tutor profile: Brian S.

## Questions

### Subject: Pre-Algebra

Add vectors v and w. v = < 3,4 > , w = < 6,2 >

Add the x parts and y parts of vectors v and w together. < 3+6 = 9 , 4+2 = 6 > Vector v + w = < 9,6 >.

### Subject: Calculus

Take the integral of 4x^3 + 2x^2 - 4x - 6 from 0 to 1.

Take on the problem in parts. First start off with the first term, 4x^3. To take the integral of this, add 1 to the exponent of the variable and then divide the coefficient by that number. 4x^3 --> 4/(3+1) x^(3+1) --> (4/4)x^4 --> x^4 Do the same process for all the terms. You'll get 2/(2+1) x^(2+1) --> (2/3)x^3 -4/(1+1) x^(1+1) --> -2x^2 -6/(0+1) x^(0+1) --> -6x Put the terms together to form the new expression. To take the integral from 0 to 1, plug in 1 into the expression. 1^4 + (2/3)1^3 - 2(1^2) - 6(1) = -19/3 Then plug in 0 into the expression. 0^4 + (2/3)0^3 - 2(0^2) - 6(0) = 0 Subtract the value of the expression that was substituted with 0 from the value of the expression that was substituted with 1. -19/3 - 0 = -19/3 The answer is -19/3 or -6.33.

### Subject: Algebra

2x + 4y = 6 3x + 2y = 8 Solve for the values of x and y.

First, solve one of the equations for the x-variable in terms of y. 2x + 4y = 6 --> 2x = -4y + 6 --> x = ( -4y + 6 ) / 2 --> x = -2y + 3 Then, substitute that into the other equation. 3 ( -2y + 3 ) + 2y = 8 Next, distribute and combine like terms. ( -6y + 9 ) + 2y = 8 -4y + 9 = 8 Isolate the y-variable to find its value. -4y = -1 --> y = 1/4 Plug the y-value into the first equation and isolate the x-variable to find its value. 2x + 4 ( 1/4 ) = 6 --> 2x + 1 = 6 --> 2x = 5 x = 5/2 x = 5/2 , y = 1/4

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