What is the supplement of 164 degrees?
2 answers:
You might be interested in
Set up the given triangle on x-y coordinates with right angle at (0,0). So the two vertices are at (5,0) and (0,2![sqrt{x} n]{3}](https://tex.z-dn.net/?f=sqrt%7Bx%7D%20n%5D%7B3%7D%20)
)
let (a,0) and (0,b) be two vertices of the<span> equilateral triangle. So the third vertex must be at </span>
for a pt (x,y) on line sx+ty=1, the minimum of
equals to
smallest value happens at
so area is
hence m=75, n=67, p=3
m+n+p = 75+67+3 = 145
let (a,0) and (0,b) be two vertices of the<span> equilateral triangle. So the third vertex must be at </span>
for a pt (x,y) on line sx+ty=1, the minimum of
equals to
smallest value happens at
so area is
hence m=75, n=67, p=3
m+n+p = 75+67+3 = 145
Answer:
- g(2.95) ≈ -1.8; g(3.05) ≈ -0.2
- A) tangents are increasing in slope, so the tangent is below the curve, and estimates are too small.
Step-by-step explanation:
(a) The linear approximation of g(x) at x=b will be ...
g(x) ≈ g'(b)(x -b) +g(b)
Using the given relations, this is ...
g'(3) = 3² +7 = 16
g(x) ≈ 16(x -3) -1
Then the points of interest are ...
g(2.95) ≈ 16(2.95 -3) -1 = -1.8
g(3.05) ≈ 16(3.05 -3) -1 = -0.2
__
(b) At x=3, the slope of the curve is increasing, so the tangent lies below the curve. The estimates are too small. (Matches description A.)
