Whenever you want to write the equation of a parallel line through some point (h, k), you can ...
- remove any added constant in the original given equation
- replace x with (x-h)
- replace y with (y-k)
- rearrange the resulting equation to the form required by the problem.
Using this formula here, we get
... 2(y +5) = 3(x -2)
Your answer form requires you solve this for y.
... 2y + 10 = 3x -6 . . . . . eliminate parentheses
... 2y = 3x -16 . . . . . . . . subtract the constant on the left (10)
... y = (3/2)x -8 . . . . . . divide by 2
Answer:
cos(θ) = 3/5
Step-by-step explanation:
We can think of this situation as a triangle rectangle (you can see it in the image below).
Here, we have a triangle rectangle with an angle θ, such that the adjacent cathetus to θ is 3 units long, and the cathetus opposite to θ is 4 units long.
Here we want to find cos(θ).
You should remember:
cos(θ) = (adjacent cathetus)/(hypotenuse)
We already know that the adjacent cathetus is equal to 3.
And for the hypotenuse, we can use the Pythagorean's theorem, which says that the sum of the squares of the cathetus is equal to the square of the hypotenuse, this is:
3^2 + 4^2 = H^2
We can solve this for H, to get:
H = √( 3^2 + 4^2) = √(9 + 16) = √25 = 5
The hypotenuse is 5 units long.
Then we have:
cos(θ) = (adjacent cathetus)/(hypotenuse)
cos(θ) = 3/5
A tangent is always perpendicular to the radius at the point of tangency.
A right angle.
Answer:
D)
Step-by-step explanation:
First we have to subtract 5 from both sides
Now, square both sides
x - 6 = 49
Now, add 6 to both sides
x - 6 +6 = 49 + 6
x = 55
