Let U = {1,2,3,4,5,6,7,8,9,10}, A = {1,3,5,7,9}, B = {2,4,6,8,10} and C = {1,2,3,4} find (i) U' ii) A∩A' iii) A – ( B U C) iv) A
Artyom0805 [142]
Answer:
(i) U' = Φ
(ii) A∩A' = Φ
(iii) A – ( B U C) = {5,7,9}
(iv) A' U ( B U C ) = {2,4,6,8,10}
(v) A' U ( B' ∩ C') = {2,4,5,6,7,8,9,10}
Step-by-step explanation:
Answer: y=-(1/8)x+(21/2)
Explanation:
The new equation’s slope needs to be perpendicular to y=8x-1
To get the new slope, we take the negative reciprocal of the slope from above
8 —> -1/8
Now we have y=-(1/8)x+b but it needs to pass through (4,10), so we need to find the value of b that makes this possible.
Since (4,10) is in the form of (x,y) we can plug in these values into the new equation to solve for b:
y=-(1/8)x+b
10=-(1/8)4+b
10=(-1/2)+b
b=(21/2)
Now put b back into the new equation
y=-(1/8)x+b
y=-(1/8)x+(21/2)
Answer: sin u = -5/13 and cos v = -15/17
Step-by-step explanation:
The nice thing about trig, a little information goes a long way. That’s because there is a lot of geometry and structure in the subject. If I have sin u = opp/hyp, then I know opp is the opposite side from u, and the hypotenuse is hyp, and the adjacent side must fit the Pythagorean equation opp^2 + adj^2 = hyp^2.
So for u: (-5)^2 + adj^2 = 13^2, so with what you gave us (Quad 3),
==> adj of u = -12 therefore cos u = -12/13
Same argument for v: adj = -15,
opp^2 + (-15)^2 = 17^2 ==> opp = -8 therefore sin v = -8/17
The cosine rule for cos (u + v) = (cos u)(cos v) - (sin u)(sin v) and now we substitute: cos (u + v) = (-12/13)(-15/17) - (-5/13)(-8/17)
I am too lazy to do the remaining arithmetic, but I think we have created a way to approach all of the similar problems.
Answer:
12 m x 10 m
Step-by-step explanation:
can you help me with my question In Paragraph 206 of "The Open Boat," Crane describes the shore the dinghy is approaching in naturalist terms as A welcoming B. indifferent C. treacherous
You could multiply it by .65 because 95.65 is equal to 61.75.
Answer is .65
