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dem82 [27]
2 years ago
9

He determines that 6 is an extraneous solution because the difference of the numerators is 6, so the 6s cancel to 0. which best

describes the reasonableness of the student’s solution? his solution for x is correct and his explanation of the extraneous solution is reasonable. his solution for x is correct, but in order for 6 to be an extraneous solution, both denominators have to result in 0 when 6 is substituted for x. his solution for x is correct, but in order for 6 to be an extraneous solution, one denominator has to result in 0 when 6 is substituted for x. his solution for x is incorrect. when solved correctly, there are no extraneous solutions.
Mathematics
1 answer:
seropon [69]2 years ago
5 0

The question is incomplete. Completed question is given below the answer.

His solution for x is correct, but in order for 6 to be an extraneous solution, one denominator has to result in 0 when 6 is substituted for x.

Given  

Then the solution follows thus:

Step 1: 8(x – 4) = 2(x + 2)

Step 2: 4(x – 4) = (x + 2)

Step 3: 4x – 16 = x + 2

Step 4: 3x = 18

Step 5: x = 6

It can be seen that his solution is correct. But 6 is not an extraneous solution.

An extraneous solution is a solution to an equation that emerges from the process of solving the problem but is not a valid solution to the original problem.

When 6 is substituted into the original equation, the original equation holds.

Therefore, his solution for x is correct, but in order for 6 to be an extraneous solution, one denominator has to result in 0 when 6 is substituted for x.

Learn more about extraneous solution here: brainly.com/question/3751209

#SPJ4

Completed question:-

A student solves the following equation for all possible values of x:His solution is as follows:

Step 1: 8(x – 4) = 2(x + 2)

Step 2: 4(x – 4) = (x + 2)

Step 3: 4x – 16 = x + 2

Step 4: 3x = 18

Step 5: x = 6

He determines that 6 is an extraneous solution because the difference of the numerators is 6, so the 6s cancel to 0.

Which best describes the reasonableness of the student’s solution?

His solution for x is correct and his explanation of the extraneous solution is reasonable.

His solution for x is correct, but in order for 6 to be an extraneous solution, both denominators have to result in 0 when 6 is substituted for x.

His solution for x is correct, but in order for 6 to be an extraneous solution, one denominator has to result in 0 when 6 is substituted for x.

His solution for x is incorrect. When solved correctly, there are no extraneous solutions.

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Can someone please explain this
Pani-rosa [81]

Answer:

x = 7

Step-by-step explanation:

There is a total of 180 degrees in a triangle. To solve you add the other angles and set = to 180 then solve for x

180 = 33 + 90 + 7x +8

172 = 33 + 90 + 7x

139 = 90 +7x

49 = 7x

7 = x

6 0
3 years ago
In a town, 15% of the population is unemployed. what is the
kow [346]

Answer:

1479 people

Step-by-step explanation:

If the amount employed is 8381, that means 8381 is 85% of the population. 8381/85 = 98.6

Every 1% of the population is 98.6, so we can just do 15 x 98.6 to get 1479

7 0
2 years ago
Read 2 more answers
What proportion could be used to solve the question?<br> What percent of 80 is 23?
Drupady [299]
The variable X (percentage) over 100 and 23 over 80.
7 0
3 years ago
A dealer sells a certain type of chair and a table for $40. He also sells the same sort of table and a desk for $83 or a chair a
vesna_86 [32]
Chair=x

Table=y

Desk=z

\begin{Bmatrix}x+y&=&40\\y+z&=&83\\x+z&=&77\end{matrix}

keep the first row as normal, then in the other ones, we can isolate Y and X

\begin{Bmatrix}x+y&=&40\\y&=&83-z\\x&=&77-z\end{matrix}

now we can replace at first row...

\begin{Bmatrix}(77-z)+(83-z)&=&40\\y&=&83-z\\x&=&77-z\end{matrix}

\begin{Bmatrix}160-2z&=&40\\y&=&83-z\\x&=&77-z\end{matrix}

\begin{Bmatrix}-2z&=&40-160\\y&=&83-z\\x&=&77-z\end{matrix}

\begin{Bmatrix}-2z&=&-120\\y&=&83-z\\x&=&77-z\end{matrix}

\begin{Bmatrix}z&=&60\\y&=&83-z\\x&=&77-z\end{matrix}

now we can replace the Z to discovery the other value

\begin{Bmatrix}z&=&60\\y&=&83-60\\x&=&77-60\end{matrix}

\boxed{\boxed{\boxed{\begin{Bmatrix}Chair&=&17\\Table&=&23\\Desk&=&60\end{matrix}}}}
6 0
3 years ago
Can someone help me please
shepuryov [24]
.15 • 50 = 7.5

25.25 - 7.5 = 17.75

thus, your answer is $17.75
6 0
3 years ago
Read 2 more answers
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