Answer:
4140
Step-by-step explanation:
use (n-2)180
n = number of sides
(25-2)180
(23)180
4140
Answer:
(x+1)(x-14)(x+3)
Step-by-step explanation:
The first step in factorising polynomials is to find the smallest factor first. The way to do this is with trial and error. Lets see if x+1 is a factor.
According to the factor theorem, (x-a) is a factor of P(x) if and only when P(a) = 0
In this case we will use f(x) = ((x)^3)-((10x)^2)-53x-42
f(-1) = 0
Therefore (x+1) is a factor of the polynomial.
Now we divide the polynomial with (x+1) via long division to get (x^2)-11x-42.
We now factorise (x^2)-11x-42 using whatever method you would like. I'm going to use the AC method, where we find a number that multiplies to AC and adds to B, In this case AC = -42, and B = -11.
Therefore (x^2)-11x-42 factosied is (x-14)(x+3)
Now merge (x+1) with (x-14)(x+3)
The final answer is (x+1)(x-14)(x+3)
If this has helped, please consider making this the Brainliest answer!
Use m.a.t.h.w.a.y. online calculator to determine the answer: (s - 1)
Answer:
how are we supposed to know what u talking about..
Step-by-step explanation:
if there is no PICTURE MWJAUBFNWAF
Answer:
10 hours at job (1)
2 hours at job (2)
Step-by-step explanation:
As per the given information, one earns ($8) dollars at one of their jobs, and ($10) hours at the other. One must earn a total of ($100) dollars, and can work no more than (12) hours. Let (x) be the hours worked at job 1 and (y) be the hours worked at job two.
Since one can work no more than (12) hours, the sum of (x) and (y) must be (12), therefore the following equation can be formed;
![x+y=12](https://tex.z-dn.net/?f=x%2By%3D12)
One earns ($8) dollars at one of their jobs and ($10) at the other, but one earns a total of (100) one can form an equation to represent this situation. Multiply the hours worked by the money earn per hour for each job, add up the result and set it equal to (100).
![8x+10y=100](https://tex.z-dn.net/?f=8x%2B10y%3D100)
Now set up these equations in a system;
![\left \{ {{x+y=12} \atop {8x+10y=100}} \right.](https://tex.z-dn.net/?f=%5Cleft%20%5C%7B%20%7B%7Bx%2By%3D12%7D%20%5Catop%20%7B8x%2B10y%3D100%7D%7D%20%5Cright.)
Use the process of elimination to solve this system. The process of elimination is a method of solving a system of equations. One must first manipulate one of the equations in the system such that one of the variable coefficients is the additive inverse of the other. That way, when one adds the equation, the variable cancels, one can solve for the other variable then back solve to find the value of the first variable,
![\left \{ {{x+y=12} \atop {8x+10y=100}} \right.](https://tex.z-dn.net/?f=%5Cleft%20%5C%7B%20%7B%7Bx%2By%3D12%7D%20%5Catop%20%7B8x%2B10y%3D100%7D%7D%20%5Cright.)
Manipulate,
![= \left \{ {{(*-8)(x+y=12)} \atop {8x+10y=100}} \right.\\\\](https://tex.z-dn.net/?f=%3D%20%5Cleft%20%5C%7B%20%7B%7B%28%2A-8%29%28x%2By%3D12%29%7D%20%5Catop%20%7B8x%2B10y%3D100%7D%7D%20%5Cright.%5C%5C%5C%5C)
Simplify,
![= \left \{ {{-8x-8y=-96} \atop {8x+10y=100}} \right.\\](https://tex.z-dn.net/?f=%3D%20%5Cleft%20%5C%7B%20%7B%7B-8x-8y%3D-96%7D%20%5Catop%20%7B8x%2B10y%3D100%7D%7D%20%5Cright.%5C%5C)
Add,
![=2y=4](https://tex.z-dn.net/?f=%3D2y%3D4)
Inverse operations,
![y=2](https://tex.z-dn.net/?f=y%3D2)
Backsolve for (x), use equation one to achieve this,
![x+y=12\\](https://tex.z-dn.net/?f=x%2By%3D12%5C%5C)
Substitute,
![x+2=12](https://tex.z-dn.net/?f=x%2B2%3D12)
Inverse operations,
![x=10](https://tex.z-dn.net/?f=x%3D10)