the tale-tell fellow is the number inside the parentheses.
if that number, the so-called "growth or decay factor", is less than 1, then is Decay, if it's more than 1, is Growth.
![\bf f(x)=0.001(1.77)^x\qquad \leftarrow \qquad \textit{1.77 is greater than 1, Growth} \\\\[-0.35em] ~\dotfill\\\\ f(x)=2(1.5)^{\frac{x}{2}}\qquad \leftarrow \qquad \textit{1.5 is greater than 1, Growth} \\\\[-0.35em] ~\dotfill\\\\ f(x)=5(0.5)^{-x}\implies f(x)=5\left( \cfrac{05}{10} \right)^{-x}\implies f(x)=5\left( \cfrac{1}{2} \right)^{-x} \\\\\\ f(x)=5\left( \cfrac{2}{1} \right)^{x}\implies f(x)=5(2)^x\qquad \leftarrow \qquad \textit{Growth} \\\\[-0.35em] ~\dotfill](https://tex.z-dn.net/?f=%5Cbf%20f%28x%29%3D0.001%281.77%29%5Ex%5Cqquad%20%5Cleftarrow%20%5Cqquad%20%5Ctextit%7B1.77%20is%20greater%20than%201%2C%20Growth%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20f%28x%29%3D2%281.5%29%5E%7B%5Cfrac%7Bx%7D%7B2%7D%7D%5Cqquad%20%5Cleftarrow%20%5Cqquad%20%5Ctextit%7B1.5%20is%20greater%20than%201%2C%20Growth%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20f%28x%29%3D5%280.5%29%5E%7B-x%7D%5Cimplies%20f%28x%29%3D5%5Cleft%28%20%5Ccfrac%7B05%7D%7B10%7D%20%5Cright%29%5E%7B-x%7D%5Cimplies%20f%28x%29%3D5%5Cleft%28%20%5Ccfrac%7B1%7D%7B2%7D%20%5Cright%29%5E%7B-x%7D%20%5C%5C%5C%5C%5C%5C%20f%28x%29%3D5%5Cleft%28%20%5Ccfrac%7B2%7D%7B1%7D%20%5Cright%29%5E%7Bx%7D%5Cimplies%20f%28x%29%3D5%282%29%5Ex%5Cqquad%20%5Cleftarrow%20%5Cqquad%20%5Ctextit%7BGrowth%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill)
now, let's take a peek at the second set.
![\bf f(x)=3(1.7)^{x-2}\qquad \leftarrow \qquad \begin{array}{llll} \textit{the x-2 is simply a horizontal shift}\\\\ \textit{1.7 is more than 1, Growth} \end{array} \\\\[-0.35em] ~\dotfill\\\\ f(x)=3(1.7)^{-2x}\implies f(x)=3\left(\cfrac{17}{10}\right)^{-2x}\implies f(x)=3\left(\cfrac{10}{17}\right)^{2x} \\\\\\ \textit{that fraction is less than 1, Decay} \\\\[-0.35em] ~\dotfill](https://tex.z-dn.net/?f=%5Cbf%20f%28x%29%3D3%281.7%29%5E%7Bx-2%7D%5Cqquad%20%5Cleftarrow%20%5Cqquad%20%5Cbegin%7Barray%7D%7Bllll%7D%20%5Ctextit%7Bthe%20x-2%20is%20simply%20a%20horizontal%20shift%7D%5C%5C%5C%5C%20%5Ctextit%7B1.7%20is%20more%20than%201%2C%20Growth%7D%20%5Cend%7Barray%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20f%28x%29%3D3%281.7%29%5E%7B-2x%7D%5Cimplies%20f%28x%29%3D3%5Cleft%28%5Ccfrac%7B17%7D%7B10%7D%5Cright%29%5E%7B-2x%7D%5Cimplies%20f%28x%29%3D3%5Cleft%28%5Ccfrac%7B10%7D%7B17%7D%5Cright%29%5E%7B2x%7D%20%5C%5C%5C%5C%5C%5C%20%5Ctextit%7Bthat%20fraction%20is%20less%20than%201%2C%20Decay%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill)
![\bf f(x)=3^5\left( \cfrac{1}{3} \right)^x\qquad \leftarrow \qquad \textit{that fraction is less than 1, Decay} \\\\[-0.35em] ~\dotfill\\\\ f(x)=3^5(2)^{-x}\implies f(x)=3^5\left( \cfrac{2}{1} \right)^{-x}\implies f(x)=3^5\left( \cfrac{1}{2} \right)^x \\\\\\ \textit{that fraction in the parentheses is less than 1, Decay}](https://tex.z-dn.net/?f=%5Cbf%20f%28x%29%3D3%5E5%5Cleft%28%20%5Ccfrac%7B1%7D%7B3%7D%20%5Cright%29%5Ex%5Cqquad%20%5Cleftarrow%20%5Cqquad%20%5Ctextit%7Bthat%20fraction%20is%20less%20than%201%2C%20Decay%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20f%28x%29%3D3%5E5%282%29%5E%7B-x%7D%5Cimplies%20f%28x%29%3D3%5E5%5Cleft%28%20%5Ccfrac%7B2%7D%7B1%7D%20%5Cright%29%5E%7B-x%7D%5Cimplies%20f%28x%29%3D3%5E5%5Cleft%28%20%5Ccfrac%7B1%7D%7B2%7D%20%5Cright%29%5Ex%20%5C%5C%5C%5C%5C%5C%20%5Ctextit%7Bthat%20fraction%20in%20the%20parentheses%20is%20less%20than%201%2C%20Decay%7D)
Answer:
1771 possible ways
Step-by-step explanation:
In this case, we need to know first how many candidates are in total:
10 + 3 + 10 = 23 candidates in total.
Now, we need to choose 3 of them to receive an award. In this case, we have several scenarios, but as it's an award we can also assume that the order in which the candidates are chosen do not matter, so, the formula to use is the following:
C = m! / n! (m - n)!
Where m is the total candidates and n, is the number of candidates to be chosen. Replacing this data we have:
C = 23! / 3! (23 - 3)!
C = 2.59x10^22 / 6(2.43x10^18)
C = 1771
So we have 1771 ways of choose the candidates.
An instructional activity that could help learners recognize fractions includes an analog clock, fraction strips, and pattern block.
<h3>What is an instructional activity?</h3>
Instructional activities are routine segments of instruction that show how the teacher and students will participate and interact with materials and content.
An instructional activity that could help learners recognize fractions includes an analog clock, fraction strips, and pattern block. For example, if one wants to teach students about fractional parts that are divided equally into twelfths, the analog clock can be ideal.
Learn more about fractions on;
brainly.com/question/78672
Answer:
10.2π
Step-by-step explanation:
Given data
Area of circle= 81 ft^2
Also, the expression for the area of a circle is
A=πr^2
substitute
81=3.142*r^2
r^2= 81/3.142
r^2= 25.77
square both sides
r= √25.77
r= 5.1
Also, the expression for the circumference is
C= 2πr
substitute
C= 2*π*5.1
C= 10.2π
Hence the circumference is 10.2π
