Aclimber on mt everest is 6000 meters from the start of his trail and at elevation 8000 meters above sea level. at x meters from the start, the elevation is h(x) meters above sea level. if h′(x)=0.6 for x near 6000 meters, what is the approximate elevation another 5 meters along the trail? chapter 2, section 2.3, supplementary question 001a since the derivative tells how fast a function is changing, use the derivative at a point to estimate values of the function at nearby points. for local linear approximation use δh=h′(x)δx. what are the values for δx and h′(x)? δx= meters and h′(x)= . click if you would like to show work for this question: open show work

h'(x) = 0.6 and Δx = 1 and 2

h(6001) = h(6000) + h'(6000)(1) = 8000 +0.6 = 8,000.6

h(6002) = h(6000) + h'(6000)(2) = 8000 + 0.6 (2) =8000 +1.2 = 8001.2

Explanation below:

Step-by-step explanation:

Based on the information given we have that when the climber is 6000 meters from the start, his elevation is 8000 meters above sea level.

We know that the function for elevation is given by h(x) where x is the number of meters from the start.

Therefore, h(6000) = 8000.

We also know that the derivative of a function tells us the reason of change at some point (or how fast a function is changing near that point). For this problem we have that h'(x) = 0.6. (meaning that the elevation increases in 0.6 meters for every meter they hike further nearby the 6000 meters)

We need to use the derivative at a point to estimate values of the function at nearby points

We're going to use h'(x)= 0.6. Δx= 1

h(6001) = h(6000) + h'(6000)(1) = 8000 +0.6 = 8,000.6

Therefore h (6001) = 8,000.6

Now for h'(x)= 0.6. Δx= 1

h(6002) = h(6000) + h'(6000)(2) = 8000 + 0.6 (2) =8000 +1.2 = 8001.2

Therefore h(6002) = 8001.2

i'm sure goog can you get that answer

(23*2)-12=x     x=34

step-by-step explanation: