Answer:
To prove that ( sin θ cos θ = cot θ ) is not a trigonometric identity.
Begin with the right hand side:
R.H.S = cot θ = 
L.H.S = sin θ cos θ
so, sin θ cos θ ≠
So, the equation is not a trigonometric identity.
=========================================================
<u>Anther solution:</u>
To prove that ( sin θ cos θ = cot θ ) is not a trigonometric identity.
Assume θ with a value and substitute with it.
Let θ = 45°
So, L.H.S = sin θ cos θ = sin 45° cos 45° = (1/√2) * (1/√2) = 1/2
R.H.S = cot θ = cot 45 = 1
So, L.H.S ≠ R.H.S
So, sin θ cos θ = cot θ is not a trigonometric identity.
Answer:
Your y intercept is 1.
Step-by-step explanation:
slope is y2-y1 over x2-x1, or 2.
slope intercept formula is y=mx+b, and if you plug values into formula you get 3=2(1)+b
and if you solve that, 2x1=2, 3-2=1.
then you get 1 as your y intercept.
Answer:
2
Step-by-step explanation:
Take any 2 points on the graph,
(50, 10) & (90, 90)
Use the slope formula:
= (90-10)/(90-50)
= (80)/(40)
= 2
Hence, the slope is 2.
<em>Feel free to mark this as brainliest! :D</em>
The perimeter of a shape is the sum of the lengths of its sides.
So, to find the perimeter of this quadrilateral, all we have to do is add the side lengths and simplify.
(x² - 6) + (2x + 5) + (x² - 3x) + (4x² + 2x)
x² + x² + 4x² + (-3x) + 2x + 2x + (-6) + 5
6x² + (-3x) + 2x + 2x + (-6) + 5
6x² + x + (-6) + 5
6x² + x + (-1)
6x² + x - 1
So, the perimeter of the quadrilateral is the quantity (6x² + x - 1).
Hope this helps!
Answer:
c
Step-by-step explanation:
