First thing to do is to change the radians to degrees so it's easier to determine our angle and where it lies in the coordinate plane.
. If we sweep out a 210 degree angle, we end up in the third quadrant, with a 30 degree angle. In this quadrant, x and y are both negative, but the hypotenuse, no matter where it is, will never ever be negative. So the side across from the 30 degree reference angle is -1, and the hypotenuse is 2, so the sine of this angle, opposite over hypotenuse, is -1/2
15/6=2.5 which is the scale factor since they are similar triangles. We divide each side of the large triangle by 2.5 to get the sides for the small triangle since they are similar. 20/2.5=8cm
25/2.5=10cm
For the second problem, you put 15/6 and set in equal to x/8 when you cross multiply you get 6x=120 then you divide x=20
For the last problem, you divide 43,75 by 5 then multiply it by 400 to get the kiloliters which equals 3,500.
Step-by-step explanation:
a). A = {x ∈ R I 5x-8 < 7}
5x - 8 < 7 <=> 5x < 8+7 <=> 5x < 15 =>
x < 3 => A = (-∞ ; 3)
A ∩ N = {0 ; 1 ; 2}
A - N* = (-∞ ; 3) - {1 ; 2}
b). A = { x ∈ R I 7x+2 ≤ 9}
7x+2 ≤ 9 <=> 7x ≤ 7 => x ≤ 1 => x ∈ (-∞ ; 1]
A ∩ N = {0 ; 1}
A-N* = (-∞ ; 1)
c). A = { x ∈ R I I 2x-1 I < 5}
I 2x-1 I < 5 <=> -5 ≤ 2x-1 ≤ 5 <=>
-4 ≤ 2x ≤ 6 <=> -2 ≤ x ≤ 3 => x ∈ [-2 ; 3]
A ∩ N = {0 ; 1 ; 2 ; 3}
A - N* = [-2 ; 3) - {1 ; 2}
d). A = {x ∈ R I I 6-3x I ≤ 9}
I 6-3x I ≤ 9 <=> -9 ≤ 6-3x ≤ 9 <=>
-15 ≤ -3x ≤ 3 <=> -5 ≤ -x ≤ 3 =>
-3 ≤ x ≤ 5 => x ∈ [-3 ; 5]
A ∩ N = {0 ; 1 ; 2 ; 3 ; 4 ; 5}
A - N* = [-3 ; 5) - {1 ; 2 ; 3 ; 4}
Without resorting to L'Hopitâl's rule,
With the rule, we get the same result:
