Answer:
so idc![\sqrt[n]{x} \sqrt{x} \alpha \pi x^{2} \\ \left \{ {{y=2} \atop {x=2}} \right. x_{123} \int\limits^a_b {x} \, dx \lim_{n \to \infty} a_n \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right]](https://tex.z-dn.net/?f=%5Csqrt%5Bn%5D%7Bx%7D%20%5Csqrt%7Bx%7D%20%5Calpha%20%5Cpi%20x%5E%7B2%7D%20%5C%5C%20%5Cleft%20%5C%7B%20%7B%7By%3D2%7D%20%5Catop%20%7Bx%3D2%7D%7D%20%5Cright.%20x_%7B123%7D%20%5Cint%5Climits%5Ea_b%20%7Bx%7D%20%5C%2C%20dx%20%20%5Clim_%7Bn%20%5Cto%20%5Cinfty%7D%20a_n%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%262%263%5C%5C4%265%266%5C%5C7%268%269%5Cend%7Barray%7D%5Cright%5D)
443
Step-by-step explanation: its 2 6\7
I believe you would have to multiply both 25 and 20 and what ever number you get dived by 100 if the numbers to high multiply aging or subtract the number (if it's wrong I'm really not good at my math I'm sorry)
Answer: f(x) = 1^(x + 1)
Step-by-step explanation:
we have that h(x) = 1^x
and h(x) = f(g(x))
This mean that we are evaluating the function f(y) in the point y = g(x)
where g(x) = x - 1
then:
f(g(x) = f(x - 1) = h(x) = 1^x
then we should have that:
f(x) = 1^(x + 1)
then:
f(x - 1) = 1^(x - 1 + 1) = 1^x
Answer:
2/9 is less than 1/3
Step-by-step explanation:
2/9= 0.222
1/3=0.333
