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Olivier P.

Tutored math for 2 years while in school

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Pre-Calculus

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Question:

Solve the following equation for x. $$3(5^{2x+1})=18(2^{5x-3})$$

Olivier P.

Answer:

divide both sides by 3 $$5^{2x+1}=6(2^{5x-3})$$ take the log of both sides $$ln(5)(2x+1)=ln(6)+ln(2)(5x-3)$$ expand the equation $$(2x)ln(5)+ln(5)=ln(6)+(5x)ln(2)-3ln(2)$$ simplify with logarithm power rule $$(x)ln(5^2)+ln(5)=ln(6)+(x)ln(2^5)-3ln(2)$$ bring x terms to the LHS $$(x)ln(5^2)-(x)ln(2^5)=ln(6)-3ln(2)-ln(5)$$ factor out the x and divide both sides by $$ln(5^2)-ln(2^5)$$ $$x=\frac{ln(6)-3ln(2)-ln(5)}{ln(5^2)-ln(2^5)}$$

Pre-Algebra

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Question:

Which of the 3 fractions is the greatest? $$3 \frac{1}{4}, \frac{18}{7}, \frac{8}{3}$$?

Olivier P.

Answer:

We need to write each fraction out with the same denominator to compare them. To do so, first we need to find a common denominator between 4, 7, and 3. This could be 42, or you could multiply 4*7*3 and get 84. Lets use 84. To convert 3 1/4, first we convert this to an improper fraction by doing (3*4)+1 = 13, giving us 13/4. Then we multiply the bottom by 3*7 to get 84 and we do the same to the top. finally giving us (3*7*13)/(3*7*4) = 273/84 then the next fraction we multiply both top and bottom by 4*3 to get 216/84 finally we multiply the last fraction by 4*7 top and bottom and get 224/84. so 273>224>216 so the largest fraction is 273/84 or 3 1/4 or 13/4

Basic Math

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Question:

A canon shoots a canon ball. The height (metres) of the canon ball's trajectory over time (seconds) can be represented by this equation $$h(t) = -3t^2 - 3t + 36$$. What is the height of the canon ball at t = 2 seconds? At what time does the ball reach a height of 30m?

Olivier P.

Answer:

1) We can simply input 2 seconds into the equation for the variable t. doing so gives us: $$h(2) = -3(2)^2 - 3(2) + 36$$ h(2) = -3 * 4 - 6 + 36 h(2) = -12 - 6 + 36 h(2) = 18 m When the canon ball reaches the 2 second mark, it is 18 m off the ground. 2) We are trying to figure out at what time t = ? does the canon ball reach a height of 30m off the ground h(t) = 30m. So then we need to solve the following quadratic equation 30 = -3t^2 - 3t + 36 we want to bring the 30 over to the right hand side so that we have a quadratic equation equal to 0 0 = -3t^2 - 3t + 6 notice each term of the quadratic equation is divisible by 3, so we can simplify this to 0 = 3(-t^2 - t + 2) remember to multiply the -1 on the first term to the 2 to give us -2! so now we ask ourselves which 2 numbers multiply to -2 and add to -1. -2 and 1 0 = 3(-t^2 - 2t + t + 2) find common factors in each set -t^2 - 2t and t + 2 (t + 4) is our common factor 0 = 3(-t(t + 2) + 1(t + 2)) 0 = 3(-t + 1)(t + 2) divide each side by 3 0/3 = 0 0 = (-t + 1)(t + 2) so our 2 solutions are t = 1, -2 we cannot have negative time, so our final solution is t = 1 second The ball reaches a height of 30m at t = 1 second

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