let's bear in mind that an absolute value expression is in effect a piece-wise expression, namely it has a ± versions of the same expression.
![\bf 5|3x-4| = x+1\implies |3x-4|=\cfrac{x+1}{5}\implies \begin{cases} +(3x-4)=\cfrac{x+1}{5}\\[1em] -(3x-4)=\cfrac{x+1}{5} \end{cases} \\\\[-0.35em] ~\dotfill\\\\ +(3x-4)=\cfrac{x+1}{5}\implies 3x-4=\cfrac{x+1}{5}\implies 15x-20=x+1 \\\\\\ 14x-20=1\implies 14x=21\implies x = \cfrac{21}{14}\implies \boxed{x=\cfrac{3}{2}} \\\\[-0.35em] ~\dotfill\\\\ -(3x-4)=\cfrac{x+1}{5}\implies -3x+4=\cfrac{x+1}{5}\implies -15x+20=x+1 \\\\\\ 20=16x+1\implies 19=16x\implies \boxed{\cfrac{19}{16}=x}](https://tex.z-dn.net/?f=%5Cbf%205%7C3x-4%7C%20%3D%20x%2B1%5Cimplies%20%7C3x-4%7C%3D%5Ccfrac%7Bx%2B1%7D%7B5%7D%5Cimplies%20%5Cbegin%7Bcases%7D%20%2B%283x-4%29%3D%5Ccfrac%7Bx%2B1%7D%7B5%7D%5C%5C%5B1em%5D%20-%283x-4%29%3D%5Ccfrac%7Bx%2B1%7D%7B5%7D%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%2B%283x-4%29%3D%5Ccfrac%7Bx%2B1%7D%7B5%7D%5Cimplies%203x-4%3D%5Ccfrac%7Bx%2B1%7D%7B5%7D%5Cimplies%2015x-20%3Dx%2B1%20%5C%5C%5C%5C%5C%5C%2014x-20%3D1%5Cimplies%2014x%3D21%5Cimplies%20x%20%3D%20%5Ccfrac%7B21%7D%7B14%7D%5Cimplies%20%5Cboxed%7Bx%3D%5Ccfrac%7B3%7D%7B2%7D%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20-%283x-4%29%3D%5Ccfrac%7Bx%2B1%7D%7B5%7D%5Cimplies%20-3x%2B4%3D%5Ccfrac%7Bx%2B1%7D%7B5%7D%5Cimplies%20-15x%2B20%3Dx%2B1%20%5C%5C%5C%5C%5C%5C%2020%3D16x%2B1%5Cimplies%2019%3D16x%5Cimplies%20%5Cboxed%7B%5Ccfrac%7B19%7D%7B16%7D%3Dx%7D)
Answer:
Step-by-step explanation:
X>X
-4+X>-2+X
<h3>✽ - - - - - - - - - - - - - - - ~<u>Hello There</u>!~ - - - - - - - - - - - - - - - ✽</h3>
➷ 12 : 40
==> 3 : 10
<h3><u>✽</u></h3>
➶ Hope This Helps You!
➶ Good Luck (:
➶ Have A Great Day ^-^
↬ ʜᴀɴɴᴀʜ ♡
Answer:
see below
Step-by-step explanation:
The formula for the sum of an infinite geometric series with first term a1 and common ratio r (where |r| < 1) is ...
sum = a1/(1 -r)
Applying this to the given series, we get ...
a. sum = 5/(1 -3/4) = 5/(1/4) = 20
b. sum = d/(1 -1/t) = d/((t-1)/t) = dt/(t-1)
_____
The derivation of the above formula is in most texts on sequences and series. In general, you write an expression for the difference of the sum (S) and the product r·S. You find all terms of the series cancel except the first and last, and the last goes to zero in the limit, because r^∞ → 0 for |r| < 1. Hence you get ...
S -rS = a1
S = a1/(1 -r)
1 is the magnitude of this question bro because it is process
