Answer:
Quotient = 18
Remainder = 10
Step-by-step explanation:
1234/68
=> 68 x 1 = 68
=> 123 - 68 = 55
=> Take the 4 down
=> 554/68
=> 68 x 8 = 544
=> 554 - 544 = 10
So, the quotient = 18.
Remainder = 10
27/8
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Answer: Choice A
y = -3(x+2)^2 + 10
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Work Shown:
y = -3x^2-12x-2 is in the form y = ax^2+bx+c with
a = -3
b = -12
c = -2
The x coordinate of the vertex is
h = -b/(2a)
h = -(-12)/(2*(-3))
h = 12/(-6)
h = -2
We'll plug this into the original equation to find the corresponding y coordinate of the vertex.
y = -3x^2-12x-2
y = -3(-2)^2-12(-2)-2
y = 10
So k = 10 is the y coordinate of the vertex.
Overall, the vertex is (h,k) = (-2,10)
Meaning that we go from this general vertex form
y = a(x-h)^2 + k
to this
y = -3(x - (-2))^2 + 10
y = -3(x+2)^2 + 10
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.
Answer:
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Step-by-step explanation:
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