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3 years ago

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Find the midpoint M of the line segment joining the points P = (-6, -8) and Q = (2, 2).

1 answer:

goldenfox [79]3 years ago

5 0

I HOPE IT WILL HELP YOU.

Have a great day ahead.

Thank you.

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I need some help please! ^_^

Degger [83]

Perpendicular lines will have negative reciprocal slopes. What that means is if u have a slope of 1/2, to find the negative reciprocal, flip the slope and change the sign.......flip 2/1, change the sign -2/1 or just -2. So the negative reciprocal for the slope of 1/2 is -2.

A. y = 1/5x + 3.....slope here is 1/5, so for a perpendicular line, u r gonna need an equation with the slope of -5.....and that would be : y + 3 = -5(x + 2).

B. Parallel lines will have the same slope. y = 5x - 2...the slope here is 5...so a parallel line will have a slope of 5.

y = mx + b
slope(m) = 5
(8,-2)...x = 8 and y = -2
now we sub, we r looking for b, the y intercept
-2 = 5(8) + b
-2 = 40 + b
-2 - 40 = b
-42 = b

so ur parallel equation is : y = 5x -42

7 0

4 years ago

4/5(20e - 5q) - 1/9(-27q + - 18e)<br><br>help

Lady bird [3.3K]

Answer: =−q+48.929073 <-- coming frm a calculator lol

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

Which of these relations on {0, 1, 2, 3} are equivalence relations? Determine the properties of an equivalence re- lation that t

slava [35]

Answer:

The relations that are equivalence relations are a) and c)

Step-by-step explanation:

A relation on a set A is called an equivalence relation if it is reflexive, symmetric, and transitive

We are going to analyze each one.

a){ (0,0), (1,1), (2,2), (3,3) }

Is an equivalence relation because it has all the properties.

b){ (0,0), (0,2), (2,0), (2,2), (2,3), (3,2), (3,3) }

Is not an equivalence relation. Not reflexive: (1,1) is missing, not transitive: (0,2) and (2,3) are in the relation, but not (0,3)

c){ (0,0), (1,1), (1,2), (2,1), (2,2), (3,3) }

Is an equivalence relation because it has all the properties.

d){ (0,0), (1,1), (1,3), (2,2), (2,3), (3,1), (3,2) (3,3) }

Is not an equivalence relation. Not transitive: (1,3) and (3,2) are in the relation, but not (1,2)

e){ (0,0), (0,1) (0,2), (1,0), (1,1), (1,2), (2,0), (2,2), (3,3) }

Is not an equivalence relation. Not symmetric: (1,2) is present, but not (2,1)Not transitive: (2,0) and (0,1) are in the relation, but not (2,1)

5 0

3 years ago

Radius of a circular ground is 21m. Find.

trapecia [35]

The circumference of a circle of radius $r$ is

$C=2\pi r$

So, in your case, the circumference is

$C=2\cdot \pi \cdot 21 = 42\pi$

The circumference is actually the length of the "perimeter", so the answer to questions 1 and 2 looks the same actually

6 0

3 years ago

Directed line segment has endpoints P(– 8, – 4) and Q(4, 12). Determine the point that partitions the directed line segment in a

larisa86 [58]

Answer:

Point (1,8)

Step-by-step explanation:

We will use segment formula to find the coordinates of point that will partition our line segment PQ in a ratio 3:1.

When a point divides any segment internally in the ratio m:n, the formula is:

$[x=\frac{mx_2+nx_1}{m+n},y= \frac{my_2+ny_1}{m+n}]$

Let us substitute coordinates of point P and Q as:

$x_1=-8$ ,

$y_1=-4$

$x_2=4$

$y_2=12$

$m=3$

$n=1$

$[x=\frac{(3*4)+(1*-8)}{3+1},y=\frac{(3*12)+(1*-4)}{3+1}]$

$[x=\frac{12-8}{4},y=\frac{36-4}{4}]$

$[x=\frac{4}{4},y=\frac{32}{4}]$

$[x=1,y=8]$

Therefore, point (1,8) will partition the directed line segment PQ in a ratio 3:1.

7 0

3 years ago