What is the solution of the quadratic equation?
2 answers:
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First, plot the points. Point R would be somewhere in the second Quadrant, point M would be in the first quadrant 1, point B would be in the fourth quadrant, and point S would be on the negative y-axis. A property of rhombi is that their diagonals are perpendicular. One would need to calculate the slopes of the diagonals and determine whether or not they are perpendicular. Lines are perpendicular if and only if their slopes are opposite reciprocals. Example: 2 and -0.5
Formulas needed:
Slope formula:
The figure would look kinda like this:
R
M
S
B
Diagonals are segment RB and segment SM
So, your slope equations would look like this:
and
Slope of RB= -1
Slope of SM=7
Not a rhombus, slopes aren't perpendicular. But this figure may very well be a parallelogram
Formulas needed:
Slope formula:
The figure would look kinda like this:
R
M
S
B
Diagonals are segment RB and segment SM
So, your slope equations would look like this:
and
Slope of RB= -1
Slope of SM=7
Not a rhombus, slopes aren't perpendicular. But this figure may very well be a parallelogram
4, 5, and 7 are mutually coprime, so you can use the Chinese remainder theorem right away.
We construct a number such that taking it mod 4, 5, and 7 leaves the desired remainders:
- Taken mod 4, the last two terms vanish and we have
so we multiply the first term by 3.
- Taken mod 5, the first and last terms vanish and we have
so we multiply the second term by 2.
- Taken mod 7, the first two terms vanish and we have
so we multiply the last term by 7.
Now,
By the CRT, the system of congruences has a general solution
or all integers ,
, the least (and positive) of which is 27.