Answer:
An extraneous solution is a root of a transformed equation that is not a root of the original equation because it was excluded from the domain of the original equation. Example 1: Solve for x , 1x − 2+1x + 2=4(x − 2)(x + 2) .Set the equation to equal zero. (this ends up being √x+4−x+2=0 )
Plug this into the y= button on your TI-83/84 calculator.
Find the value of each of your solutions (go to 2nd->Calc->Value and enter your solution for x )
You should get zero as an answer for each of them.
Step-by-step explanation:
Extraneous solutions are not solutions at all. They arise from outside the problem, from the method of solution. They are extraneous because they are not solutions of the original problem. ... To tell if a "solution" is extraneous you need to go back to the original problem and check to see if it is actually a solution.
Answer:
(2*7)*6 or 2*(7*6) = $84
Step-by-step explanation:
The associative property would always give the same value as an answer no matter the way of expression, once the expressions are right. So in this case,
Since has saves $2 every day and there are 7 days in a week,
2*7
and save this continously for 6 weeks.
(2*7)*6
This equals $84
OR
7 days in a week for 6 weeks
7*6
and saved $2 per day
2*(7*6)
This equals $84
14(12)+5(6)+16 = 168+30+16=214
This is the correct formula for the problem because you sold 12 maracas, 6 calves, and 1 single drum. You should also put the parenthesis next to the numbers that are being multiplied because you sold multiple of those items.
Answer: The percent of female commissioned officers in the army in 1940= 5.3%
Step-by-step explanation:
We are given that ,
The number of commissioned officers reported by U.S. Army in 1940 = 17,563
The number of female commissioned officers = 939
Then , the number of female commissioned officers in the army in 1940 would be :-
Hence, the percent of female commissioned officers in the army in 1940= 5.3%
