TutorMe homepage
Subjects
PRICING
COURSES
Start Free Trial
Mary K.
Experienced math, writing, and test prep tutor
Tutor Satisfaction Guarantee
SAT
TutorMe
Question:

Let $$(h,k)$$ be the vertex of the parabola $$f(x) = -2x^2-16x-32$$. Find the value of $$i^{(2h-k^2)}$$ where $$i=\sqrt{-1}$$. $$A) \space i\\ B)\space -i\\ C)\space 1\\ D)\space -1$$

Mary K.

First we find the vertex of $$f(x)$$. For a quadratic of the form $$ax^2+bx+c$$, the $$x$$-coordinate of the vertex $$(h,k)$$ can be found with the formula $$h=\frac{-b}{2a}$$. In this case, $$h =\frac{-b}{2a}= \frac{16}{2\times(-2)}=-4.$$ Therefore, $$k=f(-4)=-2(-4)^2-16(-4)-32=-32+64-32=0$$. We can then plug $$h=-4$$ and $$k=0$$ into $$i^{(2h-k^2)}$$ to get $$i^{(2h-k^2)} \\ = i^{2(-4)-0^2}\\ =i^{-8}.\\$$ Finally, we simplify using the definition $$i=\sqrt{-1}$$: $$i^{-8}\\ =\sqrt{-1}^{-8}\\ =((-1)^{1/2})^{-8}\\ =(-1)^{-4}\\ =\dfrac{1}{(-1)^4}\\ =1.$$ So our answer is C.

Geometry
TutorMe
Question:

A circle with center $$O$$ is tangent to square $$ABCD$$ at point $$M$$. If$$M$$ is the midpoint of $$\overline{AB}$$, the area of the circle is $$64\pi$$, and the area of the square is $$49$$, what is the area of $$\triangle{AOM}$$?

Mary K.

To find the area of a triangle, we use the formula $$A=\frac{1}{2}bh$$ where the base, $$b$$, and height, $$h$$, are perpendicular. First notice that $$\triangle{AOM}$$ is a right triangle. To see this, note that the radius of the circle, $$\overline{OM}$$, is perpendicular to the tangent line, $$\overline{AB}$$. Therefore, $$\overline{AM}\perp\overline{OM}$$. The area of the square is $$49$$, so each side length must be equal to $$\sqrt{49}=7$$. Because $$M$$ is the midpoint of $$\overline{AB}$$, $$AM=7$$. We also know the area of the circle is $$64\pi$$. Solving the equation for the area of a circle, $$64\pi=\pi*r^2$$ gives us $$r=8$$. Note that $$r=OM$$. Now we have the two legs of the right triangle, so we can calculate its area: $$A=\frac{1}{2}bh\\ A=\frac{1}{2}*\frac{7}{2}*8\\ A=56/4\\ A=14.$$

Algebra
TutorMe
Question:

Patty is baking a cake, but has lost the recipe. She remembers that it called for twice as much flour as sugar (in cups), that the total volume of the dry ingredients should be $$6$$ cups, and that it required a total of $$3$$ eggs. However, she only has $$2$$ eggs. How much of each ingredients should she use?

Mary K.

If we let $$f$$ represent the number of cups of flour Patty needs and $$s$$ represent the number of cups of sugar, then $$f=2*s$$. (To verify this, notice that $$1$$ cup of sugar results in $$2$$ cups of flour.). We also know that $$f+s=6$$. Because we have two unknown values and two equations, we can solve for both. Solving the second for $$f$$ gives us $$f=6-s$$. Next we substitute $$f$$ into our first equation to get $$6-s=2*s$$. Adding $$s$$ to both sides gives us $$6=3s$$. Divide both sides by $$3$$ to get $$2=s$$. Plugging that back into our original equation, we get $$f=6-2=4$$. So the original recipe called for $$4$$ cups of flour and $$2$$ cups of sugar for a total of $$6$$ cups of dry ingredients. Then we need to figure out how to scale the recipe. Since the original required $$3$$ eggs and Patty only has $$2$$, we can say $$3x=2$$ where $$x$$ is the ratio used to convert the original amounts to the scaled amounts. Dividing by $$3$$ gives $$x=\frac{2}{3}$$. Finally, we multiply $$2*\frac{2}{3}=\frac{4}{3}$$ or $$1\frac{1}{3}$$ for the amount of sugar, and $$4*\frac{2}{3}=\frac{8}{3}$$ or $$2\frac{2}{3}$$ for the amount of flour. Our final recipe is $$2$$ eggs, $$1\frac{1}{3}$$ cups of sugar, and $$2\frac{2}{3}$$ cups of flour.

Send a message explaining your
needs and Mary will reply soon.
Contact Mary