W varies partly as x and partly as the square of x. When x=3, w=18 and when x= 5, w=169. Find w in terms of x.
2 answers
I can't be able to answer it pls help me out
w=ax+bx^2+c
169=5a+25b+c
18=3a+9b+c
So, assume c=0 first
find a and b.
multiply first equation by 3, second by 5
169*3=15a+75b
90=15a+45b
subtract second equation from first
507-90=30b
b=13.9
then find a in the second equation
then, go to the original statement, check it with your a and b.
169=5a+25b+c
18=3a+9b+c
So, assume c=0 first
find a and b.
multiply first equation by 3, second by 5
169*3=15a+75b
90=15a+45b
subtract second equation from first
507-90=30b
b=13.9
then find a in the second equation
then, go to the original statement, check it with your a and b.