If two lines are perpendicular and neither one is vertical, then one of the lines has a positive slope, and the other has a negative slope. Hope this helps.

Step-by-step explanation:

6 0

2 years ago

Please help me solve this

MArishka [77]

9514 1404 393

Answer:

sin(θ) ≈ -0.92

cos(θ) ≈ 0.38

tan(θ) = -2.40

Step-by-step explanation:

Let r represent the distance of the point from the origin. The Pythagorean theorem tells us ...

r² = (5)² + (-12)²

r² = 169

r = √169 = 13

The trig relations for a point on the terminal ray are ...

(x, y) = (r·cos(θ), r·sin(θ))

Then ...

sin(θ) = y/r = -12/13 ≈ -0.92

cos(θ) = x/r = 5/13 ≈ 0.38

tan(θ) = y/x = -12/5 = -2.40

7 0

2 years ago

Find the point of intersection (if any) of the following pairs of lines: a) x – 2y +1=0 2x + 3y – 7 = 0 b) x-2y+11=0 -x+2y-13 =

pychu [463]

Answer:

a) (11/7, 9/7)

b) There's no point of intersection

Step-by-step explanation:

a) x - 2y + 1 = 0

2x + 3y - 7 = 0

To find the point of intersection, we need to solve the system of equations and the result will be the point of intersection (x,y)

$x-2y+1=0\\x= 2y-1$

Now we substitute x in the second equation:

$2x+3y-7=0\\2(2y-1)+3y-7=0\\4y-2+3y-7=0\\7y-9=0\\y=9/7$

Now we substitute y in our first equation.

$x-2y+1=0\\x-2(9/7)+1=0\\x-18/7+1=0\\x=18/7-1\\x=11/7$ .

The point of intersection is (11/7, 9/7)

b) x -2y +11 =0

-x + 2y - 13 =0

We are going to follow the same procedure:

$x-2y+11=0\\x=2y-11$

$-(2y-11)+2y-13=0\\-2y+11+2y-13=0\\0y=2\\0=2$

Since this system of equations doesn't have a solution, the system has no point of intersection.

5 0

2 years ago

(5x10,000) +(2x100) +(9x1) +(3x1/1000)

zubka84 [21]

Scientific notation:
5.0209003 x 10^4

7 0

2 years ago