Recall the double angle identity for cosine:
It follows that
Since 0° < 22° < 90°, we know that sin(22°) must be positive, so csc(22°) is also positive. Let x = 22°; then the closest answer would be C,
but the problem is that none of these claims are true; cot(32°) ≠ 4/3, cos(44°) ≠ 5/13, and csc(22°) ≠ √13/2...
Move constant to other side
add 1
4c^2-8c=1
divide by 4 to make leading coeficient 1
c^2-2c=1/4
take 1/2 of linear coeficient and square it
-2/2=-1, (-1)^2=1
add that to both sides
c^2-2c+1=1/4+1
factor perfect squaer and add
(c-1)^2=5/4
square root both sides
c-1=+/-(√5)/2
add 1
c=1+/-(√5)/2
c=2.12 or -0.12
Answer:
Step-by-step explanation:
The minimum is at the vertex of this parabola because it opens up.
Now if (-4, 2) is the minimum then all the y values on the parabola must be > 2,
But we are given that y = -3 is on the graph ( the point (6,-3) - that is y < 2 here,
Therefore (-4, 2) cannot be the vertex .
Answer:
I play football and my coach will do this: Play a "prevent* defense that guards against long gains but makes short
gains easier.
Step-by-step explanation:
hey, I'm not sure about this answer, but I think it's true, sorry
m(2g+5mk)
