# Tutor profile: Brittany B.

## Questions

### Subject: Pre-Calculus

Write the equation of a line that passes through the point (1,2) and is parallel to the line 2x + 4y = 2.

An important fact to know is the point-intercept form of a line: y = mx + b m represents the slope of the line (slope is rise over run, or change in y divided by change in x) b is the y-intercept: the point of the line that is on the y-axis (where x = 0) We also need to know what it means when two lines are parallel; parallel lines have the same slope, so their m values are the same. First, we need the line given to us in slope-intercept form, so let's do some algebra rewrite the equation to look like y = mx + b 2x + 4y = 2 subtract 2x from both sides 4y = 2 - 2x divide both sides by 4 to get y alone on the left side y = [2-2x]/4 we can simplify this a little more by dividing out a 2 from every term on the right y = [1-x]/2 we can separate this into two fractions (must keep the denominator in both!) y = 1/2 - x/2 switch the order on the right side to look like y = mx + b y = -x/2 + 1/2 here, m = -1/2, so the slope of our line is m = -1/2 So far our line looks like: y = (-1/2)x + b we must find b! x and y represent all of the many points that are on our line. We know that the point (1,2) is on our line, so we can plug this into our equation and solve for b! 2 = (-1/2)(1) + b simplify 2 = -1/2 + b add 1/2 to both sides to get b alone on the right 2 + 1/2 = b fractions must have same denominator in order to add them 4/2 + 1/2 = b 5/2 = b we have b! Now we can write the equation of our line y = (-1/2)x + 5/2 ***Note: do not leave the x-value and y-value in your equation! Equations for a line must have variables for us to be able to plug in points! So we want our equation to have 'x' and 'y' in it.

### Subject: Trigonometry

Solve for x: 2cot(x) = csc(x)

We would like to make our way to only having one trigonometric identity in our equation (sin(x), cos(x), tan(x), etc.), so let's try to rewrite cot(x) and csc(x) using trig identities. cot(x) = cos(x)/sin(x) and csc(x) = 1/sin(x). Replacing both of these will give us 2[cos(x)/sin(x)] = 1/sin(x) Now let's get all terms to one side by subtracting 1/sin(x) from both sides 2[cos(x)/sin(x)] - 1/sin(x) = 0 Because our terms have the same denominator, we can combine the fractions [2cos(x) - 1]/sin(x) = 0 In order for a fraction to equal 0, only the numerator has to equal 0 (0 divided by anything is 0). However, we don't want to denominator to equal 0 (Anything divided by 0 is undefined! Including 0/0) So: 2cos(x) - 1 = 0 Solve for cos(x) cos(x) = 1/2 What angles give us cos(x) = 1/2? The reference angle π/3 does! And cos(x) is positive in two quadrants: QI and QIV So, x = π/3 and 5π/3 *** sin(π/3) and sin(5π/3) do not equal 0, so these x-values are good!

### Subject: Statistics

If Sack A contains 3 red balls and 5 blue balls, and Sack B contains 6 red balls and 4 blue balls, what is the probability of picking one red ball out of a randomly chosen sack? Assume the likelihood of picking either sack is 0.5.

Let's first find the probability of picking a red ball from either sack. In sack A, there are 8 balls total and 3 of them are red, so the probability of picking a red ball out of sack A is 3/8. In sack B, there are 10 balls total and 6 of them are red, so the probability of picking a red ball out of sack B is 6/10. Now, we don't know which sack we are picking a ball out of. Let's pretend there are two sacks on the table in front of us and they are not labeled. If we pick one up, there is a 50% chance it is sack A and a 50% chance it is sack B. If it is sack A, the probability of pulling a red ball is 3/8. If it is sack B, the probability of pulling a red ball is 6/10. So, the probability of picking a red ball out of a randomly chosen sack is 0.5(3/8) + 0.5(6/10) = 39/80 = 0.4875

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