Answer:
what is the question send it in the comment fast
49x^2 - 9 = 0
As there is no x term, we can pretty much guess we have a situation where we factlrise by something known aa difference of two squares, so to factorise it:
49 = 7^2
9 = 3^2
x^2 = (x)^2
so...
(7x - 3)(7x + 3) = 0
7x - 3 = 0 7x + 3 = 0
x = 3/7 x = -3/7
For this case we have that by definition, the equation of the line of the slope-intersection form is given by:
![y = mx + b](https://tex.z-dn.net/?f=y%20%3D%20mx%20%2B%20b)
Where:
m: It is the slope of the line
b: It is the cut-off point with the y axis
By definition, if two lines are perpendicular then the product of their slopes is -1.
We have the following equation of the line:
![y = 2x-5](https://tex.z-dn.net/?f=y%20%3D%202x-5)
Then ![m_ {1} = 2](https://tex.z-dn.net/?f=m_%20%7B1%7D%20%3D%202)
We find ![m_ {2}:](https://tex.z-dn.net/?f=m_%20%7B2%7D%3A)
![m_ {2} = \frac {-1} {m_ {1}}\\m_ {2} = \frac {-1} {2}\\m_ {2} = - \frac {1} {2}](https://tex.z-dn.net/?f=m_%20%7B2%7D%20%3D%20%5Cfrac%20%7B-1%7D%20%7Bm_%20%7B1%7D%7D%5C%5Cm_%20%7B2%7D%20%3D%20%5Cfrac%20%7B-1%7D%20%7B2%7D%5C%5Cm_%20%7B2%7D%20%3D%20-%20%5Cfrac%20%7B1%7D%20%7B2%7D)
Thus, the perpendicular line will be of the form:
![y = - \frac {1} {2} x + b](https://tex.z-dn.net/?f=y%20%3D%20-%20%5Cfrac%20%7B1%7D%20%7B2%7D%20x%20%2B%20b)
We substitute the given point and find "b":
![-2 = - \frac {1} {2} (8) + b](https://tex.z-dn.net/?f=-2%20%3D%20-%20%5Cfrac%20%7B1%7D%20%7B2%7D%20%288%29%20%2B%20b)
![-2 = -4 + b\\-2 + 4 = b\\b = 2](https://tex.z-dn.net/?f=-2%20%3D%20-4%20%2B%20b%5C%5C-2%20%2B%204%20%3D%20b%5C%5Cb%20%3D%202)
Finally, the equation is of the form:
![y = - \frac {1} {2} x + 2](https://tex.z-dn.net/?f=y%20%3D%20-%20%5Cfrac%20%7B1%7D%20%7B2%7D%20x%20%2B%202)
ANswer:
![y = - \frac {1} {2} x + 2](https://tex.z-dn.net/?f=y%20%3D%20-%20%5Cfrac%20%7B1%7D%20%7B2%7D%20x%20%2B%202)
33/3030x4/3030=132/9180900
9180900-132=9180768
9,180,768 selections will contain no defective calculators. Hope this helps.
ok did you get what I solved I hope it's fine by you, if you have a problem with the way I solved it still ask questions ok