The first derivative gives you the formula for the slope of the tangent line to the curve at a given point.
<span>Find the first derivative of the function y=5x/x-2:
(x-2)(5) - (5x)(1)
dy/dx = ------------------------
(x-2)^2
simplify the algebra here.
(3-2)(5) - (15)(1)
Then, at the point (3,15), dy/dx = -------------------------- = -10
(3-2)^2
What is the equation of the tangent line? In other words, what is the eqn of the tan line to the given curve at when x = 3, if the slope is -10?
y-15 = -10(x-3)
This can be re-written in other forms if desired.
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Method 1
65 + 20% tip = Total cost
so we need 20% of 65
0.2 times 65 = 13
65 + 13 = $78 total cost
Method 2
65 is 100%
meal cost 120% (with 20% tip)
65 Times 1.2 = 78
(1/x.y - 1/y²) ÷ (1/x².y - 1/x.y²)
(1 - x.y)/x.y ÷ (x - y)/x.y
[(1 - x.y)/x.y].[x.y/(x - y)]
(1 - x.y)/(x- y)
Answer:The answer is D I just took the test.
Step-by-step explanation: D
w = √2 (cos(π/4) + i sin(π/4)) = √2 (1/√2 + i/√2) = 1 + i
z = 2 (cos(π/2) + i sin(π/2)) = 2i
Then
w - z = (1 + i) - 2i = 1 - i
so that
|w - z| = √(1² + (-1)²) = √2
and
arg(w - z) = -π/4
In polar form, we have
w - z = √2 (cos(-π/4) + i sin(-π/4)) = √2 (cos(π/4) - i sin(π/4))
