Solveit

Hi ! :)

\(\tan A = \frac {\textrm{opposite}} {\textrm{adjacent}} = \frac {a} {b}.\\ \text{tangent=tan.}\)

\(7(cy^2+dy-3) = 14y^2-7y+f \\ 7cy^2+7dy-21 = 14y^2-7y+f\\ 7cy^2=14y^2 \\ c=2 \\ 7dy=-7y\\ d=-1\\ -21=f \)

the sum of two numbers is 40. the larger number is 14 less than twice the smaller number. find the two numbers

x+y=40

2x-14=y

put 2x-14=y in x+y=40:

2x-14+x=40

3x=54

x=18

x+y=40      x=18

y=22

did you did everything like i sad ?

sit on number keyboard and shake your popo forever then look what you get

i think there is need to be more please write evrything