37 metres and 43'' I dont quite get.
I get a value of -410.36 instead of 37 metres
With the time part , I went (e-2) x 60 (remembering these units were initially in minutes)= which gave me approximately 8 seconds
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37 metres and 43'' I dont quite get.
I get a value of -410.36 instead of 37 metres
With the time part , I went (e-2) x 60 (remembering these units were initially in minutes)= which gave me approximately 8 seconds
| B Eng (Hons) | IB Mathematics SL | IB Mathematics HL | Australian Cricket | Casual University Statistics Tutor
| B Eng (Hons) | IB Mathematics SL | IB Mathematics HL | Australian Cricket | Casual University Statistics Tutor
^^^ Don't worry I sorted it out , with my scientific calculator I needed to type 'e1' for it to read it as 'e'
| B Eng (Hons) | IB Mathematics SL | IB Mathematics HL | Australian Cricket | Casual University Statistics Tutor
| B Eng (Hons) | IB Mathematics SL | IB Mathematics HL | Australian Cricket | Casual University Statistics Tutor
| B Eng (Hons) | IB Mathematics SL | IB Mathematics HL | Australian Cricket | Casual University Statistics Tutor
I get it now, the initial conditions set , if you want to treat the top of the building as the origin, t=0, v=25, g=10, x=0.
All good
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Yea - I noticed in your 1st post: x = -120 instead of 0.
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How can you work backwards from factorial? Thanks.
A curve is given y = x^x where x > 0. Find the gradient function of the curve y = x^x.
How do I manage to differentiate it using Logarithmic Functions? Thanks.
Screen Shot 2016-05-23 at 8.43.39 PM.png
Screen Shot 2016-05-23 at 8.44.33 PM.png
Can someone please help me with this question?? I'm really bad at reasoning with geometry!
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Similar Triangles. Circular Geo.
a)
In triangles DEG and DFC:
- angle EDG = angle FDC (same angle)
- DE/DF = DG/DC = 1/2
.: triangles DEG and DFC are similar. (pair of sides proportional and included angles equal)
.: angle DGE = angle DCF (corresp angles of similar triangles)
But these are corresponding angles
.: EG // FC (and .: AC)
b)
angle ABC (= angle EBG) = Angle ACB (AB = AC)
But angle DGE (= angle BGE) = angle ACB
.: for triangle EBG, angle EBG = angle BGE
.: EB = EG
Now EG/FC = DE/DF = 1/2
.: FC = 2 x EG = 2 x EB
QED
Last edited by Drongoski; 29 May 2016 at 6:41 PM.
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Thank you so much!!
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The best way to predict the future is to create it
Bump
"verify the result on the third column of the Pascal triangle." Does anyone have any idea on this? Or is my answer shown above good enough ?
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