Find the equation of the line that is perpendicular to y = x+1
and contains the point (9,12).

y = -x + 21

Step-by-step explanation:

Equations that are perpendicular don't normally have the same slope as the original equation..

When you graph the given equation and the point, we can see that in order to make an equation that would cross through (9,12) and the line, we would need an equation with a negative slope.

**Using desmos (online graphing calculator)

If we continue to plug in numbers for our slope, we see that the equation

y = -x + 21  goes through point (9,12) and is perpendicular (intersecting or crossing through) line y = x + 1.

We can check our work by plugging in the coordinates of the point 9, 12 into our equations to see if they equal the same values..

y = -x + 21 --> 12 = -9 + 21 --> 12 = 21 - 9 --> 12 = 12 (correct)

x=half

step-by-step explanation:

9) i can't see the angle so whatever it is just do this: cos(angle) x hypotenuse and then use that answer and they hypotenuse and plug it into this: (answer you just got)^2 + b^2 = (hypotenuse)^2. the first answer will be the answer of the adjacent side and the second answer after this equation will be the opposite side.

10) y = 14 x = √147 or 12.12435565

11) x = 4.8

12) x = 4.2

13) x = 9.3

14) pq: 13.2

qr : 17.6

angle p: 53º

15) angle c: 54.3º

angle a: 35.7º

16) h = 21040

v = 1841

step-by-step explanation:

10) 7/sin(30) = 14 = y

7^2 + b^2 = 14^2

49 + b^2 = 196

b^2 = 196 - 49

√b^2 = √147

x = √147

11) cos(70) x 14 = 4.788282007

x = 4.8

12) tan(25) x 9 = 4.196768923

x = 4.2

13) sin(31) x 18 = 9.270685348

x = 9.3

14) pq: sin(37) x 22 = 13.23993051

pq: 13.2

qr: cos(37) x 22 = 17.56998122

qr : 17.6

angle p: 180 - (37 + 90) = 53º

15) cos-1(7/12) = 54.31466529

angle c: 54.3º

angle a: 180 - (90 + 54.31466529) = 35.68533471

angle a: 35.7º

16) 4 x 5280 = 21120

cos(5) x 21120

h = 21039.63202

h = 21040

sin(5) x 21120

v = 1840.729287

v = 1841