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.
Well there are 12 players and 9 are in 8th grade so the rest is leaving 3 so 3/12 so converted to percentage is 25% hoped this helped
Answer:
Therefore x=9
Step-by-step explanation:
Give that,
log x+ log (x-3) = log 54
log x(x-3) =log 54 [
]
x(x-3) = 54
Now expand the above equation to get a quadratic equation
x²-3x=54
⇒x²-9x+6x-54=0
⇒x(x-9)+6(x-9)=0
⇒(x-9)(x+6)=0
⇒x-9=0 or, x+6=0
⇒x=9,-6
Now putting x= -6 in the given expression,
log(-6)+log(-6-9)=log 54.
Since log (-6) does not exist.
Therefore x=9
Answer:
The two iterations of f(x) = 1.5598
Step-by-step explanation:
If we apply Newton's iterations method, we get a new guess of a zero of a function, f(x), xₙ₊₁, using a previous guess of, xₙ.
xₙ₊₁ = xₙ - f(xₙ) / f'(xₙ)
Given;
f(xₙ) = cos x, then f'(xₙ) = - sin x
cos x / - sin x = -cot x
substitute in "-cot x" into the equation
xₙ₊₁ = xₙ - (- cot x)
xₙ₊₁ = xₙ + cot x
x₁ = 0.7
first iteration
x₂ = 0.7 + cot (0.7)
x₂ = 0.7 + 1.18724
x₂ = 1.88724
second iteration
x₃ = 1.88724 + cot (1.88724)
x₃ = 1.88724 - 0.32744
x₃ = 1.5598
To four decimal places = 1.5598
Answer: 2%, second option is correct.
Step-by-step explanation:
To state 1/50 in percent, divide 1 by 50, then multiply by 100
=( 1 ÷ 50) x 100
= 0.02 x 100
= 2%
I hope this helps, please mark as brainliest.