Hello,
A: roots: -1,-3
a point (-2,1)
Vertex=((-2,1)
y=k*(x+1)(x+3) using roots
but k*(-2+1)(-2+3)=1==>k*(-1)*1=1==>k=-1
eq: y=-(x+1)(x+3)
==>y=-(x²+3x+x+3)
==>y=-x²-4x-3
y=k(x+2)²+1 if x=-1,y=0 ==>k*1+1=0==>k=-1
==>y=-(x+2)²+1
Answer :A--> R,K
B)
y=k(x+4)²-2 and k=-1/2
y=-1/2(x+4)²-2
y=-1/2x²-4x-10
answer B--> I,≈W if it is written -1/2*x² (square has been forgotten)
C:
y=2x²-16x+30
y=2(x-4)²-2
answer : C-->S,J
D:
y=-(x+3)(x+1)
y=-x²-4x-3
=-(x+2)²+1
answer D--> V,L
E:
Here there is a problem: or the graph is wrong, or 2 equations are missing!
y=1(x+1)(x-3) using roots
y=x²-2x-3 ≈ T si it were -2x and not +2x.
y=(x-1)²-4 ≈H is it were -1 in place of +1 [H:y=(x+1)²-4]
Shade 4 slices out of the pie
Answer:
1/12 (i think) or 0.083
Step-by-step explanation:
well she buys you 1 and there is 12 so it would be 1 out of 12 which is written as 1/12
Number 28 would be y= -52x+148
Answers:
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Explanation:
Recall that tangent is the ratio of opposite over adjacent
tan(angle) = opposite/adjacent
So for reference angle G, we say,
tan(G) = JH/GJ = 2/1 = 2
We'll treat tan(H) in a similar fashion, but the opposite and adjacent sides swap roles. That means we'll apply the reciprocal to the result above to get 1/2 for tan(H)
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So we have this interesting property where
tan(G)*tan(H) = 2*(1/2) = 1
In general,
tan(A)*tan(B) = 1 if and only if A+B = 90 degrees
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Side note: The side sqrt(5) isn't used at all.
