First the common denominator has to be 6. So to get to 6 you multiply the 2 by 3 and the numerator by 3 giving 3 over 6 then you multiply 3 times 2 to get to 6 so the following fraction is 2 over 6. So when u add the two together it gives you 5 6ths
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:
x=−12
Step-by-step explanation:
Let's solve your equation step-by-step.
5x+6=2(2x−3)
Step 1: Simplify both sides of the equation.
5x+6=2(2x−3)
5x+6=(2)(2x)+(2)(−3)(Distribute)
5x+6=4x+−6
5x+6=4x−6
Step 2: Subtract 4x from both sides.
5x+6−4x=4x−6−4x
x+6=−6
Step 3: Subtract 6 from both sides.
x+6−6=−6−6
Subtract 60 on both sides to get -40.
You are left with 40x=-40
Divide both sides by 40 to leave x by itself.
You get x=-1.
