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NeTakaya
3 years ago
14

What causes a solution to a rational equation to be an extraneous solution

Mathematics
2 answers:
Alona [7]3 years ago
8 0
<h2>Answer with explanation:</h2>

While solving the rational equation we check the solution whether they are true solutions or are extraneous solution.

  • The true solutions are one which are solution of the rational equation as well as is a well defined solution of the equation.
  • whereas the extraneous solution are one which are solution but does not satisfy the equation i.e. these are the excluded value of the equation  which makes the denominator of the rational equation equal to zero.

mrs_skeptik [129]3 years ago
3 0
If a solution results in zero when substituted into the denominator of the equation, the solution is extraneous.
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The junior varsity football team score 23 points in last Saturday's game they score the combination of seven point touchdowns in
frutty [35]

Answer:

  2 touchdowns, 3 field goals

Step-by-step explanation:

The number of touchdowns cannot be more than 3, so it is relatively easy to find the solution by trial and error.

  23 is not divisible by 3, so 0 touchdowns is not a solution

  23 -7 = 16 is not divisible by 3, so 1 touchdown is not a solution

  23 -14 = 9 is divisible by 3, so 2 touchdowns and 3 field goals is a solution

  21 -21 = 2 is not divisible by 3, so there is only one solution.

7 0
3 years ago
In this graph what is the slope of the line?
sashaice [31]

Answer:

The slope of the line is 10

Step-by-step explanation:

To find the slope, simply identify 2 points on the line. I chose: (1,10) and (2,20)

20-10=10

2-1=1

The slope of the line is 10

To check this answer, just multiply X with 10 and you will always get Y!

7 0
3 years ago
This is just a square, help​
Kipish [7]

Answer:

Step-by-step explanation:

They are all true

4 0
3 years ago
Read 2 more answers
A(n) _____ is the set of possible values for a given attribute
Sladkaya [172]
I think this is the Domain. Since a domain is a group of numbers that can be entered in a function to create a valid output. They are set of all possible values of x which will satisfy a ffunction and output real y-values
6 0
3 years ago
Read 2 more answers
Please help me now ! Thank you
jarptica [38.1K]
First we'll do two basic steps. Step 1 is to subtract 18 from both sides. After that, divide both sides by 2 to get x^2 all by itself. Let's do those two steps now

2x^2+18 = 10
2x^2+18-18 = 10-18 <<--- step 1
2x^2 = -8
(2x^2)/2 = -8/2 <<--- step 2
x^2 = -4

At this point, it should be fairly clear there are no solutions. How can we tell? By remembering that x^2 is never negative as long as x is real. 

Using the rule that negative times negative is a positive value, it is impossible to square a real numbered value and get a negative result. 

For example
2^2 = 2*2 = 4
8^2 = 8*8 = 64
(-10)^2 = (-10)*(-10) = 100
(-14)^2 = (-14)*(-14) = 196

No matter what value we pick, the result is positive. The only exception is that 0^2 = 0 is neither positive nor negative.

So x^2 = -4 has no real solutions. Taking the square root of both sides leads to

x^2 = -4
sqrt(x^2) = sqrt(-4)
|x| = sqrt(4)*sqrt(-1)
|x| = 2*i
x = 2i or x = -2i
which are complex non-real values


5 0
4 years ago
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