Answer:
Therefore,
Option (A) is false
Option (B) is false
Option (C) is false
Step-by-step explanation:
Considering the graph
Given the vertices of the segment AB
Finding the length of AB using the formula
units
Given the vertices of the segment JK
From the graph, it is clear that the length of JK = 5 units
so
units
Given the vertices of the segment GH
Finding the length of GH using the formula
![\mathrm{Apply\:radical\:rule\:}\sqrt[n]{a^n}=a,\:\quad \mathrm{\:assuming\:}a\ge 0](https://tex.z-dn.net/?f=%5Cmathrm%7BApply%5C%3Aradical%5C%3Arule%5C%3A%7D%5Csqrt%5Bn%5D%7Ba%5En%7D%3Da%2C%5C%3A%5Cquad%20%5Cmathrm%7B%5C%3Aassuming%5C%3A%7Da%5Cge%200)
units
Thus, from the calculations, it is clear that:
Thus,
Therefore,
Option (A) is false
Option (B) is false
Option (C) is false
if ur talking about like this then here
1. 354
2. 453 If this is what ur talking about i hope this helps :) :) :)
This is an exponential equation. We will solve in the following way. I do not have special symbols, functions and factors, so I work in this way
2 on (2x) - 5 2 on x + 4=0 =>. (2 on x)2 - 5 2 on x + 4=0 We will replace expression ( 2 on x) with variable t => 2 on x=t =. t2-5t+4=0 => This is quadratic equation and I solve this in the folowing way => t2-4t-t+4=0 => t(t-4) - (t-4)=0 => (t-4) (t-1)=0 => we conclude t-4=0 or t-1=0 => t'=4 and t"=1 now we will return t' => 2 on x' = 4 => 2 on x' = 2 on 2 => x'=2 we do the same with t" => 2 on x" = 1 => 2 on x' = 2 on 0 => x" = 0 ( we know that every number on 0 gives 1). Check 1: 2 on (2*2)-5*2 on 2 +4=0 => 2 on 4 - 5 * 4+4=0 => 16-20+4=0 =. 0=0 Identity proving solution.
Check 2: 2 on (2*0) - 5* 2 on 0 + 4=0 => 2 on 0 - 5 * 1 + 4=0 =>
1-5+4=0 => 0=0 Identity provin solution.
A quarter, two dimes, and two pennies.
Answer:
do we do the x veriable as the numerator? if so it's not a or c if not vice versa
Step-by-step explanation:
