Hey can any one tell me if 5/√5 is rational or irrational
2 answers:
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Solution :
We observe that :
But BA is the perpendicular.
From the center B and WX is a chord.
Therefore, TW = TX (perpendicular from the centre of a circle to a chord bisects it)
Consider Δ BTX,
∠BTX = 90° (BA ⊥ WX)
BT = XT (Δ BTX is isosceles)
Since the angles opposite to equal sides are equal of a triangle arc are equal.
∠BTX = ∠BXT
But in the triangle,
∠TBX + ∠TXB + ∠BTX = 180°
∠TBX + ∠TBX + 90° = 180°
2 ∠TBX = 90°
∠TBX = 45°
From trigonometry, we get
...............(1)
WX = 10
i.e., TX + TW = 10
But TX = TW
2 TX = 10
Tx = 5
BX = radius of circle.
∴
= 7
Therefore, the radius of the circle is 7 units.
Okay so I attatched pictures of my work. I would just graph it by plugging in x-values on either side of the vertex into the equation. For example, the vertex is at (-2,3). So try plugging x=-5 and x=1 into the equation (y=a(x-h)^2 +k).
y=(x+2)^2 +3
y=((1)+2)^2 +3=(3)^2+3= 9+3= 12
Now you can plot the vertex (-2,3), the point (1,12), and (-5,12). From there just connect the dots and you should get a graph that looks similar to the last picture I attached. Hope this all made sense sorry it was so wordy I just wasn't sure how much you knew or didn't know. So, I thought it would be better to explain it all rather than not enough.
y=(x+2)^2 +3
y=((1)+2)^2 +3=(3)^2+3= 9+3= 12
Now you can plot the vertex (-2,3), the point (1,12), and (-5,12). From there just connect the dots and you should get a graph that looks similar to the last picture I attached. Hope this all made sense sorry it was so wordy I just wasn't sure how much you knew or didn't know. So, I thought it would be better to explain it all rather than not enough.