Answer:
![\large\boxed{-\bigg[(x-3)^2+2x\bigg]+1}\\\\\boxed{-x^2+4x-8}](https://tex.z-dn.net/?f=%5Clarge%5Cboxed%7B-%5Cbigg%5B%28x-3%29%5E2%2B2x%5Cbigg%5D%2B1%7D%5C%5C%5C%5C%5Cboxed%7B-x%5E2%2B4x-8%7D)
Step-by-step explanation:

![f(x)=(x-3)^2+2x\\\\g(x)=-x+1\\\\g\ \circ\ f\to\text{put f(x) instead of x in the function g(x)}:\\\\(g\ \circ\ f)(x)=-\bigg[\underbrace{(x-3)^2+2x}_{x}\bigg]+1](https://tex.z-dn.net/?f=f%28x%29%3D%28x-3%29%5E2%2B2x%5C%5C%5C%5Cg%28x%29%3D-x%2B1%5C%5C%5C%5Cg%5C%20%5Ccirc%5C%20f%5Cto%5Ctext%7Bput%20f%28x%29%20instead%20of%20x%20in%20the%20function%20g%28x%29%7D%3A%5C%5C%5C%5C%28g%5C%20%5Ccirc%5C%20f%29%28x%29%3D-%5Cbigg%5B%5Cunderbrace%7B%28x-3%29%5E2%2B2x%7D_%7Bx%7D%5Cbigg%5D%2B1)
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![-\bigg[(x-3)^2+2x\bigg]+1=-(x-3)^2-2x+1\\\\\text{use}\ (a-b)^2=a^2-2ab+b^2\\\\=-(x^2-(2)(x)(3)+3^2)-2x+1=-(x^2-6x+9)-2x+1\\\\=-x^2-(-6x)-9-2x+1=-x^2+6x-9-2x+1\\\\\text{combine like terms}\\\\=-x^2+(6x-2x)+(-9+1)=-x^2+4x-8](https://tex.z-dn.net/?f=-%5Cbigg%5B%28x-3%29%5E2%2B2x%5Cbigg%5D%2B1%3D-%28x-3%29%5E2-2x%2B1%5C%5C%5C%5C%5Ctext%7Buse%7D%5C%20%28a-b%29%5E2%3Da%5E2-2ab%2Bb%5E2%5C%5C%5C%5C%3D-%28x%5E2-%282%29%28x%29%283%29%2B3%5E2%29-2x%2B1%3D-%28x%5E2-6x%2B9%29-2x%2B1%5C%5C%5C%5C%3D-x%5E2-%28-6x%29-9-2x%2B1%3D-x%5E2%2B6x-9-2x%2B1%5C%5C%5C%5C%5Ctext%7Bcombine%20like%20terms%7D%5C%5C%5C%5C%3D-x%5E2%2B%286x-2x%29%2B%28-9%2B1%29%3D-x%5E2%2B4x-8)
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Answer: A) .1587
Step-by-step explanation:
Given : The amount of soda a dispensing machine pours into a 12-ounce can of soda follows a normal distribution with a mean of 12.30 ounces and a standard deviation of 0.20 ounce.
i.e.
and 
Let x denotes the amount of soda in any can.
Every can that has more than 12.50 ounces of soda poured into it must go through a special cleaning process before it can be sold.
Then, the probability that a randomly selected can will need to go through the mentioned process = probability that a randomly selected can has more than 12.50 ounces of soda poured into it =
![P(x>12.50)=1-P(x\leq12.50)\\\\=1-P(\dfrac{x-\mu}{\sigma}\leq\dfrac{12.50-12.30}{0.20})\\\\=1-P(z\leq1)\ \ [\because z=\dfrac{x-\mu}{\sigma}]\\\\=1-0.8413\ \ \ [\text{By z-table}]\\\\=0.1587](https://tex.z-dn.net/?f=P%28x%3E12.50%29%3D1-P%28x%5Cleq12.50%29%5C%5C%5C%5C%3D1-P%28%5Cdfrac%7Bx-%5Cmu%7D%7B%5Csigma%7D%5Cleq%5Cdfrac%7B12.50-12.30%7D%7B0.20%7D%29%5C%5C%5C%5C%3D1-P%28z%5Cleq1%29%5C%20%5C%20%5B%5Cbecause%20z%3D%5Cdfrac%7Bx-%5Cmu%7D%7B%5Csigma%7D%5D%5C%5C%5C%5C%3D1-0.8413%5C%20%5C%20%5C%20%5B%5Ctext%7BBy%20z-table%7D%5D%5C%5C%5C%5C%3D0.1587)
Hence, the required probability= A) 0.1587
Https://www.illustrativemathematics.org/content-standards/tasks/703 check it out it might help
Step-by-step explanation:
believe it or not, the area of any triangle is
baseline × height / 2
we have here 2 triangles, both with the same baseline (8 cm), but with 2 different heights : 6.5 cm and 4.6 cm.
so, their areas are
8 × 6.5 / 2 = 4 × 6.5 = 26 cm²
8 × 4.6 / 2 = 4 × 4.6 = 18.4 cm²
so, in total, the shaded region is
26 + 18.4 = 44.4 cm²