If you know that
, then you know right away

###
Otherwise, you can derive the same result. Let
, so that
.
is bounded, so we know
. For these values of
, we always have
.
So, recalling the Pythagorean theorem, we find

Then

as expected.
15,785 and well try to do your best haha
Answer:
24
Step-by-step explanation:
6+6 =12
12+6 = 18
18+6 = 24
The speed of wind and plane are 105 kmph and 15 kmph respectively.
<u>Solution:</u>
Given, it takes 6 hours for a plane to travel 720 km with a tail wind and 8 hours to make the return trip with a head wind.
We have to find the air speed of the plane and speed of the wind.
Now, let the speed wind be "a" and speed of aeroplane be "b"
And, we know that, distance = speed x time.

Now at head wind → 
So, solve (1) and (2) by addition
2a = 210
a = 105
substitute a value in (1) ⇒ 105 + b = 120
⇒ b = 120 – 105 ⇒ b = 15.
Here, relative speed of plane during tail wind is 120 kmph and during head wind is 90 kmph.
Hence, speed of wind and plane are 105 kmph and 15 kmph respectively.
Answer:
We can see that this is dependent probability. We can find dependent probability of happening event A then event B by multiplying probability of event A by probability of event B given that event A already happened.
Step-by-step explanation:
In our case event A is pirate hitting captain's ship and event B is captain missing pirate's ship. We have been given that pirate shoots first so pirate's ship can't be hit before pirate shoots his cannons. So probability of hitting captain's ship is 1/3. We have been given that if Captain Ben's ship is already hit then Captain Ben will always miss. So the probability of Captain missing the dread pirate's ship given the pirate Luis hitting the Captain ship is 1. Now to find probability that pirate hits Captain, but Captain misses we will multiply our both probabilities.