Could u please help in solving the missing functions in this machine . ignore my answer in blue ink

ashebiny

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20191008_191021.jpg
 

LCKurtz

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For 6, try plotting the points \(\displaystyle (7,24),~(8,19),~(10,9)\) to scale and see if that gives you any ideas.
 

ashebiny

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IMG-20191008-WA0005.jpg
 

ashebiny

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Solved thank you
 

HallsofIvy

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Given n data points there exist a unique polynomial of degree at most n-1 giving those points. Here we have three data points, (7, 24), (8, 19), and (10, 9), so we can fit a quadratic, \(\displaystyle y= ax^2+ bx+ c\), to them. We must have 24= 49a+ 7b+ c, 19= 64a+ 8b+ c, and 9= 100a+ 10b+ c. Subtract the first equation from the second to get -5= 35a+ b, eliminating c. Subtract the first equation from the third to get -15= 6a+ 3b, also eliminating c. Subtract 3 times the first of those equations from the second to get 0= 99a so a= 0 (and the formula is actually linear, not quadratic). Then 6a+ 3b= 3b= -15 so b= -5. Finally 24= 0- 35+ c so c= 24+ 35= 59. The formula is y= -5x+ 59.
 
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