Answer:
Step-by-step explanation:
a.) the maximum is just the vertex
-35/(2*-4.9)= it takes 3.5714 seconds to hit the maximum
which in height is
-4.9*3.5714^2+35*3.5714 +1= 63.5 metres
b.) for this just plug in 2
-4.9*2^2+35*2+1= 51.4 metres
c.) just plug in f(0) =1 meter
d.) find the x intercepts (using the quadratic formula)
seconds
(there is another answer but it doesn't make sense in the context of this question)
Answer:
C
Step-by-step explanation:
x^2/12.25 + y^2/4 = 1
Let's work on the left side first. And remember that
the<u> tangent</u> is the same as <u>sin/cos</u>.
sin(a) cos(a) tan(a)
Substitute for the tangent:
[ sin(a) cos(a) ] [ sin(a)/cos(a) ]
Cancel the cos(a) from the top and bottom, and you're left with
[ sin(a) ] . . . . . [ sin(a) ] which is [ <u>sin²(a)</u> ] That's the <u>left side</u>.
Now, work on the right side:
[ 1 - cos(a) ] [ 1 + cos(a) ]
Multiply that all out, using FOIL:
[ 1 + cos(a) - cos(a) - cos²(a) ]
= [ <u>1 - cos²(a)</u> ] That's the <u>right side</u>.
Do you remember that for any angle, sin²(b) + cos²(b) = 1 ?
Subtract cos²(b) from each side, and you have sin²(b) = 1 - cos²(b) for any angle.
So, on the <u>right side</u>, you could write [ <u>sin²(a)</u> ] .
Now look back about 9 lines, and compare that to the result we got for the <u>left side</u> .
They look quite similar. In fact, they're identical. And so the identity is proven.
Whew !
854 it’s the only one I can think of
