What is the approximate area of a circle with a radius of 6 m?
2 answers:
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Answer:
Step-by-step explanation:
F(x) = x² - 2x + 1
= (x - 1)²
By comparing this equation with the vertex form of the quadratic equation,
y = (x - h)² + k
Here, (h, k) is the vertex
Vertex of the parabola → (1, 0)
x-intercepts → (x - 1)² = 0
x = 1
y-intercepts → y = (0 - 1)²
y = 1
Now we can draw the graph of the given function,
From this graph,
As x → 0,
f(0) = (0 - 1)²
= 1
Since,
Therefore, given function is continuous at x = 0.
Answer:
Step-by-step explanation:
2 cos x + √ 2 = 0
2 cos x = -√ 2
cos x = -√ 2 / 2
x = arcCos( -√ 2 / 2 )
so to solve we have to use "co-terminal " angles .. do you know what I'm saying? do you understand the words coming out of my mouth :DDDDD OKay back to math and not movie lines .. :P
x = arcCos( √ 2 / 2 )
x = 45 °
now find the "co terminal" angle that is on 45 ° but in the correct quadrant... since the -√ 2 is negative.. we now that we go down the y axis.. but also positive on the x axis.. soooo.. that put the angle in the 4th quadrant... so this is an angle of 315° if we go in the CCW ( counter clock wise ) direction but it's also -45° in the CW (clock wise ) direction
below is the table to remember the trig special angles
notice how it's 1,2,3,4 .. so it's super easy to remember.. the trig books don't show you this "trick" :P
copy and paste this to your computer some where handy
Sin(0) = 0/2 =0
Sin(30)= /2 = 1/2
Sing(45) =/2 =
/2
Sin(60)=/2 =
/2
Sin(90)=/2 = 1
Cos is exactly the same but counts backwards from 90°
Cos(90) = 0/2 = 0
Cos(60) = /2 = 1/2
Cos(45) = /2 =
/2
Cos(30) = /2 =
/2
Cos(0) = /2 = 1