The functions f(x), g(x) and h(x) are all linear functions
- The function value that is largest is h(-100)
- The function value that is largest is h(-100)
- The function value that is largest is f(1/100)
<h3>How to determine the function with the largest values</h3>
The functions are given as:
At x = 100, we have:
Hence, the function value that is largest is f(100)
At x = -100, we have:
Hence, the function value that is largest is h(-100)
At x = 1/100, we have:
Hence, the function value that is largest is f(1/100)
Read more about linear functions at:
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(-1)+(-6)= -7
-7 + 6 = -1
therefore, the answer is 6.
Answer:
∠1 = 48
∠2 = 132
∠4 - 132
Step-by-step explanation:
∠3 and ∠1 are vertical angles
vertical angles are congruent ( equal to each other )
So if ∠3 = 48° then ∠1 also = 48°
∠3 and ∠4 are supplementary angles
supplementary angles add up to equal 180°
Hence, ∠3 + ∠4 = 180
48 + ∠4 = 180
180 - 48 = 132
Hence, ∠4 = 132
∠4 and ∠2 are vertical angles
Like stated previously vertical angles are congruent
Hence, ∠2 = 132
Check the picture below.
so the parabola looks more or less like so, bearing in mind that the directrix is a horizontal line, thusu the parabola is a vertical one, so the squared variable is the "x".
The vertex is always half-way between the focus point and the directrix, as you see there, and the distance from the vertex to the focus is "p" distance, since the parabola is opening downwards, "p" is negative, in this case -2.
![\bf \textit{parabola vertex form with focus point distance} \\\\ \begin{array}{llll} 4p(x- h)=(y- k)^2 \\\\ \stackrel{\textit{we'll use this one}}{4p(y- k)=(x- h)^2} \end{array} \qquad \begin{array}{llll} vertex\ ( h, k)\\\\ p=\textit{distance from vertex to }\\ \qquad \textit{ focus or directrix} \end{array} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \begin{cases} h=0\\ k=0\\ p=-2 \end{cases}\implies 4(-2)(y-0)=(x-0)^2\implies -8y=x^2\implies y=-\cfrac{1}{8}x^2](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Bparabola%20vertex%20form%20with%20focus%20point%20distance%7D%20%5C%5C%5C%5C%20%5Cbegin%7Barray%7D%7Bllll%7D%204p%28x-%20h%29%3D%28y-%20k%29%5E2%20%5C%5C%5C%5C%20%5Cstackrel%7B%5Ctextit%7Bwe%27ll%20use%20this%20one%7D%7D%7B4p%28y-%20k%29%3D%28x-%20h%29%5E2%7D%20%5Cend%7Barray%7D%20%5Cqquad%20%5Cbegin%7Barray%7D%7Bllll%7D%20vertex%5C%20%28%20h%2C%20k%29%5C%5C%5C%5C%20p%3D%5Ctextit%7Bdistance%20from%20vertex%20to%20%7D%5C%5C%20%5Cqquad%20%5Ctextit%7B%20focus%20or%20directrix%7D%20%5Cend%7Barray%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5C%20%5Cbegin%7Bcases%7D%20h%3D0%5C%5C%20k%3D0%5C%5C%20p%3D-2%20%5Cend%7Bcases%7D%5Cimplies%204%28-2%29%28y-0%29%3D%28x-0%29%5E2%5Cimplies%20-8y%3Dx%5E2%5Cimplies%20y%3D-%5Ccfrac%7B1%7D%7B8%7Dx%5E2)
You know that the square root of 144 is 12 and 169 is 13. 150 is in between 144 and 169 so its square root must be between 12 and 13.
