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# Tutor profile: Tucker H.

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Tucker H.
Tutor for 3 years, student pursuing Chemistry degree
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## Questions

### Subject:Pre-Calculus

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Question:

Solve the following equation for x: 3x^2 + 7x +6= 4

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Tucker H.

To solve this problem, we will use the quadratic formula. Since the quadratic formula requires an equation in the form of ax^2 + bx +c = 0, we will subtract 4 from both sides and get the equation 3x^2 + 7x + 2 = 0. The quadratic formula is [-b +/- sqrt(b^2 - 4*a*c)]/(2a) Looking at our equation, we see that a = 3, b = 7, and c = 2. Plugging these values in, we get: [-7 +/- sqrt(7^2 - 4*3*18)]/(2*3) Simplifying some of the values yields: [-7 +/- sqrt (49 - 24)]/6 Further simplifying gives: [-7 +/- sqrt(25)]/6 = (-7 +/- 5)/6 Thus, we know x will either equal (-7+5)/6 = -1/3 or (-7-5)/6 = -2. We will plug both of these values back into our original equation to make sure we don't have any extraneous values. Using x = -1/3: 3(-1/3)^2 + 7(-1/3) + 2 = 0 3(1/9) - 7/3 + 2 = 0 3/9 - 7/3 + 2 = 0 1/3 - 7/3 + 6/3 = 0 0 = 0 We see that x = - 1/3 is a valid solution. Checking x = -2: 3(-2)^2 + 7(-2) +2 =0 3(4) -14 + 2 = 0 12 - 14 + 2 = 0 0 = 0 x = -2 is also a valid solution. Therefore, x = -1/3 or x = -2

### Subject:Basic Math

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Question:

Consider a movie theater with 300 seats in it. The movie theater sells children's tickets for \$5.00, adult tickets for \$7.00, and senior tickets for \$6.00. Answer the following parts of this question: 1. One night, the movie theater sold tickets for 75% of its seats. How many tickets did it sell? 2. In that same night, the movie theater sold 70 senior tickets and 55 children's tickets. How many adult tickets did the theater sell? 3. How much money did the theater bring in in ticket sales that night?

Inactive
Tucker H.

We will start by solving part one of the problem. If the movie theater sold tickets for 75% of its 300 seats, we can solve for the number of tickets sold by multiplying 300 by the decimal form of 75%: 1. 300 x 0.75 = 225 We see that the movie theater sold 225 of its tickets that night. Moving on to part two, we can find the number of adult tickets sold relatively easily. Since we know there are only three types of tickets - adult, senior, child - if we add the number of tickets sold in each category together, they will add to the total. The total is 225, the number of senior tickets is 70, and the number of children's tickets is 55. As such, we have: 2. 70 senior tickets + 55 children's tickets + x adult tickets = 225 tickets. We solve for the number of adult tickets by subtracting the amount of senior and children's tickets from the total, so: 3. # of adult tickets = 225 - 55 children's tickets - 70 senior tickets 4. # of adult tickets = 100 We now know the total numbers for each type of ticket sale: 100 adult tickets, 55 children's tickets, and 70 senior tickets. Since we know the price for each ticket, we can find the total amount in ticket sales using simple multiplication. We will multiply each ticket by the amount it costs and add all of the costs together: 5. 100 x 7 + 55 x 5+ 70 x 6 = ? By order of operations, we do our multiplications first, so we now have: 6. 700 + 275 + 420 = 1395. We have now found that the movie theater has brought in \$1395 in this one night of ticket sales.

### Subject:Algebra

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Question:

Solve the following system of equations for x, y, and z. If necessary, round to 2 decimal places. 1. 6x + 4y +21z = 387 2. 6x + 8y = 194 3. y + z = 33

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Tucker H.

We will be using the method of substitution to solve this problem. Our first step will be to solve equation 3 for y in terms of z: 4. y = 33 - z Next, we will use equation 4 to solve equation 2 for x in terms of z by substituting 33-z in for y: 5. 6x + 8(33-z) =194 Solving for x yields: 6. 6x + 264 - 8z = 194 7. 6x = 194 - 264 -(-8z) 8. 6x = -70 + 8z 9. x = -70/6 + 8z/6 We now have x and y in terms of z, so we will use these values in equation 1 to solve for z. We will note that equation 1 included the term 6x and that equation 8 solved for 6x. We begin with equation 1: 1. 6x + 4y + 21z = 387 Substituting in equation 8 and 4: 10. -70 + 8z + 4(33 - z) + 21z = 387 We distribute the 4 through the values grouped in parentheses: 11. -70 + 8z + 132 - 4z + 21z = 387 Adding like terms together and simplifying: 12. 8z - 4z + 21z = 387 - (-70) -132 13. 25z = 325 Solving for z: 14. z = 325/25 = 13 Now that we have a value for z, we can find values for x and y since we solved for them in terms of z. Referring back to equation 4: 4. y = 33 - z 15. y = 33 - 13 = 20 And referring back to equation 9: 9. x = -70/6 + 8z/6 16. x = -70/6 + 8(13)/6 = 5.666666 Rounding to two decimal places: x = 5.67 We have now solved for x, y, and z: x = 5.67, y = 20, and z = 13

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