Find the limit multiple choice
Find the limit: limx->0[(√x+4)-2]/xa. 0b.1/9c.1/4d.1/6
10/03/20
precal question
Solve the logarithmic equation for x, as in Example 7. (Enter your answers as a comma-separated list.)2 log(x) = log(2) + log(3x − 4)
fixed the problem becaise i submitted it wrongf beforehand
The point P(1/2,10) lies on the curve y=5x. If is the point (x,5/x), find the slope of the secant line PQ for the values below of .If x=0.6, the slope of PQ is: and if x=0.51, the slope...
more
10/01/20
Calculus Question
Given that g(5) = −6 and g'(5) = 3 find f '(5), if possible. (If not possible, enter IMPOSSIBLE.) f(x) = [g(x)]3
09/30/20
Use the Binomial Theorem to expand and simplify the expression.
Use the Binomial Theorem to expand and simplify the expression (x2 + y2)3
09/26/20
Find the value of k so that the differential equation is exact.
(2xy2 + yex)dx + (2x2y + kex − 1)dy = 0
09/24/20
linear function
Find the parameters a and b included in the linear function f(x) = a x + b so that f -1 (2) = 3 and f -1 (-3) = 6, where f -1 (x) is the inverse of function f.
09/24/20
Consider the function for f (x)
Consider the function for f (x) = 9/(x-6) for x > 6 .(A) Find f^-1(7) = ?B) Use Theorem 7, page 156 of the Stewart Essential Calculus textbook to find (f^-1) (7) (f^-1)' (7) = ?
09/23/20
naming new points in transformations of linear equations
Use the transformation to identify the new point.g(x) = f(x-8) + 5original point: (-4, 12)answer key says new point is (4, 17)I think new point is (-4, -7)Please help!
09/22/20
Peter the Plumber charges his customers a fee of $40 to look at the problem and then $60 per hour to fix the problem. What's the independent and dependent variable? What's the function rule?
Instruction: Find the independent and dependent variable, then write a function rule using the problem.Independent variable:Dependent variable:Function Rule:
Still looking for help? Get the right answer, fast.
Ask a question for free
Get a free answer to a quick problem.
Most questions answered within 4 hours.
OR
Find an Online Tutor Now
Choose an expert and meet online. No packages or subscriptions, pay only for the time you need.