The answer is 10 because 5×2 is 10 and 10 times one is 10. So that's the lowest common denominator her.
Hello :
P(x) + P(x') =1 ...( x' is the <span>complement of x)
0.3 +</span> P(x') =1
P(x') = 1 - 0.3 = 0.7
<span>x = 9
Since ZP bisects â OZQ, that means that the measurements for â OZP and â PZQ are the same. So create an equation with their respective values set to each other.
8x - 9 = 5x + 18
Now solve for x
8x - 9 = 5x + 18
Subtract 5x from both sides
3x - 9 = 18
Add 9 to both sides
3x = 27
Divide both sides by 3
x = 9</span>
<span>N(t) = 16t ; Distance north of spot at time t for the liner.
W(t) = 14(t-1); Distance west of spot at time t for the tanker.
d(t) = sqrt(N(t)^2 + W(t)^2) ; Distance between both ships at time t.
Let's create a function to express the distance north of the spot that the luxury liner is at time t. We will use the value t as representing "the number of hours since 2 p.m." Since the liner was there at exactly 2 p.m. and is traveling 16 kph, the function is
N(t) = 16t
Now let's create the same function for how far west the tanker is from the spot. Since the tanker was there at 3 p.m. (t = 1 by the definition above), the function is slightly more complicated, and is
W(t) = 14(t-1)
The distance between the 2 ships is easy. Just use the pythagorean theorem. So
d(t) = sqrt(N(t)^2 + W(t)^2)
If you want the function for d() to be expanded, just substitute the other functions, so
d(t) = sqrt((16t)^2 + (14(t-1))^2)
d(t) = sqrt(256t^2 + (14t-14)^2)
d(t) = sqrt(256t^2 + (196t^2 - 392t + 196) )
d(t) = sqrt(452t^2 - 392t + 196)</span>