Find the distance between the points (2, –3) and (6, 1).?

Question

Problem 2: Find the real solution set to the following equation.
(w – 1) –2/5 = 4

note: -2/5 is a fraction

Answer 1

The distance formula is

d = √(x2 - x1) ² + (y2 - y1)²

d = √(2 - 6)² + (-3 - 1)²

d = √(-4)² + (-4)²

d = √16 + 16

d = √32

d = 5.656854249

d = 5.66 rounded to two decimal places

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(w - 1) - 2/5 = 4

The fraction must be cleared by multiplying both sides of the equation by 5.

5(w - 1) - (5)2/5 = (5)4

5w - 5 - 2 = 20

The distributive property.

5w - 7 = 20
+ 7 +7

add 7 to both sides

5w = 27

5w/5 = 27/5

divide 5 from both sides

w = 27/5

w = 5.4

Answer 2

use pythagoras
subtract the x-coordinates, and the y-coordinates these are the lengths of you triangle.
thus the hypotenusa of this triangle is :
sqrt( (2-6)^2 + (-3-1)^2 ) = sqrt(32)

prob2)
the trich is to move all the unknowns to the left side, and the knowns(that are the numbers) to the right side
when moving from one side to the other the sign changes

w = 4 + (2/5) + 1

Answer 3

mom exchange into proper, yet why is it Plus? the gap formulation is ?[(X2 - X1)² + (Y2 - Y1)²] = ?[(4 - -6)² + (2 - -3)^2] = ?[(10)² + (5)²] = ?[one hundred + 25] = ?[a hundred twenty five] = 5?5 Subtracting unfavourable numbers is often the reason for blunders, be careful. it seems this consequence isn't between the recommendations. Did you write the recommendations properly?

Answer 4

distance between (2,-3) and (6,1), it is

(6-2)^2 + (1+3)^2

=16 + 16 = 32

distance = sqrt of 32 = 5.66

Answer 5

(2,-3) and (6,1)
D = sqrt((x2 - x1)^2 + (y2 - y1)^2)
D = sqrt((6 - 2)^2 + (1 - (-3))^2)
D = sqrt(4^2 + (1 + 3)^2)
D = sqrt(16 + 4^2)
D = sqrt(16 + 16)
D = sqrt(32)
D = sqrt(16 * 2)
D = 4sqrt(2)

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(w - 1) - (2/5) = 4
5(w - 1) - 2 = 20
5(w - 1) = 22
w - 1 = (22/5)
w = 1 + (22/5)
w = (5/5) + (22/5)
w = (5 + 22)/5
w = 27/5
w = 5.4

Answer 6

1. 6-2=4 and 1-(-3)=4
4^2+4^2=32
32^.5=5.65685

2. w-1=4+2/5
w=5+2/5

Answer 7

6-2 =4
1-(-3)=4
sqrt(4^2+4^2)=4*1.41=5.6
w=22/5+1
w=27/5
w=5.4

Answer 8

distance between 2 pts (x1,y1) and (x2,y2) is given by

distance = ( (x2-x1)^2 + (y2-y1)^2 )^(1/2)
i.e distance = squareroot of : (x2-x1)^2 + (y2-y1)^2