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Kathryn M.

Tutor for five years, Computer Engineering Student

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SAT

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Question:

For a triangle $$PQR$$, $$∠Q$$ has a measure of $$90°$$, $$PQ=30$$, and $$PR=50$$. There is another triangle $$XYZ$$ that is similar to triangle $$PQR$$, where vertices $$X$$, $$Y$$, and $$Z$$ correspond to vertices $$P$$, $$Q$$, and $$R$$, respectively, and each side of the triangle $$XYZ$$ is one tenth the length of the corresponding side of triangle $$PQR$$. What is the value of $$\cos(Z)$$?

Kathryn M.

Answer:

Triangle $$PQR$$, a right triangle, has a right angle at the vertex $$Q$$. As a result, the triangle has a hypotenuse on $$PQ$$, with the other side lengths acting as the legs of the triangle. To find the length of leg $$QR$$, we can use the Pythagorean theorem, as $$QR=\sqrt(50^{2}-30^{2})=\sqrt(2500-900)=40$$ Because of similar triangles, if triangle $$XYZ$$ is similar to triangle $$PQR$$, with angle $$Z$$ corresponding to angle $$R$$, the measure of $$∠Z$$ will be the same as the measure of $$∠R$$. Thus, $$\cos(Z)=\cos(R)$$. We can use the sides of the triangle $$PQR$$ and the equation of the sine function to compute $$\cos(R)=\frac{adjacent side}{hypotenuse}=\frac{QR}{PR}=\frac{40}{50}=\frac{4}{5}$$, so that $$\cos(Z)=\frac{4}{5}$$.

Calculus

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Question:

You have 12 square centimeters of cardboard to build a box with a square base and an open top. Calculate the largest possible volume of a box that you can build.

Kathryn M.

Answer:

You are building a box with a square base and an open top. You can give the side lengths of the square base a value of b, and the height of the box a value of h. The surface area is the amount of material to make the box, and you are supplied with 12 $$ cm^{2}$$ of cardboard. The surface area of the box, which we can give the variable S, will be equal to the combination of the area of the square base and the four sides of the box. The area of the base will be the side length squared, $$b^{2}$$, and the area of a side of the box will be equal to the side length multiplied by the heigh, bh. The equation for the surface area is thus S = $$ b^{2}$$ + $$ 4*bh$$. Solving this equation for the variable h, we get a new equation, h= $$ \frac{12-b^{2}}{4b}$$. The volume of the box, V, will be equal to the area of the base multiplied by the height, giving an equation V= $$b^{2}*h$$. If we insert the equation for h into the equation for V, we get V= $$b^{2}*\frac{12-b^{2}}{4b}$$, which is equal to V= $$3b-\frac{b^{3}}{4}$$. Now, we can use critical points to find the extrema of this equation, which will allow us to discern the maximum volume that can be reached. Critical points will occur when the derivative of our equation for V is either 0 or Does Not Exist. V' = $$3-\frac{3b^{2}}{4}$$, and there are no points of V' that don't exist, so we must find the solutions of this equation by setting it equal to 0 and solving for b. We find solutions at b= -2 and b=2. Because a side length can't be negative, the only extrema of the equation for V is when a side length b has a measurement of 2 cm. To see whether b=2 indicated a maxima or a minima of the equation V, we must undergo a First Derivative Test. As V(b) is continuous and differentiable on (0,12), the size constraints of the box, and at point b=2, a critical point, V'(b) changes from positive to negative, indicating that a local maximum is present according the First Derivative Test. The largest possible volume of the box is thus at b=2, when the volume is maximized, so at V(2) = $$b^{2}*\frac{12-2^{2}}{4*2}$$, when b=2 and h=1, $$V=4cm^{3}$$.

ACT

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Question:

You are given the functions f(x) = $$ \sqrt{3x + 4} $$ and k(x) = 8x + h. in the Cartesian (x,y) plane, y = f(k(x)) intersects the point (2,8). What is the value of h?

Kathryn M.

Answer:

If f(x) = $$ \sqrt{3x + 4} $$ and k(x) = 8x + h, then f(k(x)) = $$ \sqrt{3(8x+h) + 4} $$, which is f(k(x)) = $$ \sqrt{24x+3h + 4} $$. With y= f(k(x), this must intersect the point (2,8), so 8 will equal f(k(x)), and the x value will be 2. Thus, 8 = $$ \sqrt{24*2+3h + 4} $$ = $$ \sqrt{48+3h + 4} $$. Squaring both sides to remove the square root yields 64 = 48 +3h +4. Combining the integers on the right brings 64 = 52 + 3h. We can then subtract 52 from both sides to get 12 = 3h. Now, we divide both sides by 3 to get 4 = h. Our final answer is that h has a value of 4.

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