# Tutor profile: Yuria U.

## Questions

### Subject: SAT II Mathematics Level 2

The number n is doubled, then tripled, then reduced by 5, then cube rooted. The value is now -5. What was the value of n?

Step 1: Rewrite the problem as an expression "doubled" -> $$2n $$ "then tripled" -> $$(3)(2n) $$ "then reduced by 5" -> $$(6n)-5$$ "then cube rooted" -> $$\sqrt[3]{6n-5}$$ "value is now -5" -> $$\sqrt[3]{6n-5}=-5$$ Step 2: Work backwards and solve $$\sqrt[3]{6n-5}=-5$$ $$6n-5 = (-5)^3$$ $$6n = -125+5 $$ $$n = -120/6 $$ $$n = -20 $$ Answer: $$n = -20 $$

### Subject: Calculus

An open box is made by cutting congruent squares from the corners of a 12" by 10" cardboard and folding up the sides. What are the dimensions of the box such that volume is maximized?

Step 1: Define variables $$x$$ = side length of square that is cut out, height of box $$12-2x$$ = side length of box $$10-2x$$ = width of box Step 2: Set up equation $$Volume = (x)(12-2x)(10-2x) $$ $$Volume = 4x(x^2-11x+30)$$ Step 3: Minimize volume $$V' = (4)(x^2-11x+30)+(4x)(2x-11)$$ $$V' = 3x^2-22x+30$$ $$0 = 3x^2-22x+30$$ $$x = 5.52, 1.81$$ (Note: 5.52 is outside the domain) Step 4: Calculate dimensions $$height = 1.81$$ $$length = 12-2(1.81) = 8.38$$ $$width = 10-2(1.81) = 6.38$$ Answer: $$1.81''\times 8.38''\times 6.38'' $$

### Subject: ACT

A rectangular box has a square base. The box's height is 3 inches more than twice its width. Which of the following gives the box's volume in terms of its width (w)?

Step 1: Find the expression for the box's volume in terms of height ($$h$$) and width ($$w$$) Because the box has a "SQUARE base" we know its dimensions are $$h$$ by $$w$$ by $$w$$. $$Volume = h\cdot w\cdot w$$ Step 2: Find the relationship between the height ($$h$$) and width ($$w$$) "box's height is 3 inches MORE than TWICE its width" $$h = 3+2w$$ Step 3: Rewrite volume equation in terms of "w" $$Volume = (3+2w)\cdot w\cdot w$$ Answer: $$(3+2w)\cdot w\cdot w$$

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