The value of the x is 2 and the value of the y is 0 in the system of equation 3y=x-2, y=-2x+4.
<h3>What is a linear equation?</h3>
It is defined as the relation between two variables, if we plot the graph of the linear equation we will get a straight line.
If in the linear equation, one variable is present, then the equation is known as the linear equation in one variable.
We have a system of the equation:
3y=x-2 ..(1)
y=-2x+4 ...(2)
From the equation (2) take the value of y and plug in the equation (1)
3(-2x+4) = x -2
-6x + 12 = x - 2
7x = 14
x = 14/7
x = 2
Plug this value in the equation (2)
y = -2(2) + 4
y = 0
Thus, the value of the x is 2 and the value of the y is 0 in the system of equation 3y=x-2, y=-2x+4.
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Answer:
24 combinations
Step-by-step explanation:
For this problem, you use factorials. Factorials are when you multiply every number below it except zero. So like the factorial of 3 or 3! is 3*2*1 is 6.
Factorials are used to figure out how many combinations there are of something. So for you, we have 4 different quadrants so we write that as 4! (! is the sign for factorials.)
By the way, welcome to brainly.
Y1 is the simplest parabola. Its vertex is at (0,0) and it passes thru (2,4). This is enough info to conclude that y1 = x^2.
y4, the lower red graph, is a bit more of a challenge. We can easily identify its vertex, which is (-4,0), and several points on the grah, such as (2,-3).
Let's try this: assume that the general equation for a parabola is
y-k = a(x-h)^2, where (h,k) is the vertex. Subst. the known values,
-3-(-4) = a(2-0)^2. Then 1 = a(2)^2, or 1 = 4a, or a = 1/4.
The equation of parabola y4 is y+4 = (1/4)x^2
Or you could elim. the fraction and write the eqn as 4y+16=x^2, or
4y = x^2-16, or y = (1/4)x - 4. Take your pick! Hope this helps you find "a" for the other parabolas.
It is Irrational because is can not be written as a fraction
