1.
the x value of the vertex in form
ax^2+bx+c=y
is
-b/2a
so
-2x^2+8x-18
x value of vertex is
-8/(2*-2)=-8/-4=2
plug it in to get y value
-2(2)^2+8(2)-18
-2(4)+16-18
-8-2
-10
vertex is at (2,-10)
or you could complete the square to get into y=a(x-h)^2+k, where the vertex is (h,k)
so as follows
y=(-2x^2+8x)-18
y=-2(x^2-4x)-18
y=-2(x^2-4x+4-4)-18
y=-2((x-2)^2-4)-18
y=-2(x-2)^2+8-18
y=-2(x-2)^2-10
vertex is (2,-10)
5.
vertex is the time where the speed is the highest
at about t=10, the speed is at its max
Answer: 2^14
Step-by-step explanation:
Answer:
The answer is 88.
Step-by-step explanation:
68+81+74+89+x/5=80
312+x/5=80
312+x=80*5
312+x=400
x=400-312
x=88
T=-1
sinA=sin(π/2-3A), A=2nπ+π/2-3A, 4A=2nπ+π/2, A=nπ/2+π/8 where n is an integer.
Also, π-A=2nπ+π/2-3A, 2A=2nπ-π/2, A=nπ-π/4.
The hard way:
cos3A=cos(2A+A)=cos(2A)cosA-sin(2A)sinA.
Let s=sinA and c=cosA, then s²+c²=1.
cos3A=(2c²-1)c-2c(1-c²)=c(4c²-3).
s=c(4c²-3) is the original equation.
Let t=tanA=s/c, then c²=1/(1+t²).
t=4c²-3=4/(1+t²)-3=(4-3-3t²)/(1+t²)=(1-3t²)/(1+t²).
So t+t³=1-3t², t³+3t²+t-1=0=(t+1)(t²+2t-1).
So t=-1 is a solution.
t²+2t-1=0 is a solution, t²+2t+1-1-1=0=(t+1)²-2, so t=-1+√2 and t=-1-√2 are solutions.
Therefore tanA=-1, -1+√2, -1-√2 are the three solutions from which:
A=-π/4, π/8, -3π/8 radians and these values +2πn where n is an integer.
Replacing π by 180° converts the solutions to degrees.
