TutorMe homepage
Subjects
PRICING
COURSES
Start Free Trial
Linda C.
Math Tutor from Middle School Math to College Calculus and GRE Math
Tutor Satisfaction Guarantee
Trigonometry
TutorMe
Question:

What quadrant contains the terminal side of the angle 10pi/3?

Linda C.

Calculus
TutorMe
Question:

Find the slop of the line tangent to y(X) when X is equal to 6. y(X)=X^4-X^3+4X^2

Linda C.

The slop of the line tangent to a function f(x) at point x is actually the derivative of the function at that point. For example, the slope of the line tangent to f(X)=2X is the derivative of f(X), df(X)/dX=2. This means function f(X)=2X has a constant slope 2, no matter the value of x. Another example, the slope of the line tangent to f(X)=X^2 is the derivative of f(X), d(X^2)/dx=2X. When x is 6, the slope is 12. So the slope of the line tangent to y(X)=x^4-X^3+4X^2 is d(X^4-X^3+4X^2)/dx=4X^3-3X^2+8X When x is 6, plug 6 into the slope, which is: 4*6^3-3*6^2+8*6=804

Algebra
TutorMe
Question:

If x is the average (arithmetic mean) of 2m and 7, y is the average of 4m and 19, and z is the average of 6m and 22, what is the average of x, y, and z in terms of m?

Linda C.

First, let's break down the problem one segment of sentence by one. 1)" If x is the average (arithmetic mean) of 2m and 7" In mathematic terms, this sentence can be written as: x=(2m+7)/2 2) "y is the average of 4m and 19" In mathematic terms, this sentence can be written as: y=(4m+19)/2 3) "z is the average of 6m and 22" In mathematic terms, this sentence can be written as: z=(6m+22)/2 The question asks "what is the average of x, y, and z in terms of m?" So what is the average of x, y, and z written in mathematic terms? It should be (x+y+z)/3 We know x, y, and z, as x=(2m+7)/2, y=(4m+19)/2 and z=(6m+22)/2. Therefore, (x+y+z)/3={(2m+7)/2+(4m+19)/2 +(6m+22)/2}/3={(12m+48)/2}/3=2m+8. Since (x+y+z)/3 is the average of x, y, and z, and it equals to 2m+8 Therefore, the average of x, y, and z in terms of m is 2m+8

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