6x,5y and 3 those are terms
4b=a then plus 10 to see if the age holds true to 3b=a in ten years. Then since the question is how old is she today? You must either add 6 to young age (of20) or minus 4 from the older (of 30).
So Betty would be 26 today.
To my best knowledge the slop that is increased is number 3 it is moving up in a slop and it is sloped.
Answer:
D is correct option
Step-by-step explanation:
The correct option is D.
The standard quadratic equation is ax²+bx+c=0
Where a and b are coefficients and c is constant.
It means that constant are on the L.H.S and there is 0 on the right hand side.
Therefore to make it a quadratic equation first of all you have to add 11 at both sides so that the R.H.S becomes 0.
The given equation is:
2x2-x+ 2 = -11
If we add 11 on both sides the equation will be:
2x2-x+ 2 +11= -11+11
2x^2-x+13=0
Thus the correct option is D
You can further solve it by applying quadratic formula....
