Enable contrast version

Tutor profile: Jeenal C.

Inactive
Jeenal C.
Inofrmation Technology (IT) Engineer
Tutor Satisfaction Guarantee

Questions

Subject: Trigonometry

TutorMe
Question:

If Sin(theta) = 1 , find cos(theta) and tan (theta).

Inactive
Jeenal C.
Answer:

Step 1 : Read the problem and gather the given information sin (theta) = 1 Step 2 : Determine the unknowns cos (theta) and tan (theta) Step 3: Develop equations ******************************************************************************************************************** We know the relation between sine and cosine, [sin (theta) ]^2 + [cos (theta) ] ^2 =1 Rearranging the equation to shift unknown terms on the left-hand side. [cos (theta) ] ^2 = 1 - [sin (theta) ]^2 [cos (theta) ]^2 = 1 - [1]^2 ..................................................... { Given information } cos(theta) = 0 ........................................................................ { Result 1 } ******************************************************************************************************************** Relation between sine, cosine, and tan tan (theta) = sin(theta) / cos (theta) tan (theta) = 1 / 0 ......................................................................{ From given information and Result 1 } Any number divided by zero is undefined. tan (theta) = undefined .......................................................{ Result 2}

Subject: Statistics

TutorMe
Question:

The arithmetic mean of the 5 consecutive integers starting with 's' is 'a'. What is the arithmetic mean of 9 consecutive integers that start with s + 2 ?

Inactive
Jeenal C.
Answer:

Step 1 - Read the problem and fetch the given information The arithmetic mean of 5 consecutive integers starting with 's' is 'a' Step 2 : Assigning variables M1 = Mean of 5 consecuitve integers starting with 's' = a M2 = Mean of 9 consecutive integers starting with 's+2' (To be determined) For the problem at hand, Mean = (Sum of consecutive integers) / (Number of consecuitve integers) Therefore- **************************************************************************************************************** M1 =[ (s) + (s+1) + (s+2) + (s+3) + (s+4) ] / 5 a = [ 5s + 10 ] / 5 We will now try to find the value of 's' in terms of known value 'a'. So rearranging the equation. a = s + 2 s = a -2 .......................................................................{ Result 1 } ***************************************************************************************************************** Now, determining the unknown. M2 =[ (s+2) + (s+3) + (s+4) + (s+5) + (s+6) + (s+7) + (s+8) +(s+9) + (s+10) ] / 9 M2 = [ 9s + 54 ] / 9 M2 = s + 6 M2 = ( a - 2) + 6 ....................................................... { Using value from Result 1 } M2 = a + 4 = Mean of 9 consecutive integers starting with 's+2' ...... { Result 2 }

Subject: Algebra

TutorMe
Question:

The braking distance for a car traveling at 70 miles per hour is 5 feet less than twice the braking distance at 50 miles per hour. If the braking distance is 381 feet at 70 miles per hour, what is the braking distance at 50 miles per hour?

Inactive
Jeenal C.
Answer:

This is a very good example of a word problem. Solving these type of questions involve steps such as reading the problem, determining the knowns and unknowns of the question, and generating equations that relate them. Step 1 - Assigning variables d1= braking distance at 70 miles/hours = 381 feet d2=braking distance at 50 miles/hour ( To be determined) Step 2 - Creating equations Braking distance @ 70 miles/hr = twice the braking distance at 50 miles/hr - 5 feet less d1 = 2 * d2 - 5 Step 3 - Solving the equations Rearranging the equation to shift the unknown variable on the left-hand side of the equation. d1 = 2 * d2 - 5 2*d2 = d1 + 5 d2 = (d1+5)/2 d2 = (381+5)/2 ...................................... { Using value from step 1 } d2 = 193 feet = stopping distance at 50 miles/hr.

Contact tutor

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

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.
BEST IN CLASS SINCE 2015
TutorMe homepage
Made in California by Zovio
© 2013 - 2022 TutorMe, LLC
High Contrast Mode
On
Off