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3 years ago

Which equation has no real solutions?

Mathematics

1 answer:

Sladkaya [172]3 years ago

5 0

Answer:

$B. 4x^2+76=32$

Step-by-step explanation:

$In ~ option~ B$

$4x^2+76=32\\$

$4x^2=-76+32$

$4x^2=-44$

$x^2=-44/4=-11$

$x^2=-11$

A square of any real number can't be a negative integer.

So, Option B is your answer

$--------------$

hope it helps..

have a great day!!

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bezimeni [28]

Answer:

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Step-by-step explanation:

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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

Nostrana [21]

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":

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

4 0

3 years ago