Determine the domain and the range for this relationship.

Mathematics

2 answers:

Jlenok [28]3 years ago

7 0

Answer:

I can't drag duder.

Step-by-step explanation:

Vedmedyk [2.9K]3 years ago

6 0

Answer:

how are we supposed to know the answer if we don't know the problem?

Step-by-step explanation:

It makes no sense

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1. L(15. 1) is the midpoint of the straight line joining point (p. - 2) to point D(-1. q) find p and q.

kkurt [141]

1. The values of p and q are: p=31 and q= 4

2. B(11, 29/5)

Further explanation:

<u>1. L(15. 1) is the midpoint of the straight line joining point (p. - 2) to point D(-1. q) find p and q.</u>

Given:

M = (15. 1)

(x1, y1) = (p, -2)

(x2, y2) = (-1, q)

The formula for mid-point is:

$(\frac{x_1+x_2}{2} , \frac{y_1+y_2}{2}) = M \\Putting\ the\ values\\(\frac{p-1}{2} , \frac{-2+q}{2}) = (15,1)\\Putting\ realtive\ values\ equal\\\frac{p-1}{2} = 15\\p-1 = 15(2)\\p-1 = 30\\p = 30+1\\p = 31\\\frac{-2+q}{2} =1\\-2+q = 2(1) \\-2+q = 2\\q = 2+2 \\q =4$

Hence,

p=31

q=4

<u>2. M is the midpoint of the straight line joining point A (3. 1/5) to point B.If m has coordinates (7. 3), find the coordinates of B.</u>

Here,

(x1,y1) = (3, 1/5)

(x2, y2) = ?

M(x,y) = (7,3)

Putting values in the formula of mid-point

$(\frac{x_1+x_2}{2} , \frac{y_1+y_2}{2}) = M\\(\frac{3+x_2}{2} , \frac{1/5+y_2}{2}) = (7,3)\\\frac{3+x_2}{2} = 7\\3+x_2 = 7*2\\3+x_2 = 14\\x_2 = 14-3\\x_2 = 11\\\frac{\frac{1}{5}+y_2}{2} = 3\\{\frac{1}{5}+y_2} = 3*2\\{\frac{1}{5}+y_2} = 6\\y_2 = 6 - \frac{1}{5}\\y_2 = \frac{30-1}{5}\\y_2 = \frac{29}{5}$