Answer:
Applying Quotient and power rule. Graphically, for x>0 both curves coincide.
Step-by-step explanation:
1) Firstly, by applying the Quotient and then the Power Rule, we have a subtraction of the argument and, finally the exponent turns to be the coefficient. As it follows:
2) Check the graph below to see this equivalence. to see x>0 they both coincide.
Neutrons are in the nucleus with the protons.
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![\bf \cfrac{(x-2)(x+3)}{2x+2}\implies \cfrac{x^2+x-6}{2x+2}~~ \begin{array}{llll} \leftarrow \textit{2nd degree polynomial}\\ \leftarrow \textit{1st degree polynomial} \end{array} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{vertical asymptote}}{2x+2=0}\implies 2x=-2\implies x=-\cfrac{2}{2}\implies x=-1](https://tex.z-dn.net/?f=%5Cbf%20%5Ccfrac%7B%28x-2%29%28x%2B3%29%7D%7B2x%2B2%7D%5Cimplies%20%5Ccfrac%7Bx%5E2%2Bx-6%7D%7B2x%2B2%7D~~%20%5Cbegin%7Barray%7D%7Bllll%7D%20%5Cleftarrow%20%5Ctextit%7B2nd%20degree%20polynomial%7D%5C%5C%20%5Cleftarrow%20%5Ctextit%7B1st%20degree%20polynomial%7D%20%5Cend%7Barray%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Cstackrel%7B%5Ctextit%7Bvertical%20asymptote%7D%7D%7B2x%2B2%3D0%7D%5Cimplies%202x%3D-2%5Cimplies%20x%3D-%5Ccfrac%7B2%7D%7B2%7D%5Cimplies%20x%3D-1)
when the degree of the numerator is greater than the denominator's, then it has no horizontal asymptotes.
quick note:
when the degree of the numerator is 1 higher than the degree of the denominator, then it has an slant-asymptote, so this one has a slant-asymptote.
The x-value increases by a factor of 5 from 1.6 to 8 between the two points. Because the variation is inverse, the y-value must decrease by a factor of 5 between the two points.
6/5 = 1.2
The second point is (8, 1.2), corresponding to the 4th selection.
Answer:
12
Step-by-step explanation:
PEMDAS
9/3=3
3+9=12
