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fenix001 [56]
2 years ago
5

Solve for linear inequality of x - 1 ≥ 4 + x​

Mathematics
1 answer:
defon2 years ago
8 0

\huge\bf{ANSWER:}

\huge\mathrm{No\:solution.}\checkmark

\huge\bf{SOLUTION}:

Hey there, hope you are having a wonderful day! :)

Let's solve this inequality.

First of all, let's move all the numbers to the right, using the \bf{\underline{opposite\:\:operation\:}:

\longmapsto\sf{x-1\geq 4+x

\longmapsto\sf{x\geq 4+x+1

\longmapsto\sf{x\geq x+5

Now, let's move the variables to the left:

\longmapsto\sf{x-x\geq 5

\longmapsto\sf{0\geq 5

Now, as you can see, we ended up with a false statement. 0 is NOT greater than 5.

Thus, there are no values of x that make the inequality true, and the inequality  has

\bold{No\:solutions.}

Hope you find it helpful.

Feel free to ask if you have any doubts.

\bf{-MistySparkles^**^*

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Step-by-step explanation:

x+4-5x-2

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3 years ago
A professor has recorded exam grades for 10 students in his​ class, but one of the grades is no longer readable. If the mean sco
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Answer:

unreadable score = 35

Step-by-step explanation:

We are trying to find the score of one exam that is no longer readable, let's give that score the name "x". we can also give the addition of the rest of 9 readable s scores the letter "R".

There are two things we know, and for which we are going to create equations containing the unknowns "x", and "R":

1) The mean score of ALL exams (including the unreadable one) is 80

so the equation to represent this statement is:

mean of ALL exams = 80

By writing the mean of ALL scores (as the total of all scores added including "x") we can re-write the equation as:

\frac{R+x}{10} =80

since the mean is the addition of all values divided the total number of exams.

in a similar way we can write what the mean of just the readable exams is:

\frac{R}{9} = 85\\ (notice that this time we don't include the grade x in the addition, and we divide by 9 instead of 10 because only 9 exams are being considered for this mean.

Based on the equation above, we can find what "R" is by multiplying both sides by 9:

\frac{R}{9} = 85\\R=85*9= 765

Therefore we can now use this value of R in the very first equation we created, and solve for "x":

\frac{R+x}{10} =80\\\frac{765+x}{10} =80\\765+x=80*10=800\\765+x=800\\x=800-765=35

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