TutorMe homepage

SIGN IN

Start Free Trial

Terry M.

Electrical Engineering PhD Student with a love of math

Tutor Satisfaction Guarantee

Basic Math

TutorMe

Question:

Add the two fractions $$5/14$$ and $$7/6$$.

Terry M.

Answer:

To add fractions first you must find the equivalent fractions that have a common denominator. One easy way to do this would be to make the denominator of both fractions the product of the two fractions. So for this example we would want the denominator to be $$14\times6=84$$. But there is another number that is smaller that will give us easier numbers to multiply by. The number $$42$$ has factors 14 and 6 and thus is a lower common denominator than 84. So $$42=14\times3$$ and $$42= 6\times7$$ thus we want to multiply the fraction $$5/14$$ by $$5/5$$ and $$7/6$$ by $$7/7$$: $$\frac{5}{14}\times\frac{3}{3}=\frac{15}{42}$$ $$\frac{7}{6}\times\frac{7}{7}=\frac{49}{42}$$ Now we can add the two fractions with the common denominator: $$\frac{15}{42}+\frac{49}{42}=\frac{64}{42}$$. Now finally we can simplify the fraction by dividing both top and bottom by 2 $$\frac{64}{42}\rightarrow\frac{32}{21}$$.

Calculus

TutorMe

Question:

What is the average value of the function $$\frac{1}{x^2+1}$$ from $$x=0$$ to $$x=1$$?

Terry M.

Answer:

The average value formula for a function $$f(x)$$ on the interval $$x=a$$ to $$x=b$$ is $$\frac{1}{b-a}\int_a^b f(x)dx$$. So to solve the problem we will use $$\frac{1}{1-0}\int_0^1\frac{1}{x^2+1}dx = tan^{-1}(x) |^1_0= tan^{-1}(1)-tan^{-1}(0)=\frac{\pi}{4}$$

Algebra

TutorMe

Question:

A concert venue charges $12.00 per ticket for children and $15.00 per ticket for adults for the JayZ concert. The venue makes $330,660 in ticket sales and 22,304 people attended. How many adults and how many children went to the JayZ concert?

Terry M.

Answer:

Because the child ticket is worth $12.00 the concert makes $$12c$$ child tickets where $$c$$ is the number children. Similarly, the venue makes $$15a$$ dollars off adult ticket sales, where $$a$$ is the number of adults. So the we get the following system of equations: $$a+c=22,304$$ $$15a+12c=330660$$ Now we can use the substitution method to solve the system of equations. First, solve one equation for one variable: $$a+c=22,304 \rightarrow a = 22,304-c$$ Then we plug the equation for $$a$$ into the remaining equation: $$15a+12c=330,660 \rightarrow 15(22,304-c)+12c = 330,660$$ Now we can solve for $$c$$. First, simplify the equation: $$15(22,304-c)+12c = 330,660 \rightarrow 334560-15c+12c = 330660$$ Then combine like terms: $$334560-3c =330660$$ Then we can subtract 334560 from both sides to get only the term with the $$c$$ by itself. $$334560-3c-334,560=330660-334,560 \rightarrow -3c=-3900$$ Then we can divide both sides by $$-15$$ to get $$c$$ by itself: $$\frac{-3c}{-3} =\frac{-3900}{3} \rightarrow c=1300$$ Now we can plug $$c=1300$$ back into the first equation to find $$a$$. $$a+c=22304 \rightarrow a+1300=22304$$ And finally we can solve for $$a$$ by subtracting 1300 from both sides: $$a+1300=22304 \rightarrow a=22304-1300=21000$$ Thus our answer should be there were 21,000 adults and 1,300 children at the concert. We can do one last check to make sure everything is true: $$21000+1300=22304 $$ is true $$15(21000)+12(1300)=330660$$ is true.

Send a message explaining your

needs and Terry will reply soon.

needs and Terry will reply soon.

Contact Terry

Ready now? Request a lesson.

Start Session

FAQs

What is a lesson?

A lesson is virtual lesson space on our platform where you and a tutor can communicate.
You'll have the option to communicate using video/audio as well as text chat.
You can also upload documents, edit papers in real time and use our cutting-edge virtual whiteboard.

How do I begin a lesson?

If the tutor is currently online, you can click the "Start Session" button above.
If they are offline, you can always send them a message to schedule a lesson.

Who are TutorMe tutors?

Many of our tutors are current college students or recent graduates of top-tier universities
like MIT, Harvard and USC.
TutorMe has thousands of top-quality tutors available to work with you.

Made in California

© 2019 TutorMe.com, Inc.