Lines that are perpendicular to another line has an opposite reciprocal slope. (opposite meaning to negate the number).
Lines that are parallel have the same slope.
But the line you provided is a vertical line, and vertical lines are undefined. There is no possible perpendicular or parallel line. Maybe you gave the wrong equation and didn't put it in <em>y=mx+b</em> form. If so, you can correct it in the comments, and if not, that's you answer!
:)
Answer:
The answer would be 20 mi
Step-by-step explanation:
a^2+b^2=c^2
a^2=144
b^2=256
C^2=400
Then you find the square root of ABC to find that C is 20
![16{ x }^{ 4 }-41{ x }^{ 2 }+25=0](https://tex.z-dn.net/?f=16%7B%20x%20%7D%5E%7B%204%20%7D-41%7B%20x%20%7D%5E%7B%202%20%7D%2B25%3D0)
![{ x }^{ 4 }={ ({ x }^{ 2 }) }^{ 2 }\\ \\ 16{ ({ x }^{ 2 }) }^{ 2 }-41{ x }^{ 2 }+25=0](https://tex.z-dn.net/?f=%7B%20x%20%7D%5E%7B%204%20%7D%3D%7B%20%28%7B%20x%20%7D%5E%7B%202%20%7D%29%20%7D%5E%7B%202%20%7D%5C%5C%20%5C%5C%2016%7B%20%28%7B%20x%20%7D%5E%7B%202%20%7D%29%20%7D%5E%7B%202%20%7D-41%7B%20x%20%7D%5E%7B%202%20%7D%2B25%3D0)
First of all to make our equation simpler, we'll equal
![x^{2}](https://tex.z-dn.net/?f=%20x%5E%7B2%7D%20)
to a variable like 'a'.
So,
![{ x }^{ 2 }=a](https://tex.z-dn.net/?f=%7B%20x%20%7D%5E%7B%202%20%7D%3Da)
Now let's plug
![x^{2}](https://tex.z-dn.net/?f=%20x%5E%7B2%7D%20)
's value (a) into the equation.
![16{ ({ x }^{ 2 }) }^{ 2 }-41{ x }^{ 2 }+25=0\\ \\ { x }^{ 2 }=a\\ \\ 16{ (a) }^{ 2 }-41{ a }+25=0](https://tex.z-dn.net/?f=16%7B%20%28%7B%20x%20%7D%5E%7B%202%20%7D%29%20%7D%5E%7B%202%20%7D-41%7B%20x%20%7D%5E%7B%202%20%7D%2B25%3D0%5C%5C%20%5C%5C%20%7B%20x%20%7D%5E%7B%202%20%7D%3Da%5C%5C%20%5C%5C%2016%7B%20%28a%29%20%7D%5E%7B%202%20%7D-41%7B%20a%20%7D%2B25%3D0)
Now we turned our equation into a quadratic equation.
(The variable 'a' will have a solution set of two solutions, but 'x' , which is the variable of our first equation will have a solution set of four solutions since it is a quartic equation (<span>fourth-degree <span>equation) )
Let's solve for a.
The formula used to solve quadratic equations ;
![\frac { -b\pm \sqrt { { b }^{ 2 }-4\cdot t\cdot c } }{ 2\cdot t }](https://tex.z-dn.net/?f=%5Cfrac%20%7B%20-b%5Cpm%20%5Csqrt%20%7B%20%7B%20b%20%7D%5E%7B%202%20%7D-4%5Ccdot%20t%5Ccdot%20c%20%7D%20%20%7D%7B%202%5Ccdot%20t%20%7D%20)
The formula is used in an equation formed like this :
</span></span>
![t{ x }^{ 2 }+bx+c=0](https://tex.z-dn.net/?f=t%7B%20x%20%7D%5E%7B%202%20%7D%2Bbx%2Bc%3D0)
In our equation,
t=16 , b=-41 and c=25
Let's plug the values in the formula to solve.
![t=16\quad b=-41\quad c=25\\ \\ \frac { -(-41)\pm \sqrt { -(41)^{ 2 }-4\cdot 16\cdot 25 } }{ 2\cdot 16 } \\ \\ \frac { 41\pm \sqrt { 1681-1600 } }{ 32 } \\ \\ \frac { 41\pm \sqrt { 81 } }{ 32 } \\ \\ \frac { 41\pm 9 }{ 32 }](https://tex.z-dn.net/?f=t%3D16%5Cquad%20b%3D-41%5Cquad%20c%3D25%5C%5C%20%5C%5C%20%5Cfrac%20%7B%20-%28-41%29%5Cpm%20%5Csqrt%20%7B%20-%2841%29%5E%7B%202%20%7D-4%5Ccdot%2016%5Ccdot%2025%20%7D%20%20%7D%7B%202%5Ccdot%2016%20%7D%20%5C%5C%20%5C%5C%20%5Cfrac%20%7B%2041%5Cpm%20%5Csqrt%20%7B%201681-1600%20%7D%20%20%7D%7B%2032%20%7D%20%5C%5C%20%5C%5C%20%5Cfrac%20%7B%2041%5Cpm%20%5Csqrt%20%7B%2081%20%7D%20%20%7D%7B%2032%20%7D%20%5C%5C%20%5C%5C%20%5Cfrac%20%7B%2041%5Cpm%209%20%7D%7B%2032%20%7D%20)
So the solution set :
![\frac { 41+9 }{ 32 } =\frac { 50 }{ 32 } \\ \\ \frac { 41-9 }{ 32 } =\frac { 32 }{ 32 } =1\\ \\ a\quad =\quad \left\{ \frac { 50 }{ 32 } ,\quad 1 \right\}](https://tex.z-dn.net/?f=%5Cfrac%20%7B%2041%2B9%20%7D%7B%2032%20%7D%20%3D%5Cfrac%20%7B%2050%20%7D%7B%2032%20%7D%20%5C%5C%20%5C%5C%20%5Cfrac%20%7B%2041-9%20%7D%7B%2032%20%7D%20%3D%5Cfrac%20%7B%2032%20%7D%7B%2032%20%7D%20%3D1%5C%5C%20%5C%5C%20a%5Cquad%20%3D%5Cquad%20%5Cleft%5C%7B%20%5Cfrac%20%7B%2050%20%7D%7B%2032%20%7D%20%2C%5Cquad%201%20%5Cright%5C%7D%20)
We found a's value.
Remember,
![{ x }^{ 2 }=a](https://tex.z-dn.net/?f=%7B%20x%20%7D%5E%7B%202%20%7D%3Da)
So after we found a's solution set, that means.
![{ x }^{ 2 }=\frac { 50 }{ 32 }](https://tex.z-dn.net/?f=%7B%20x%20%7D%5E%7B%202%20%7D%3D%5Cfrac%20%7B%2050%20%7D%7B%2032%20%7D%20)
and
![{ x }^{ 2 }=1](https://tex.z-dn.net/?f=%7B%20x%20%7D%5E%7B%202%20%7D%3D1)
We'll also solve this equations to find x's solution set :)
![{ x }^{ 2 }=\frac { 50 }{ 32 } \\ \\ \frac { 50 }{ 32 } =\frac { 25 }{ 16 } \\ \\ { x }^{ 2 }=\frac { 25 }{ 16 } \\ \\ \sqrt { { x }^{ 2 } } =\sqrt { \frac { 25 }{ 16 } } \\ \\ x=\quad \pm \frac { 5 }{ 4 }](https://tex.z-dn.net/?f=%7B%20x%20%7D%5E%7B%202%20%7D%3D%5Cfrac%20%7B%2050%20%7D%7B%2032%20%7D%20%5C%5C%20%5C%5C%20%5Cfrac%20%7B%2050%20%7D%7B%2032%20%7D%20%3D%5Cfrac%20%7B%2025%20%7D%7B%2016%20%7D%20%5C%5C%20%5C%5C%20%7B%20x%20%7D%5E%7B%202%20%7D%3D%5Cfrac%20%7B%2025%20%7D%7B%2016%20%7D%20%5C%5C%20%5C%5C%20%5Csqrt%20%7B%20%7B%20x%20%7D%5E%7B%202%20%7D%20%7D%20%3D%5Csqrt%20%7B%20%5Cfrac%20%7B%2025%20%7D%7B%2016%20%7D%20%20%7D%20%5C%5C%20%5C%5C%20x%3D%5Cquad%20%5Cpm%20%5Cfrac%20%7B%205%20%7D%7B%204%20%7D%20)
![{ x }^{ 2 }=1\\ \\ \sqrt { { x }^{ 2 } } =\sqrt { 1 } \\ \\ x=\quad \pm 1](https://tex.z-dn.net/?f=%7B%20x%20%7D%5E%7B%202%20%7D%3D1%5C%5C%20%5C%5C%20%5Csqrt%20%7B%20%7B%20x%20%7D%5E%7B%202%20%7D%20%7D%20%3D%5Csqrt%20%7B%201%20%7D%20%5C%5C%20%5C%5C%20x%3D%5Cquad%20%5Cpm%201)
So the values x has are :
![\frac { 5 }{ 4 }](https://tex.z-dn.net/?f=%5Cfrac%20%7B%205%20%7D%7B%204%20%7D%20)
,
![-\frac { 5 }{ 4 }](https://tex.z-dn.net/?f=-%5Cfrac%20%7B%205%20%7D%7B%204%20%7D%20)
,
![1](https://tex.z-dn.net/?f=1)
and
![-1](https://tex.z-dn.net/?f=-1)
Solution set :
![x=\quad \left\{ \frac { 5 }{ 4 } \quad ,\quad -\frac { 5 }{ 4 } \quad ,\quad 1\quad ,\quad -1 \right\}](https://tex.z-dn.net/?f=x%3D%5Cquad%20%5Cleft%5C%7B%20%5Cfrac%20%7B%205%20%7D%7B%204%20%7D%20%5Cquad%20%2C%5Cquad%20-%5Cfrac%20%7B%205%20%7D%7B%204%20%7D%20%5Cquad%20%2C%5Cquad%201%5Cquad%20%2C%5Cquad%20-1%20%5Cright%5C%7D%20)
I hope this was clear enough. If not please ask :)
Answer:
The trainers are 44.71% off the original price i think.
Step-by-step explanation:
So the formula for price change is amount of change / original price * 100
50-40
= 10
10/50*100
= 1/5*100
= 20%