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il63 [147K]
3 years ago
6

Find the distance between the points given. (-10,3) and (-10,12)

Mathematics
1 answer:
aksik [14]3 years ago
5 0

Answer:

The distance is 9.

Step-by-step explanation:

Just look, that  these points have got the same x-coordinate, 

so they lie  on the same line, which is parallel to the OY axis, 

So  your taks is only  find different  in y-coordinate. That is

12-3 = 9 

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Put the values of x to the equations of the functions:

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2. f(-2) → x = -2; f(x) = 4x - 1

f(-2) = 4(-2) - 1 = -8 - 1 = -9

3. f(-5) → x = -5; f(x) = -2x + 8

f(-5) = -2(-5) + 8 = 10 + 8 = 18

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2 years ago
X-y+z=-4<br><br> 3x+2y-z=5<br><br> -2x+3y-z=15<br><br> How do I solve this?
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1)\ \ \ x-y+z=-4\ \ \ \Rightarrow\ \ \ z=-4-x+y\\\\2)\ \ \ 3x+2y-z=5 \\.\ \ \ \  \Rightarrow\ \ 3x+2y-(-4-x+y)=5 \\.\ \ \ \  \Rightarrow\ \ 3x+2y+4+x-y=5 \\.\ \ \ \  \Rightarrow\ \ 4x+y=1\\\\3)\ \ \ -2x+3y-z=15\\.\ \ \ \  \Rightarrow\ \ -2x+3y-(-4-x+y)=15\\.\ \ \ \  \Rightarrow\ \ -2x+3y+4+x-y=15\\.\ \ \ \  \Rightarrow\ \ -x+2y=11\\--------------------\\

z=-4-x+y\\.\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ and\\ \left \{ {{4x+y=1\ \ \ \ } \atop {-x+2y=11\ /\cdot4}} \right. \\\\\left \{ {{4x+y=1\ \ \ \ } \atop {-4x+8y=44}} \right. \\-------\\y+8y=1+44\\9y=45\ /:9\\y=5\\\\-x+2y=11\ \ \ \Rightarrow\ \ \ x=2y-11\ \ \ \Rightarrow\ \ \ x=2\cdot5-11=-1\\\\z=-4-x+y=-4-(-1)+5=-4+1+5=2\\\\Ans.\ x=-1\ \ \ and\ \ \ y=5\ \ \ and\ \ \ z=2
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3 years ago
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