False because you should always make sure a company has medical and detal procedures
Answer:
A. 29i
Step-by-step explanation:
Step 1: Plug in given variables
(2 - 5i)(2 + 5i)i
Step 2: Difference of squares (expand)
(4 - 25i²)i
Step 3: Imaginary numbers rules
(4 - 25(-1))i
Step 4: Combine like terms
(4 + 25)i
(29)i
Your final answer will be 29i
Step-by-step explanation:
<h3>Appropriate Question :-</h3>
Find the limit
![\rm \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x^2-x}-\dfrac{1}{x^3-3x^2+2x}\right]](https://tex.z-dn.net/?f=%5Crm%20%5C%3A%20%5Csf%20%7B%5Cdisplaystyle%7B%5Clim_%7Bx%5Cto%201%7D%7D%7D%20%5C%3A%20%5Cleft%5B%5Cdfrac%7Bx-2%7D%7Bx%5E2-x%7D-%5Cdfrac%7B1%7D%7Bx%5E3-3x%5E2%2B2x%7D%5Cright%5D)

Given expression is
![\rm \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x^2-x}-\dfrac{1}{x^3-3x^2+2x}\right]](https://tex.z-dn.net/?f=%5Crm%20%5C%3A%20%5Csf%20%7B%5Cdisplaystyle%7B%5Clim_%7Bx%5Cto%201%7D%7D%7D%20%5C%3A%20%5Cleft%5B%5Cdfrac%7Bx-2%7D%7Bx%5E2-x%7D-%5Cdfrac%7B1%7D%7Bx%5E3-3x%5E2%2B2x%7D%5Cright%5D)
On substituting directly x = 1, we get,


which is indeterminant form.
Consider again,
![\rm \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x^2-x}-\dfrac{1}{x^3-3x^2+2x}\right]](https://tex.z-dn.net/?f=%5Crm%20%5C%3A%20%5Csf%20%7B%5Cdisplaystyle%7B%5Clim_%7Bx%5Cto%201%7D%7D%7D%20%5C%3A%20%5Cleft%5B%5Cdfrac%7Bx-2%7D%7Bx%5E2-x%7D-%5Cdfrac%7B1%7D%7Bx%5E3-3x%5E2%2B2x%7D%5Cright%5D)
can be rewritten as
![\rm \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x(x - 1)}-\dfrac{1}{x( {x}^{2} - 3x + 2)}\right]](https://tex.z-dn.net/?f=%5Crm%20%5C%3A%20%5Csf%20%7B%5Cdisplaystyle%7B%5Clim_%7Bx%5Cto%201%7D%7D%7D%20%5C%3A%20%5Cleft%5B%5Cdfrac%7Bx-2%7D%7Bx%28x%20-%201%29%7D-%5Cdfrac%7B1%7D%7Bx%28%20%7Bx%7D%5E%7B2%7D%20-%203x%20%2B%202%29%7D%5Cright%5D)
![\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x(x - 1)}-\dfrac{1}{x( {x}^{2} - 2x - x + 2)}\right]](https://tex.z-dn.net/?f=%5Crm%20%5C%3A%20%3D%20%5C%3A%20%5Csf%20%7B%5Cdisplaystyle%7B%5Clim_%7Bx%5Cto%201%7D%7D%7D%20%5C%3A%20%5Cleft%5B%5Cdfrac%7Bx-2%7D%7Bx%28x%20-%201%29%7D-%5Cdfrac%7B1%7D%7Bx%28%20%7Bx%7D%5E%7B2%7D%20-%202x%20-%20x%20%2B%202%29%7D%5Cright%5D)
![\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x(x - 1)}-\dfrac{1}{x( x(x - 2) - 1(x - 2))}\right]](https://tex.z-dn.net/?f=%5Crm%20%5C%3A%20%3D%20%5C%3A%20%5Csf%20%7B%5Cdisplaystyle%7B%5Clim_%7Bx%5Cto%201%7D%7D%7D%20%5C%3A%20%5Cleft%5B%5Cdfrac%7Bx-2%7D%7Bx%28x%20-%201%29%7D-%5Cdfrac%7B1%7D%7Bx%28%20x%28x%20-%202%29%20-%201%28x%20-%202%29%29%7D%5Cright%5D)
![\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x(x - 1)}-\dfrac{1}{x(x - 2) \: (x - 1))}\right]](https://tex.z-dn.net/?f=%5Crm%20%5C%3A%20%3D%20%5C%3A%20%5Csf%20%7B%5Cdisplaystyle%7B%5Clim_%7Bx%5Cto%201%7D%7D%7D%20%5C%3A%20%5Cleft%5B%5Cdfrac%7Bx-2%7D%7Bx%28x%20-%201%29%7D-%5Cdfrac%7B1%7D%7Bx%28x%20-%202%29%20%5C%3A%20%28x%20-%201%29%29%7D%5Cright%5D)
![\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{ {(x - 2)}^{2} - 1}{x(x - 2) \: (x - 1))}\right]](https://tex.z-dn.net/?f=%5Crm%20%5C%3A%20%3D%20%5C%3A%20%5Csf%20%7B%5Cdisplaystyle%7B%5Clim_%7Bx%5Cto%201%7D%7D%7D%20%5C%3A%20%5Cleft%5B%5Cdfrac%7B%20%7B%28x%20-%202%29%7D%5E%7B2%7D%20-%201%7D%7Bx%28x%20-%202%29%20%5C%3A%20%28x%20-%201%29%29%7D%5Cright%5D)
![\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{ (x - 2 - 1)(x - 2 + 1)}{x(x - 2) \: (x - 1))}\right]](https://tex.z-dn.net/?f=%5Crm%20%5C%3A%20%3D%20%5C%3A%20%5Csf%20%7B%5Cdisplaystyle%7B%5Clim_%7Bx%5Cto%201%7D%7D%7D%20%5C%3A%20%5Cleft%5B%5Cdfrac%7B%20%28x%20-%202%20-%201%29%28x%20-%202%20%2B%201%29%7D%7Bx%28x%20-%202%29%20%5C%3A%20%28x%20-%201%29%29%7D%5Cright%5D)
![\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{ (x - 3)(x - 1)}{x(x - 2) \: (x - 1))}\right]](https://tex.z-dn.net/?f=%5Crm%20%5C%3A%20%3D%20%5C%3A%20%5Csf%20%7B%5Cdisplaystyle%7B%5Clim_%7Bx%5Cto%201%7D%7D%7D%20%5C%3A%20%5Cleft%5B%5Cdfrac%7B%20%28x%20-%203%29%28x%20-%201%29%7D%7Bx%28x%20-%202%29%20%5C%3A%20%28x%20-%201%29%29%7D%5Cright%5D)
![\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{ (x - 3)}{x(x - 2)}\right]](https://tex.z-dn.net/?f=%5Crm%20%5C%3A%20%3D%20%5C%3A%20%5Csf%20%7B%5Cdisplaystyle%7B%5Clim_%7Bx%5Cto%201%7D%7D%7D%20%5C%3A%20%5Cleft%5B%5Cdfrac%7B%20%28x%20-%203%29%7D%7Bx%28x%20-%202%29%7D%5Cright%5D)



Hence,
![\rm\implies \:\boxed{ \rm{ \:\rm \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x^2-x}-\dfrac{1}{x^3-3x^2+2x}\right] = 2 \: }}](https://tex.z-dn.net/?f=%5Crm%5Cimplies%20%5C%3A%5Cboxed%7B%20%5Crm%7B%20%5C%3A%5Crm%20%5C%3A%20%5Csf%20%7B%5Cdisplaystyle%7B%5Clim_%7Bx%5Cto%201%7D%7D%7D%20%5C%3A%20%5Cleft%5B%5Cdfrac%7Bx-2%7D%7Bx%5E2-x%7D-%5Cdfrac%7B1%7D%7Bx%5E3-3x%5E2%2B2x%7D%5Cright%5D%20%3D%202%20%5C%3A%20%7D%7D)

Answer:
-6x+12
-9+5a
7x+63
8y-4/5x-12
Step-by-step explanation:
To solve these all you have to do is take the number outside of the parenthesis and multiply them by each term.
-3(2x)-3(-4) All I did here was expand the problem to show what i mean by terms.
-6x+12
Do this for each problem.
0.1(-90)+0.1(50a)
-9+5a
-7(-x)-7(-9) For this one you are taking negatives times negatives so each answer will be positive.
7x+63
4/5(10y)+4/5(-x)+4/5(-15)
8y-4/5x-12
A. radical 76 is8.6 so false
B. 3 radical 10 is 9.4 true
C. 4 radical 2 is 5.6 so false
D. radical 29 is 5.3 so true
E. 2 radical 50 is 14.14 so false