If ∆JKL ≅ ∆PQR, which one of these is not a pair of corresponding parts? ∠J and ∠P and ∠L and ∠R and
1 answer:
You might be interested in
Answer:
B.
Step-by-step explanation:
Option B is the correct answer
The slope of this line is: m = 3/4
Answer:
See the explanation for the answer.
Step-by-step explanation:
Given function:
The n-th order Taylor polynomial for function f with its center at a is:
As n = 3 So,
= 3 + 0.0092592593 (x - 81) + 1/2 ( - 0.000085733882) (x - 81)² + 1/6
(0.0000018522752) (x-81)³
= 0.0092592593 x - 0.000042866941 (x - 81)² + 0.00000030871254
(x-81)³ + 2.25
Hence approximation at given quantity i.e.
x = 94
Putting x = 94
= 0.0092592593 (94) - 0.000042866941 (94 - 81)² +
0.00000030871254 (94-81)³ + 2.25
= 0.87037 03742 - 0.000042866941 (13)² + 0.00000030871254(13)³ +
2.25
= 0.87037 03742 - 0.000042866941 (169) +
0.00000030871254(2197) + 2.25
= 0.87037 03742 - 0.007244513029 + 0.0006782414503 + 2.25
= 3.113804102621
Compute the absolute error in the approximation assuming the exact value is given by a calculator.
Compute as
using calculator
Exact value:
(94) = 3.113737258478
Compute absolute error:
Err = | 3.113804102621 - 3.113737258478 |
Err (94) = 0.000066844143
If you round off the values then you get error as:
|3.11380 - 3.113737| = 0.000063
Err (94) = 0.000063
If you round off the values up to 4 decimal places then you get error as:
|3.1138 - 3.1137| = 0.0001
Err (94) = 0.0001
Differentiate both sides of the equation of the circle with respect to , treating
as a function of
:
This gives the slope of any line tangent to the circle at the point .
Rewriting the given line in slope-intercept form tells us its slope is
In order for this line to be tangent to the circle, it must intersect the circle at the point such that
In the equation of the circle, we have
If , then
, so we omit this case.
If , then
, as expected. Therefore
is a tangent line to the circle
at the point (1, -2).