Step 1) Draw a dashed line through the points (0,6) and (4,7). These two points are on the line y = (1/4)x+6. To find those points, you plug in x = 0 to get y = 6. Similarly, plug in x = 4 to get y = 7. The dashed line indicates that none of the points on this line are part of the solution set.
Step 2) Draw a dashed line through (0,-1) and (1,1). These two points are on the line y = 2x-1. They are found in a similar fashion as done in step 1.
Step 3) Shade the region that is above both dashed lines. We shade above because of the "greater than" sign. This is shown in the attached image I am providing below. The red shaded region represents all of the possible points that are the solution set. Once again, any point on the dashed line is not in the solution set.
Answer:
x=5
Step-by-step explanation:
Congruent triangles mean they have the same side lengths, so the equation here is x^2-7=4x-2. Then, solve the quadratic equation:
x^2-7=4x-2
x^2-4x-5=0
(x-5)(x+1)=0
x=5,-1
Since a side of a triangle can't be negative, the answer is 5.
The rule of the linear equation which is 'y is three more than twice x' can be translated into y = 2x + 3. Thus, we simply use the points and subtitute it to the y and x variables.
1st point: (-1, 1)
1 = 2(-1) + 3
1 = -2 + 3
1=1
2nd point: (0, 3)
3 = 2(0) + 3
3 =3
3rd point: (-1/3, 7/3)
7/3 = 2(-1/3) + 3
7/3 = -2/3 + 3
7/3 = 7/3
All points prove the equation satifsfies the set of coordinates.
It would become a larger bar that is all I can do without the graph
Rational is all real numbers. Irrational isnt