Enable contrast version

Tutor profile: Partha Sarathi Reddy K.

Inactive
Partha Sarathi Reddy K.
Tutor for nine years.
Tutor Satisfaction Guarantee

Questions

Subject: Pre-Calculus

TutorMe
Question:

Find the vertical and horizontal asymptotes of the rational function f(x)=(2x^2-5)/(x^2-9)

Inactive
Partha Sarathi Reddy K.
Answer:

Vertical asymptotes: To find the vertical asymptotes, we equate the denominator to 0 and solve for x. x^2-9=0 We factor it by using the difference of square formula a^2-b^2=(a+b)(a-b) So, x^2-3^2=0 (x+3)(x-3)=0 Equate each term to 0 and solve for x. x+3=0.........so, x=-3 x-3=0..........so, x=3 The vertical asymptotes are x=-3 and x=3. Horizontal asymptotes: To find the horizontal asymptotes, we look at the degree(exponent of the variable) of the numerator and the denominator. If the degree of the numerator and the denominator are equal, then we just divide the co-efficients of the highest degree term to get the horizontal asymptote. Degree of the numerator 2x^2-5 is 2. Degree of the denominator x^2-9 is 2. As the degree of the numerator and denominator is same, we divide the co-efficient of the highest degree term. So, y=2/1=2 y=2 is the horizontal asymptote.

Subject: Trigonometry

TutorMe
Question:

Find the amplitude, period and phase shift of the function f(x)=2sin(3(x-pi/3))

Inactive
Partha Sarathi Reddy K.
Answer:

First we start by comparing given function with the general sine function f(x)=Asin(B(x-C))+D where A=amplitude Period=(2pi)/B C is the phase shift and D is the vertical shift Comparing the given function f(x)=2sin(3(x-pi/3)) with the general sine function, we get Amplitude A=2 Period=(2pi)/B=(2pi)/3 Phase shift C=pi/3

Subject: Algebra

TutorMe
Question:

The admission fee at a fair is $3 for children and $5 for adults. On a certain day, 2200 people enter the fair. The fair made $8400 from admission tickets that day. How many children and adults bought a ticket?

Inactive
Partha Sarathi Reddy K.
Answer:

First we start by defining the variables for the number of tickets bought by the children and adults. Let number of tickets bought by the children be "x" and number of tickets bought by the adults be "y". Given that, 2200 people attended the fair. So, x+y=2200 Subtract x on both sides to get y by itself y=2000-x...........equation(1) The admission fee is $3 for children, $5 for adults and the fair made $8400 from the tickets. So, 3x+5y=8400......equation(2) Now substitute y=2000-x in the second equation 3x+5(2200-x)=8400 Distribute 5 to 2200-x term. 3x+10000-5x=8400 Combine like terms (3x-5x)+11,000=8400 -2x+11000=8400 Subtract 11,000 on both sides -2x=8400-11000 -2x=-2600 Divide by -2 on both sides x=1300 Now substitute the value of x in the first equation y=2200-1300 y=900 So, 1300 children and 900 adults attended the fair.

Contact tutor

Send a message explaining your
needs and Partha Sarathi Reddy will reply soon.
Contact Partha Sarathi Reddy

Request lesson

Ready now? Request a lesson.
Start Lesson

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 Lesson" 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.