Answer: x=3
Step-by-step explanation:
To solve this problem you must apply the proccedure shown below:
- Multiply both sides of the equation by 3:
![3(2x-3)=3(\frac{x^2}{3})\\6x-9=x^{2}](https://tex.z-dn.net/?f=3%282x-3%29%3D3%28%5Cfrac%7Bx%5E2%7D%7B3%7D%29%5C%5C6x-9%3Dx%5E%7B2%7D)
- Now you must make the equation equal to zero as following:
![6x-9=x^{2}\\x^{2}-6x+9=0](https://tex.z-dn.net/?f=6x-9%3Dx%5E%7B2%7D%5C%5Cx%5E%7B2%7D-6x%2B9%3D0)
- Factor it, as you can see below. Therefore, you obtain the following result:
![(x-3)(x-3)=0\\(x-3)^2=0\\x=3](https://tex.z-dn.net/?f=%28x-3%29%28x-3%29%3D0%5C%5C%28x-3%29%5E2%3D0%5C%5Cx%3D3)
Ok, remember that half a circle is always 180 degrees. so..
1. 141
2. 23
3. 88
4. 120
5. 119
6. 152.
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Answer: There is probability of 0.367 that there will be no breakdown on his car in the trip.
Step-by-step explanation:
Since we have given that
Mean (λ) = 0.5 breakdown per week
Number of weeks the owner of the car is planning to have a trip on his car for = 2 weeks
So, mean for 2 weeks would be
![0.5\times 2=1.0](https://tex.z-dn.net/?f=0.5%5Ctimes%202%3D1.0)
We need to find the probability that there will be no breakdown on his car in the trip.
Probability that there will be no breakdown on his car in the trip is given by
P(X=0) is given by
![\dfrac{e^{-\lambda}\lambda^k}{k!}\\\\=\dfrac{e^{-1}1^0}{0!}\\\\=0.367](https://tex.z-dn.net/?f=%5Cdfrac%7Be%5E%7B-%5Clambda%7D%5Clambda%5Ek%7D%7Bk%21%7D%5C%5C%5C%5C%3D%5Cdfrac%7Be%5E%7B-1%7D1%5E0%7D%7B0%21%7D%5C%5C%5C%5C%3D0.367)
Hence, there is probability of 0.367 that there will be no breakdown on his car in the trip.
Answer:
an infinite number
Step-by-step explanation:
This equation has a graph that is a line. Every point on the line is a solution.
There are an infinite number of solutions.
Answer:
Rhombus
Step-by-step explanation:
This undoubtedly has to be a rhombus. When you look at other quadrilaterals for example and want to judge and or differentiate, we see that in the case of a square, the diagonals are equal to one another and they are not congruent. Same can be applied to that of a rectangle as well, it's diagonals aren't congruent, even if they appear perpendicular.
On the other hand, the diagonals of a rhombus is both perpendicular, and at the same time congruent. And thus, that's our answer