<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>
Answer:
Plot the points (0,0) and (1,3) to make the line.
Step-by-step explanation:
We know that a linear equation is y=mx+b, m is the slope and b is the y-intercept. We can see that the slope is m=3, and there is no b meaning b=0. Since the y-intercept is 0 we can start from the origin (0,0). The slope is 3/1 or known as rise/run. You go up 3 and go right 1, making the second point (1,3). This should work.
D , because 28-9=21 and 21=x
Given that
The cost of 1 m ribbon = Rs.75
The cost of 7/5 m ribbon
→ (7/5)×75
→ (7×75)/5
→7×15
→₹ 105
The cost of 7/5 m ribbon is ₹105.
Answer:
Step-by-step explanation:
In the diagram below we have
ABCD is a parallelogram. K is the point on diagonal BD, such that
And AK meets BC at E
now in Δ AKD and Δ BKE
∠AKD =∠BKE ( vertically opposite angles are equal)
since BC ║ AD and BD is transversal
∠ADK = ∠KBE ( alternate interior angles are equal )
By angle angle (AA) similarity theorem
Δ ADK and Δ EBK are similar
so we have
( ABCD is parallelogram so AD=BC)
( BC= BE+EC)
( subtracting 1 from both side )
taking reciprocal both side
