3.5555555556 / the third one
The last one and the first one
hope you got it
The square of a binomial expands as follows:
![(a+b)^2=a^2+2ab+b^2](https://tex.z-dn.net/?f=%28a%2Bb%29%5E2%3Da%5E2%2B2ab%2Bb%5E2)
Expanding all the binomials, we have
![4k^2-36k+81+k^2+4k+4=4k^2+4k+1+k^2-16k+64](https://tex.z-dn.net/?f=4k%5E2-36k%2B81%2Bk%5E2%2B4k%2B4%3D4k%5E2%2B4k%2B1%2Bk%5E2-16k%2B64)
Sum like terms:
![5k^2-32k+85=5k^2-12k+65](https://tex.z-dn.net/?f=5k%5E2-32k%2B85%3D5k%5E2-12k%2B65)
Simplify terms appearing on both sides:
![-32k+85=-12k+65](https://tex.z-dn.net/?f=-32k%2B85%3D-12k%2B65)
Move all terms involving k to the left and all numbers to the right:
![-20k=-20](https://tex.z-dn.net/?f=-20k%3D-20)
Divide both sides by -20:
![k=1](https://tex.z-dn.net/?f=k%3D1)
Answer:
x = 6
Step-by-step explanation:
move all terms to the left:
-1.5x+12-(2.5x-12)=0
get rid of parentheses
-1.5x-2.5x+12+12=0
add all the numbers together, and all the variable.
-4x+24=0
move all terms containing x to the left, all other terms to the right
-4x=-24
x=-24/-4 - divide
x=+6
Answer: The minimum reliability for the second stage be 0.979.
Step-by-step explanation:
Since we have given that
Probability for the overall rocket reliable for a successful mission = 97%
Probability for the first stage = 99%
We need to find the minimum reliability for the second stage :
So, it becomes:
P(overall reliability) = P(first stage ) × P(second stage)
![0.97=0.99\times x\\\\\dfrac{0.97}{0.99}=x\\\\0.979=x](https://tex.z-dn.net/?f=0.97%3D0.99%5Ctimes%20x%5C%5C%5C%5C%5Cdfrac%7B0.97%7D%7B0.99%7D%3Dx%5C%5C%5C%5C0.979%3Dx)
Hence, the minimum reliability for the second stage be 0.979.