Answer:
j
Step-by-step explanation:
Answer:
0.83333333
Step-by-step explanation:
The 3 is repeating
1. solutions are intersection points
they intersect at x=4 and x=1 or the points (4,0) and (1,0)
select 4 and 1
2.
sub -x for x
if you get the same function, it is even
if you get the exact negative of the function, it is odd
if neither, then neither
basically
f(-x)=f(x) is even
f(-x)=-f(x) is odd
and neither is neighter
so
f(-x)=-2(-x)²+3(-x)
f(-x)=-2(x)²-3x
f(-x)=-2x²-3x
that is not the same function nor is it the exact oposite
neither even nor odd
3.
domain is the time you can use
obvioulsy start at time=0
and stop when it hits the ground because it shouldn't go underground
so domain is all real numbers from 0 to 5 including 0 and 5
that is [0,5]
4. (the (f+g)(x) one)
(f+g)(x)=f(x)+g(x)
so that is just
x²-36+x³+2x²-10=
x³+3x²-46
5. just mulitply them
x⁷+9x⁴-9x³-81
6. minus them
(3x⁵+6x²-5)-(2x⁴+7x²-x+16)
3x⁵+6x²-5-2x⁴-7x²+x-16
3x⁵-2x⁴-x²+x-21
I’m assuming ‘dives further’ means to go directly down
the angle of elevation of the ship from the submarine is equal to the angle of depression of the submarine from the ship, if we assume the sea level is perpendicular to ‘directly down’.
let both of these angles to be = $ when the submarine is at A and ¥ when the submarine is at B (excuse the lack of easily accessible variables as keys)
then this become a simple trig problem:
A)
Let O be the position of of the ship, and C be the original position of the submarine.
therefore, not considering direction
|OC| = 1.78km = 1780m
|CA| = 45m
these are the adjacent and opposite sides of a right angled triangle.
But tan($) = opp/adj = |CA|/|OC| = 45/1780
therefore $ = arctan(45/1780) which is roughly 1.45 degrees,
B)
similarly, noting that |CB| = |CA| + |AB| = 45 + 62 = 107m
tan(¥) = 107/1780
¥ = arctan(107/1780) which is roughly 3.44 degrees