I am pretty sure you can only make one.
If two sides are known, and one of the angles, then the other bits can be deduced and are fixed.
Given the function :
![f(x)=\sqrt[]{x+2}+1](https://tex.z-dn.net/?f=f%28x%29%3D%5Csqrt%5B%5D%7Bx%2B2%7D%2B1)
We need to find each missing value
Given x = -3 , -2 , -1 , 2 , 7
So, substitute with each value of x to find the corresponding value of f(x)
![x=-3\rightarrow f(x)=\sqrt[]{-3+2}+1=\sqrt[]{-1}+1](https://tex.z-dn.net/?f=x%3D-3%5Crightarrow%20f%28x%29%3D%5Csqrt%5B%5D%7B-3%2B2%7D%2B1%3D%5Csqrt%5B%5D%7B-1%7D%2B1)
So, there is no value for f(x) at x = -3 (the function undefined because the square root of -1)
![\begin{gathered} x=-2\rightarrow f(x)=\sqrt[]{-2+2}+1=\sqrt[]{0}+1=0+1=1 \\ \\ x=-1\rightarrow f(x)=\sqrt[]{-1+2}+1=\sqrt[]{1}+1=1+1=2 \\ \\ x=2\rightarrow f(x)=\sqrt[]{2+2}+1=\sqrt[]{4}+1=2+1=3 \\ \\ x=7\rightarrow f(x)=\sqrt[]{7+2}+1=\sqrt[]{9}+1=3+1=4 \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20x%3D-2%5Crightarrow%20f%28x%29%3D%5Csqrt%5B%5D%7B-2%2B2%7D%2B1%3D%5Csqrt%5B%5D%7B0%7D%2B1%3D0%2B1%3D1%20%5C%5C%20%20%5C%5C%20x%3D-1%5Crightarrow%20f%28x%29%3D%5Csqrt%5B%5D%7B-1%2B2%7D%2B1%3D%5Csqrt%5B%5D%7B1%7D%2B1%3D1%2B1%3D2%20%5C%5C%20%20%5C%5C%20x%3D2%5Crightarrow%20f%28x%29%3D%5Csqrt%5B%5D%7B2%2B2%7D%2B1%3D%5Csqrt%5B%5D%7B4%7D%2B1%3D2%2B1%3D3%20%5C%5C%20%20%5C%5C%20x%3D7%5Crightarrow%20f%28x%29%3D%5Csqrt%5B%5D%7B7%2B2%7D%2B1%3D%5Csqrt%5B%5D%7B9%7D%2B1%3D3%2B1%3D4%20%5Cend%7Bgathered%7D)
the graph of the function and the points will be as shown in the following image :
Answer:
No ABC is not a triangle
Step-by-step explanation:
Perimeter = 22 cm
AB = 8 cm ; BC = 8 cm
AC = 22 - ( 8 + 8 ) = 6 cm
But all those sides doesn't follow Pythagorean Theorem(AB^2 + BC^2 = AC^2) ....because 8^2 + 8^2 isn't equal to 6^2 ......
Had it followed the theorem , it could have been a right angle triangle...
The answer is B, the answer is 9,318,000= 9.318•10 to the 6th power