<span>if we take square root (x - 3)^2 + y^2
y = x^2
(x - 3)^2 + x^4
f(x) = x^4 + x^2 - 6x + 9.
first derivation:
f'(x) = 4x^3 + 2x - 6.
4x^3 + 2x - 6 = 0.
one solution is x = 1
second derivation
f"(x) = 12x^2 + 2
it is positive for x = 1
hope it helps
</span>
So you can do this multiple ways, I'll do this the way that I think makes sense the l most easily.
Cos (0) = 1
Cos (pi/2)=0
Cos (pi) =-1
Cos (3pi/2)=0
Cos (2pi)=1
Now if you multiply the inside by 4, the graph oscillates more violently (goes up and down more in a shorter period).
But you can always reduce it.
Cos (0)= 1
Cos (4pi/2) = cos (2pi)=1
Cos (4pi) =Cos (2pi) =1 (Any multiple of 2pi ==1)
etc...
the pattern is that every half pi increase is now a full period as apposed to just a quarter of one. That's in theory.
Now that you know that, the identities of Cosine are another beast, but mathematically.
You have.
Cos (2×2t) = Cos^2 (2t)-Sin^2 (2t)
Sin^2 (t)=-Cos^2 (t)+1..... (all A^2+B^2=C^2)
Cos (2×2t) = Cos^2 (t)-(-Cos^2 (t)+1)
Cos (2×2t)= 2Cos^2 (2t) - 1
2Cos^2 (2t) -1= 2 (Cos^2(t)-Sin^2(t))^2 -1
(same thing as above but done twice because it's cos ^2 now)
convert sin^2
2Cos^2 (2t)-1 =2 (Cos^2 (t)+Cos^2 (t)-1)^2 -1
2 (2Cos^2(t)-1)^2 -1
2 (2Cos^2 (t)-1)(2Cos^2 (t)-1)-1
2 (4Cos^4 (t) - 2 (2Cos^2 (t))+1)-1
Distribute
8Cos^4 (t) -8Cos^2 (t) +1
Cos (4t) =8Cos^4-8Cos^2 (t)+-1
If it took 4 miles per hour you multiply that twice cuss its traveling back then forth so you repeat the same process 30 miles times 2 which is 60 hope this helps.
Answer:
The equation of the straight line is x - 2y +6 =0
Step-by-step explanation:
<u>Explanation:</u>-
Given a point ( 2, 4) and slope m = 
The equation of the straight line passing through the point and having slope 'm'
y - y₁ = m ( x - x₁)
y - 4 = ( x - 2)
2( y -4) = ( x-2)
2 y - 8 = x -2
x - 2 y - 2 + 8 =0
x - 2y + 6 =0
The equation of the straight line is x - 2y +6 =0
Answer:
Step-by-step explanation:
