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anonymous
 4 years ago
The point P(16 , 9 ) lies on the curve y = \sqrt{ x } + 5. Let Q be the point (x, \sqrt{ x }+ 5 ).
a.) Find the slope of the secant line PQ for the following values of x. (Answers here should be correct to at least 6 places after the decimal point.)
If x= 16.1, the slope of PQ is: ?
If x= 16.01, the slope of PQ is:?
If x= 15.9, the slope of PQ is:?
If x= 15.99, the slope of PQ is:?
b.) Based on the above results, estimate the slope of the tangent line to the curve at P(16 , 9 ).
Answer: ??
anonymous
 4 years ago
The point P(16 , 9 ) lies on the curve y = \sqrt{ x } + 5. Let Q be the point (x, \sqrt{ x }+ 5 ). a.) Find the slope of the secant line PQ for the following values of x. (Answers here should be correct to at least 6 places after the decimal point.) If x= 16.1, the slope of PQ is: ? If x= 16.01, the slope of PQ is:? If x= 15.9, the slope of PQ is:? If x= 15.99, the slope of PQ is:? b.) Based on the above results, estimate the slope of the tangent line to the curve at P(16 , 9 ). Answer: ??

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0And then (x, \sqrt{ x }+ 5 ) is dw:1328845351408:dw

Mertsj
 4 years ago
Best ResponseYou've already chosen the best response.1slope at x= 16.9= .1248053 slope at x= 16.01=.1249805 slope at x = 15.9 = .12519593 slope at x = 15.99 = .1250195

Mertsj
 4 years ago
Best ResponseYou've already chosen the best response.1Well based on looking at those slopes I would expect it to be between .1249 and .1251 so I guess my estimate would be .1250 What would yours be?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yeah, not quite mine is .1255, but i believe yours is right!
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