Plot the points d (-6,-2) and t (7,-8). describe how to find the distance between d and t using the pythagorean theorem

Step-by-step explanation:

* Lets explain how to solve the problem

- The coordinates of point D are (-6 , -2)

- The coordinates of point T are (7 , -8)

- We want to find the distance between D and T using Pythagorean

 Theorem

- Pythagorean Theorem is:

 in any right angle triangle the length of the hypotenuse (opposite

 side to the right angle) = square root of the sum of the squares of

 the other two sides

- So to make the right angle with points D and T, we must to draw a

  horizontal segment from point D and vertical segment from point T

  cross each other at point N whose coordinates are (x , y)

- Look to the attached graph for more understand

- D is the red point , N is the green point and T is the blue point

* Lets find x and y

- The y-coordinates of all points lie on a horizontal segment are equal

∵ DN is a horizontal segment

∴ y-coordinate of point D = y-coordinate of point N

∵ y-coordinate of point D is -2

∵ y- coordinate of point N = -2

∴ y = -2

- The x-coordinates of all points lie on a vertical segment are equal

∵ TN is a vertical segment

∴ x-coordinate of point T = x-coordinate of point N

∵ x-coordinate of point T is 7

∵ x- coordinate of point N = 7

∴ x = 7

∴ The coordinates of point N are (7 , -2)

* Lets find the length of DN and TN

∵ The length of the horizontal segment is the difference between

   the x-coordinates of its endpoints

∵ x-coordinate of D is -6 and x-coordinate of N is 7

∴ DN = 7 - (-6) = 7 + 6 = 13

∵ The length of the vertical segment is the difference between

   the y-coordinates of its endpoints

∵ y-coordinate of T is -8 and y-coordinate of N is -2

∴ TN = (-2) - (-8) = -2 + 8 = 6

* Lets use Pythagorean Theorem to find DT

- In Δ DNT

∵ m∠N = 90°

∵ DT is opposite to ∠N

∴

∴

∴

option 2 is correct.

step-by-step explanation:

we are given two inequality equation

x+y> 5 and -3x+2y< -2

first we draw the graph of both function using t-table method

equation: x+y> 5

for x=0, y=5

for x=-5, y=0

table:

x   :   0     5

y   :   5     0

test point (0,0) , 0> 5

false

equation: -3x+2y< -2

for x=0, y=-1

for x=2, y=4

table:

x   :   0     2

y   : -1       4

test point (0,0) , 0< -2

true

now we draw the graph. see the attachment for graph.

common region would be solution set.

option 2 is correct.