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3 years ago

Right angle triangle bisector divorce hypotenuse to two segments 15 and 20. Find right angle triangle's perimeter.

Mathematics

1 answer:

fenix001 [56]3 years ago

5 0

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Find the equation of the line that is parallel to the given line and passes through the given point

Sati [7]

Answer:

$\huge\boxed{y=-\dfrac{1}{4}x-1\to x+4y=-4}$

Step-by-step explanation:

$\text{Let}\ k:y=m_1x+b_1;\ l:y=m_2x+b_2\\\\l\ ||\ k\iff m_1=m_2\\\\l\ \perp\ k\iff m_1m_2=-1\to m_2=-\dfrac{1}{m_1}\\==========================\\\\\text{We have}\ x+4y=-6.\\\text{Convert to the slope-intercept form:}\\\\x+4y=-6\qquad\text{subtract}\ x\ \text{from both sides}\\\\4y=-x-6\qquad\text{divide both sides by 4}\\\\y=\dfrac{-x}{4}-\dfrac{6}{4}\\\\y=-\dfrac{1}{4}x-\dfrac{3}{2}\to m_1=-\dfrac{1}{4}$

$\text{Lines are to be parallel. Therefore}\ m_2=-\dfrac{1}{4}.\\\\\text{We initially have the form equation}\ y=-\dfrac{1}{4}x+b.\\\\\text{The line passes through the point}\ (9,\ -3).\\\\\text{Substitute the coordinates of the point to the equation of a line:}\\\\x=9,\ y=-3\\\\-3=-\dfrac{1}{4}(9)+b\\\\-3=-\dfrac{9}{4}+b\qquad\text{add}\ \dfrac{9}{4}\ \text{to both sides}\\\\-\dfrac{12}{4}+\dfrac{9}{4}=b\to b=-\dfrac{3}{4}$

$x=8,\ y=-3\\\\-3=-\dfrac{1}{4}(8)+b\\\\-3=-2+b\qquad\text{add 2 to both sides}\\\\-1=b\to b=-1$

$\text{Therefore the equation is:}\ y=-\dfrac{1}{4}x-1.\\\\\text{Convert to the standard form}\ Ax+By=C:\\\\y=-\dfrac{1}{4}x-1\qquad\text{multiply both sides by 4}\\\\4y=-x-4\qquad\text{add}\ x\ \text{to both sides}\\\\x+4y=-4$

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3 years ago

Need help fastt plss will give brainliest too who is corect plssss helppp

ehidna [41]

Answer:

0-5

Step-by-step explanation:

The Range Is 0-5 Atleast what i understand Gl Passing!

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3 years ago

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PLEASE HELP ME!! There is a photo attached. Tysm ❤️❤️

LiRa [457]

You have to show us the image Sherlock

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3 years ago

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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.