Oh yes, there is! Here: \(\frac{n!}{r!(n-r)!}\)- combinations; order does not matter
The order does matter( Permutations): \(\frac{n!}{(n-r)!}\)
If you take AoPS, they give you wonderful examples!
Also, a quote from CPhill's explanation: "We have C(16, 4) = 1820 possible sets of any four of the points."
We can express this as \(C\binom{16}{3}\) , where it is combinations.
It's not a fraction. However, it means 16 choose 4; it computes the number of ways to choose items from a collection of items.
Let \(x \) be the cost of the food. Then \(1.2(1.06x)=25.44\implies x=\boxed{\$20}\).
Another way to approach it is realization!
While we can solve for \(a\) and \(b\) individually, it is simpler to note that \(f(1) = a + b\). Thus, substituting \(1\) into the given equation, we obtain\(g(f(1)) = 3 \cdot 1 + 4 = 7.\)Thus, \(g(f(1)) = 2 \cdot f(1) - 5 = 7 \Longrightarrow f(1) = \boxed{6}.\)
Also, \(e\) if you want to know is named after Euler. Read more information, here: https://www.mathsisfun.com/numbers/e-eulers-number.html
Welcome!
Sorry if my explanation was confusing. Read more about the quadratic formula https://www.purplemath.com/modules/quadform.htm and https://www.khanacademy.org/math/algebra/quadratics/solving-quadratics-using-the-quadratic-formula/a/quadratic-formula-review, and I'm sure that you'll get a deeper understanding.
Nope, not quite. The answer is actually, (C), $5069.96. First, you multiply \(\:0.045\cdot \:7\) , which is \(0.315\) . Now, \(e^{0.315}=1.37026\dots \), so we have \(3700\cdot \:1.37026\dots =5069.95945\dots \approx=5069.96\) or (C).
Anytime! Glad to help out users!
