First let’s find the equation:
a = (42-21)/(26-13)
a= 21/13
y-y1 = a(x-x1)
y-21=(21/13)(x-13)
y-21=(21/13)x - 21
y=(21/13)x-21+21
y=(21/13)x
Now we will find the other point:
X = 2 >> y =(21/13)*2 = 3.23 (not)
X = 3 >> y =(21/13)*3 = 4.84 (not)
X = 4 >> y =(21/13)*4 = 6.46 (not)
X = 5 >> y =(21/13)*5 = 8.07 (yes) find the answer
Given the graph, we need to look for the equation of the line using the linear equation since the plot is obviously a line.
To do that, we need two points. The simplest points to locate are the two points that converges the x and y axes which are (3,0) and (4,0), respectively. Now we apply the linear equation formulation:
y = mx+b
First, we need to get the slope.
m = (y2-y1)/(x2-x1)
m = (4-0)/(0-3)
m = -4/3
Since we know the slope, we need to look for the b which is also the y-intercept or the point at which x = 0.
So at x = 0, which point or value at y-axis hits with the line? It is the point (0,4).
Therefore, b = 4.
Completing the equation,
y = -(4/3)x + 4
or we can write it as
4x + 3y = 12.
Now to know which of the choices is correct, we simply assign values and see for ourselves which is true.
Let x = 2, so y = 2
4(2) + 3y = 12
8 < 12
If you assign more values, you will notice that 4x + 3y is always less than or equal to 12. Thus, the answer is letter a.
Answer:
x = 6 , y = -2
Step-by-step explanation:
2 x + 3 y = 6 (Equation 1)
x + y = 4 (Equation 2)
<u>Solving by substitution</u> :
(Equation 2) ⇒ y = 4 - x
<u>Substitute y by 4 - x in (equation 1)</u> :
2 x + 3 (4 - x) = 6
Then
2 x + 12 - 3x = 6
Then
12 - x = 6
Then
12 - 6 = x
Then
x = 6
<u>We obtain</u> :
y = 4 - x and x = 6
Then
y = 4 - 6
= -2
9514 1404 393
Answer:
C. (10, 5), (5, 0), (0, 5)
Step-by-step explanation:
In order for the square to have an area of 50 square units, its side lengths must be √50 = 5√2 units long. The only way to get such a length with integer coordinates is for the sides of the square to be themselves the diagonals of a square 5 units on a side.
That is, two of the other vertices of the square must be among ...
(5±5, 10±5) ⇒ (10, 15), (0, 15), (10, 5), (0, 5)
Only choice C contains two of the vertices on this list.
