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Avary K.
Tutor for 7 years. Graduate Teaching Assistant for 4 years.
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Basic Math
TutorMe
Question:

During a back to school sale an instrument store is giving a 15% is discount for students. Julie, a student at the local high school, purchases a flute for $306. How much did it originally cost?

Avary K.
Answer:

To start any math problem, your first step should be to write down what we know and what we want to know: back to school sale- 15% paid for flute- $306 original cost- ?? Now, since we don't know what our original cost of the flute is, let that be $$x$$ original cost=$$x$$ so our goal is to find $$x$$. To do so, we are going to set up an equation. This equation can be used for problems similar to this. It reads: (original cost)-(%)(original cost)=final cost or in words "the original price minus the percent discount (as a decimal) times the original price is equal to the final price". In terms of our problem, the back to school discount is a total of 15% (or 0.15 as a decimal), the final cost is $306, and $$x$$ represents the original cost which leads to our equation $$x-0.15x=306$$ now we solve for $$x$$. Notice that on the left side of the equation, both terms have $$x$$ so we can combine them: $$x-0.15x = 1x-0.15x = 0.85x$$ so our equation becomes $$0.85x=306$$ Now, since we are multiplying $$x$$ by $$0.85$$ to get $$x$$ by itself we have to divide both sides by $$0.85$$ $$\dfrac{0.85x}{0.85}=\dfrac{306}{0.85}$$ $$x=360$$ which tells us that the original price ($$x$$) of the flute was $360. It should be noted that the (original)-(%)(original)=final isn't just used for discounts or sales. It can also be used for any type of percent decrease. Additionally, if we have + (instead of -) we can use this equation for taxes, increases, etc. (where % is ALWAYS written as a decimal).

Calculus
TutorMe
Question:

Find the relative maximum and minimum of the function $$f(x)=x^3−9x^2+24x$$.

Avary K.
Answer:

Before we start solving, lets talk about the maximum and minimum of a function. If you draw any graph, notice that at the max/min the slope of the tangent line is either 0 or does not exist. For example, looking at $$x^2$$; the minimum of this function occurs at $$x=0$$ where the tangent line is a horizontal line which we know has a slope 0 (all horizontal lines have slope 0). Another example is $$|x|$$; the minimum of this function occurs at $$x=0$$ where there is a sharp corner which we know has no tangent line. Thinking about this, we know at every max/min either the slope of the tangent line is 0 or DNE. The slope of the tangent line is defined as the derivative so it would make sense, if we want to find the max/min, to first find where the derivative is 0 or DNE. Step 1) Find $$f'(x)$$. By the power rule we have $$f'(x)=3x^2-18x+24.$$ Step 2) Find the $$x$$'s that make $$f'(x)=0$$ or DNE. Lets first find the $$x$$'s that make $$f'(x)=0$$: $$0=f'(x)=3x^2-18x+24\implies 0=3(x^2-6x+8)\implies 0=3(x-2)(x-4)\implies x-2=0 \quad \text{or}\quad x-4=0\implies x=2 \quad \text{or}\quad x=4$$ so we found two values of $$x$$ that make $$f'(x)=0$$: $$x= 2,4$$. Now, to find the $$x$$'s that make $$f'(x)$$ DNE aka the values of $$x$$ where $$f'(x)$$ is discontinuous. Since $$f'(x)$$ in this example is a polynomial, it is never discontinuous, so we don't get any $$x$$ values from this. Now we need to figure out whether the $$x$$ values found in Step 2 are max or min. There are a few ways we can go about this. In this example, we are going to use the Second Derivative Test. To use this test, notice that when we have a minimum, the graph is concave up and when we have a maximum, the graph is concave down. We know that the second derivative is what tells us about concavity. Specifically if $$f''(x)>0$$ we say $$f(x)$$ is concave up; if $$f''(x)<0$$ we say $$f(x)$$ is concave down. So using this logic, we can determine whether the points found in Step 2 are max or min's by looking at whether the second derivative at those points are positive or negative. This leads up to Step 3 and 4. Step 3) Find $$f''(x)$$. By using the power rule on $$f'(x)$$ we get $$f''(x)=6x-18.$$ Step 4) Determine the sign of $$f''(x)$$ at the $$x$$ values found in Step 2. That is, evaluate: $$f''(2)=6(2)-18=12-18=-6<0$$. We can now say that at $$x=2$$, $$f'(2)=0$$ and $$f''(2)<0$$ so $$f(x)$$ has a relative maximum at $$x=2$$. Similarly $$f''(4)=6(4)-18=24-18=6>0$$. We can now say that at $$x=4$$, $$f'(4)=0$$ and $$f''(4)>0$$ so $$f(x)$$ has a relative minimum at $$x=4$$. To find the $$y$$ values of these points, plug in the $$x$$ value into the ORIGINAL function: $$f(2)=2^3−9(2^2)+24(2)=8-36+48=20$$ $$f(4)=4^3−9(4^2)+24(4)=64-144+96=16$$ To conclude, $$f(x)$$ has a relative maximum at the point $$(2,20)$$ and a relative minimum at the point $$(4,16)$$

Algebra
TutorMe
Question:

Two cyclists leave towns 128 miles apart at the same time and travel toward each other. One cyclist travels 4 mph slower than the other. If they meet in 2 hours, what is the rate of each cyclist?

Avary K.
Answer:

First lets start with what we want to know. Usually in math, the "what we want to know" will be our variable. The questions asks "what is the rate of each cyclist" so, we let x = rate (miles per hour) of cyclist 1 then since the other cyclist is traveling "4 mph slower" we know x-4 = rate (miles per hour) of cyclist 2. Now, the only other information given in the problem is the distance and time. The two cyclists start 128 miles apart and after 2 hours, they meet. So we somehow need to use the rate we have above and relate it to the distance and time. Remember the formula Distance = Rate x Time. Using this and remembering that the time is 2 hours, we can say y = distance traveled by cyclist 1 = (rate of cyclist 1) x (time) = (x)(2) = 2x then using this, if cyclist 1 traveled distance y then cyclist 2 must have traveled 128-y miles. So 128-y = distance traveled by cyclist 2 = (rate of cyclist 2) x (time) = (x-4)(2) = 2x-8 We now have a system of two equations y = 2x 128 - y = 2x-8 To solve for x (the rate of the cyclists) let's use the Substitution Method. So taking 2x (from the first equation) and substituting it in for y in the second equation we get 128 - 2x = 2x - 8 128 = 4x - 8 136 = 4x x = 34 So by this we know that x = 34 mph = rate of cyclist 1 x-4 = 30 mph = rate of cyclist 2.

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