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3 years ago

Which is true about the solution to the system of inequalities shown?

Mathematics

1 answer:

EleoNora [17]3 years ago

8 0

they don't share any points

that's one thing, but I don't know what your options are of course.

see screenshot for illustration of the inequalities

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Please round to nearest hundredth and give explanation.

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Solve |y - 2|<10<br><br> {y|-8 < y < 12}<br> {y|-12 < y < 8}<br> {y|-8 < y < 8}

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3 years ago

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3 years ago

Segment PH is graphed on the coordinate grid where P is (-31,8) and H is (29,-10). Point E is the midpoint of segment PH. If poi

maks197457 [2]

Answer:

The coordinates of point A are (-11 , 2)

Step-by-step explanation:

- The coordinates of point P are (-31 , 8)

- The coordinates of point H are (29 , -10)

- Point E is the mid-point of segment PH

- The coordinates of the the mid-point are:

→ $x=\frac{x_{1}+x_{2}}{2}$ and $y=\frac{y_{1}+y_{2}}{2}$

∵ E is the mid-point of segment PH

∴ The x-coordinate of point E is $x=\frac{-31+29}{2}=-1$

∴ The x-coordinate of point E is -1

∴ The y-coordinate of point E is $y=\frac{8+-10}{2}=-1$

∴ The y-coordinate of point E is -1

∴ The coordinates of point E are (-1 , -1)

- Point A is graphed $\frac{2}{3}$ of the way along the segment from P to E

- That mean the distances from points P to A is 2 parts and from P to E

is 3 parts, then the distance from A to E is "3 - 2" = 1 part

- So point A divides the line segment PE at ratio 2 : 1 from P

- The coordinates of the division point are:

→ $x=\frac{x_{1}m_{2}+x_{2}m_{1}}{m_{1}+m_{2}}$ and $y=\frac{y_{1}m_{2}+y_{2}m_{1}}{m_{1}+m_{2}}$

∵ Point A divides PE at ratio 2 : 1

∵ The coordinates of point P are (-31 , 8)

∵ The coordinates of point E are (-1 , -1)

∴ The x-coordinate of point A is $x=\frac{(-31)(1)+(-1)(2)}{2+1}= -11$

∴ The x-coordinate of point A is -11

∴ The y-coordinate of point A is $x=\frac{(8)(1)+(-1)(2)}{2+1}=2$

∴ The y-coordinate of point A is 2

∴ The coordinates of point A are (-11 , 2)

* The coordinates of point A are (-11 , 2)

3 0

3 years ago