so, this is a quadratic equation, meaning two solutions, and we have a factored form of it, meaning you can get the solutions by simply zeroing out the f(x).
![\bf \stackrel{f(x)}{0}=-(x-3)(x+11)\implies 0=(x-3)(x+11)\implies x= \begin{cases} 3\\ -11 \end{cases} \\\\\\ \boxed{-11}\stackrel{\textit{\large 7 units}}{\rule[0.35em]{10em}{0.25pt}}-4\stackrel{\textit{\large 7 units}}{\rule[0.35em]{10em}{0.25pt}}\boxed{3}](https://tex.z-dn.net/?f=%5Cbf%20%5Cstackrel%7Bf%28x%29%7D%7B0%7D%3D-%28x-3%29%28x%2B11%29%5Cimplies%200%3D%28x-3%29%28x%2B11%29%5Cimplies%20x%3D%20%5Cbegin%7Bcases%7D%203%5C%5C%20-11%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5C%5C%20%5Cboxed%7B-11%7D%5Cstackrel%7B%5Ctextit%7B%5Clarge%207%20units%7D%7D%7B%5Crule%5B0.35em%5D%7B10em%7D%7B0.25pt%7D%7D-4%5Cstackrel%7B%5Ctextit%7B%5Clarge%207%20units%7D%7D%7B%5Crule%5B0.35em%5D%7B10em%7D%7B0.25pt%7D%7D%5Cboxed%7B3%7D)
so the zeros/solutions are at x = 3 and x = -11, now, bearing in mind the vertex will be half-way between those two, checking the number line, that midpoint will be at x = -4, so the vertex is right there, well, what's f(x) when x = -4?
![\bf f(-4)=-(-4-3)(-4+11)\implies f(-4)=7(7)\implies f(-4)=49 \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill \stackrel{vertex}{(-4~~,~~49)}~\hfill](https://tex.z-dn.net/?f=%5Cbf%20f%28-4%29%3D-%28-4-3%29%28-4%2B11%29%5Cimplies%20f%28-4%29%3D7%287%29%5Cimplies%20f%28-4%29%3D49%20%5C%5C%5C%5C%5B-0.35em%5D%20%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5C%20~%5Chfill%20%5Cstackrel%7Bvertex%7D%7B%28-4~~%2C~~49%29%7D~%5Chfill)
Given that the sides of the acute triangle are as follows:
21 cm
x cm
2x cm
Stated that 21 cm is one of the shorter sides of the triangle2x is greater than x, so it follows that 2x MUST be the longest side
For acute triangles, the longest side must be less than the sum of the 2 shorter sides
Therefore, 2x < x + 21cm
2x – x < 21cm
x < 21cm
If x < 21cm, then 2x < 42cm
Therefore, the longest possible length for the longest side is 42cm
Answer:
169
Step-by-step explanation:
1 parallelogram
2 parallelogram
3 rhombus
4 rectangle
5 rhombus
I got 3x^2 + 12x + 4. I recommend you verify my answer, so you have to combine like terms in the first expression then add all the expressions.
