Question 129723
1) By data it is given that price is related to diameter by a quadratic equation.


A general quadratic equation is given by f(x) = {{{ax^2 + bx + c}}} 


We shall write it as y = f(x) = {{{ax^2 + bx + c}}} 


==> y = {{{ax^2 + bx + c}}} 


So now lets plug in all the given set of values, in the form of co-ordinates. This implies that it forms 3 equations.


3.39 = 64a + 8b + c  ------------(1)


5.12 = 144a + 12b + c  -----------(2)


7.29 = 256a + 16b + c -----------(3) 


Let solve them simlutaneously.


Subtracting (1) & (2), we get:


1.73 = 80a + 4b --------------(4) 



subtracting (2) & (3), we get:


2.17 = 112a + 4b ------------(5) 


subtracting (4) & (5) 


0.44 = 32a 


==> a = 0.01375 


Now by back substituion we find b


1.73 - 1.1 = 4b 


0.63 = 4b


==> b = 0.1575 


substitute the a and b value in one of the equations. So we get:


3.39 = 0.88 + 1.26 + c


3.39 - 2.14 = c


c = 1.25


Thus the quadratic equation would be 


y = {{{0.01375x^2 + 0.1575x + 1.25}}}



Price of a 14 inch pizza would be just plug in x = 14 in the above equation we get:



{{{y = 0.01375(14)^2 + 0.1575(14) + 1.25}}}



y = $6.15 


==============================================================================


2) Solve for x. {{{(x-6)^2 + (x+1)^2}}} = 0


The given equation can be written as:


{{{x^2 + 36 - 12x + x^2 + 1 + 2x}}} = 0 



{{{2x^2 - 10x + 37 }}} = 0 


Using the quadratic formula we find the roots.


{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}} 


Substituting for the values we get:


{{{x = (10 +- sqrt( 100-4*2*37 ))/(2*2) }}} 



{{{x = (10 +- sqrt(100 - 296))/4}}} 



{{{x = (10 +- sqrt(-196))/4}}} 



{{{x = (10 +- 14i)/4}}} 


Taking 2 as a common factor we get:


So the roots are: 5 + 7i/2 and 5 - 7i/2 


The solutions are 5 + 7 i /2, 5-7 i/2


=============================================================================


3) What is the maximum product of two numbers whose sum is -6? What numbers yield this product? The max product is?


Let x and y be the two numbers.


It is given that the product is maximum and the sum is negative six.


so lets find the multiples of 6. They are + -1, +- 2, +-3, +-6 


Taking them 2 at a time we find that -3 and -3 when added gives us -6 


And their product gives us 9 which is the maximum number.


Hence the solution.


===========================================================================

4) solve for "n" 

N = {{{n^2 – 19n}}}/6


This can be written as:


6N = {{{n^2 - 19n}}} 


{{{n^2 - 19n - 6N = 0 }}}


Using the quadratic formula we get:


{{{n = (-b +-sqrt( b^2-4*a*c ))/(2*a) }}} 


Substituing for the values, we get:



{{{n = (19 +- sqrt(361 + 4*1*6N))/2}}} 



{{{n = (19 +- sqrt(361 + 24N))/2}}} 




Thus the solution.


Regards


If you stil have any doubts you can get back to me.. I'm a online teacher.. I'll help you..