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.
the value of c that makes the expression a perfect square binomial is c=4 .
<u>Step-by-step explanation:</u>
Here we have , an expression x2 + 4x + c or , 
. We need to find the value of c that makes the expression a perfect square binomial. Let's find out:
We have ,
⇒ 
⇒ 
Now , we know that 
Comparing above equation , to we get ;
⇒ { 
}
⇒ 
⇒ 
Therefore , the value of c that makes the expression a perfect square binomial is c=4 .
Answer:
C. Graph c
Step-by-step explanation:
I got it correct
Answer: 145 cans
Step-by-step explanation:
arithmetic sequence
aₙ = a₁ + (n-1).r
aₙ → last term
a₁ → 1st term
n → quantity of terms
r → common difference
a₁ = 1 (one can at the top)
aₙ₋₁ = 25
aₙ = 28
To find out How many cans are in the entire display, we need the SUM of the arithmetic sequence: S = (a₁+aₙ)n/2
∴
S = (1+28).n/2
n = ?
aₙ = a₁ + (n - 1).r
r = 28 - 25 = 3
28 = 1 + (n - 1).3
27 = (n - 1).3
27/3 = (n - 1)
9 = n - 1
n = 9 + 1 = 10
S = (1+28).n/2
S = (1+28).10/2 = 29.10/2 = 29.5 = 145
