If ray np bisects angle mnq and measure of angle mnq= 8x + 12 and measure of pnq= 78 degrees and measure of angle rnm= 3y -9, find the values of x and y

9514 1404 393

Step-by-step explanation:

Given:

  ∠MNP +∠PNQ = ∠MNQ = (8x+12)°

  ∠MNP = ∠PNQ = 78° = (3y-9)°

Find:

  x, y

Solution:

The overall angle measure is ...

  78° +78° = (8x +12)°

  78 = 4x +6 . . . . . . . . . divide by 2°

  72 = 4x . . . . . . . . . subtract 6

  18 = x . . . . . . . . divide by 4

__

One angle measure is ...

  78° = (3y -9)°

  26 = y -3 . . . . . . divide by 3°

  29 = y . . . . . . . . add 3

The values of x and y are 18 and 29, respectively.

Comment on the question

We have interpreted the statement "angle RNM = 3y-9" to mean "angle PNM = 3y-9". This interpretation was chosen because point R is not mentioned or described anywhere else in the problem statement.

there are lots of ways.   pick's theorem tells us if the vertices are lattice points the area a is

a = b/2 + i - 1

where b is the number of boundary lattice points and i is the number of interior lattice points.

in the figure i count b=16, i=50 so

a = 8 + 50 - 1 = 57

let's try the shoelace formula.   the area is half the sum of the cross products of the sides.  

a(-2,6)   -2(2)-6(-5) = 26

b(-5,2)   -5(-3) - 2(-2) = 19

c(-2,-3) -2(-3) - (-3)(4) = 18

d(4,-3)   4(3) - (-3)(5) = 27

e(5,3)   5(3) - 3(1) = 12

f(1,3)     1(6) - 3(-2) = 12

a(-2,6)

each of the cross products is twice the area of the triangle with the associated side and one vertex the origin. so the area of abo=26/2=13.

area = (1/2) (26 + 19 +   18 + 27 + 12 + 12) = 57

that checks!

c

step-by-step explanation:

there is vertical asymptote at x = 2 and x = 0

vertical asymptote occurs at the points where the denominator is 0. so we can say that

x = 2 occurs when there is a factor (x-2) in the denominator, and

x = 0 occurs when there is a factor "x" in the denominator

thus, the denominator of the equation would have "x * (x-2)", this is option c only. that's the correct answer.