well, we'll be using the distance formula twice here.
that point, is on the x-axis, what's the value of "y" at the x-axis? well y = 0, since it's way down at the 0 level, so the point must be (x , 0).
the distance then from (x,0) to (-3,-1) is the same distance as from (x,0) to 1,6).
Step-by-step explanation:
im not going to go in to detail but the answer is c
Answer:
b
Step-by-step explanation:
Because y is squared, or, actually, because (y + 6) is squared, we know immediately that the graph is horizontal. Because both (y + 6)^2 and (x - 7) are positive, we know that the graph opens to the right. (b) is correct.
9514 1404 393
Answer:
- EF = DE = 44
- FG = DG = 36
- FH = DF = 31
Step-by-step explanation:
Since EH is the perpendicular bisector of DF, ∆DEF is isosceles and sides DE and EF have the same length.
DE = EF
(9x -1) = (7x +9)
2x = 10 . . . . . . . add 1-7x
x = 5 . . . . . . . . . divide by 2
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Similarly, marked sides GD and GF are the same length, so ...
GD = GF
(10y -4) = (7y +8)
3y = 12 . . . . . . . . . . add 4-7y
y = 4 . . . . . . . . . divide by 3
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Now, we have what we need to calculate the side lengths.
EF = 7x+9 = 7·5 +9 = 44
DE = 9x-1 = 9·4 -1 = 44
FG = 7y+8 = 7·4 +8 = 36
DG = 10y-4 = 10·4 -4 = 36
FH = 3x+4y = 3·5 +4·4 = 31
DF = FH = 31
Answer:
60°
Step-by-step explanation:
sum of interior angles of quadrilateral is 360°
∠C = 70°
∠D = 50°
∠C + ∠D = 120°
∴ ∠A + ∠B = 360° - 120° =240°
AO and BO bisect ∠A and ∠B
∠OAB + ∠OBA = 1/2 x ( ∠A + ∠B) = 120°
∠AOB = 180° - (∠OAB + ∠OBA) = 180° - 120° = 60°
