Use substitution
From second equation:-
x = 13 - 5y, Substitute for x in the first eqaution:
(13 - 5y)^2 + y^2 = 13
169 -130y + 25y^2 + y^2 - 13 = 0
26y^2 - 130y + 156 = 0
26(y^2 - 5y + 6) = 0
26(y - 3)(y - 2) = 0
y = 2 , y = 3
From the second equation when y = 2, x = 13 - 5(2) = 3
and when y = 3, x = 13 - 5(3) = -2
So the solution is [3, 2} and {-2, 3}
Answer: First option, the graph of the relation passes the vertical line test.
Step-by-step explanation:
A function is a relation that maps elements from a set (the domain, the x-values) into elements from another set (the range, the y-values)
Where the only condition we have is that each element from the domain can be mapped into only one element from the range.
So if we draw a vertical line at any point of the graph, it can intersect the graph of the function a maximum of one time. (this is known as the vertical line test)
Then the correct option is that the graph of the relation passes the vertical line test.
The length
the width
the height
so the amount of small cubes is
I believe that 4608 is the answer
good luck
Answer:
(x-4) (x-5)
Step-by-step explanation:
x^2 -9x+20
What two number multiply to 20 and add to -9
-4*-5 = 20
-4+-5 = -9
(x-4) (x-5)
