5/10 divide each number by 5, 5 divided by 5 is 1 and 10 divided by 5 is two so it would be 1/2
4/12 divide each number by 4, four divided by 4 is 1, and 12 divided by 4 is 3 so it would be 1/3
3/9 divide each number by 3, 3 divided by 3 is 1, and 9 divided by 3 is 3 or it would also be 1/3
Hope that helped you
The distributive property is a simplification rule that does not change the value of an expression. Nothing need be done to the other side of the equation.
q = -4(3+x)
q = -4*3 -4*x . . . . . the -4 is distributed. The right side has not changed value.
Any real number line ranges from negative infinity to positive infinity. A real number number line consists of all the rational and irrational numbers. Let us take three intervals which contain both the rational and irrational numbers.
First interval: [3,4]
Since every integer is a rational number, 3 and 4 are both rational. In this interval there occurs the value of π (3.14159..) which is an irrational number.
Second interval : [0,2]
0 and 2 are integers and hence are rational. In this interval, occurs √2 (1.41421...) is an irrational number.
Third interval : [2,3]
In this interval, Eulers number 'e' lie whose value is (2.718281.. )
Hence we can conclude that, there occurs an irrational number between any two rational number.
we know all it's doing is adding 6 over again to each term to get the next one, so then
now for the explicit one
![\bf n^{th}\textit{ term of an arithmetic sequence} \\\\ a_n=a_1+(n-1)d\qquad \begin{cases} n=n^{th}\ term\\ a_1=\textit{first term's value}\\ d=\textit{common difference}\\[-0.5em] \hrulefill\\ a_1=7\\ d=6 \end{cases} \\\\\\ a_n=7+(n-1)6\implies a_n=7+6n-6\implies \stackrel{\textit{Explicit Formula}}{\stackrel{f(n)}{a_n}=6n+1} \\\\\\ therefore\qquad \qquad f(10)=6(10)+1\implies f(10)=61](https://tex.z-dn.net/?f=%5Cbf%20n%5E%7Bth%7D%5Ctextit%7B%20term%20of%20an%20arithmetic%20sequence%7D%20%5C%5C%5C%5C%20a_n%3Da_1%2B%28n-1%29d%5Cqquad%20%5Cbegin%7Bcases%7D%20n%3Dn%5E%7Bth%7D%5C%20term%5C%5C%20a_1%3D%5Ctextit%7Bfirst%20term%27s%20value%7D%5C%5C%20d%3D%5Ctextit%7Bcommon%20difference%7D%5C%5C%5B-0.5em%5D%20%5Chrulefill%5C%5C%20a_1%3D7%5C%5C%20d%3D6%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5C%5C%20a_n%3D7%2B%28n-1%296%5Cimplies%20a_n%3D7%2B6n-6%5Cimplies%20%5Cstackrel%7B%5Ctextit%7BExplicit%20Formula%7D%7D%7B%5Cstackrel%7Bf%28n%29%7D%7Ba_n%7D%3D6n%2B1%7D%20%5C%5C%5C%5C%5C%5C%20therefore%5Cqquad%20%5Cqquad%20f%2810%29%3D6%2810%29%2B1%5Cimplies%20f%2810%29%3D61)
Answer:
Answer:f(2)=-3(2)^2+2*2+11
Answer:f(2)=-3(2)^2+2*2+11 =-3*4+4+11
Answer:f(2)=-3(2)^2+2*2+11 =-3*4+4+11 =-12+4+11
Answer:f(2)=-3(2)^2+2*2+11 =-3*4+4+11 =-12+4+11 =3
hence, f(2)=3
