When a graph is transformed, it could be translated, reflected, dilated, or rotated. However, no matter what kind of transformation that is, it does not change the nature of the graph. A line would still be a line; a curve would still be a curve. So, that is my basis for my solution. There is no need to graph the problem. You only need to find the equation of g(x) through the two points given: (2,0) and (3,2)
m = Δy/Δx = (2-0)/(3-2) = 2
Use point (2,0) to find b:
y = mx + b
0 = 2(2) + b
b = -4
So, <em>g(x) = 2x - 4</em>
Answer:
97°
It would be 97° because anything higher than 90 is obtuse and it couldnt be 180 because 180 would be a straight angle
hope this helps
Answer:
The zeros are x1=4.45 and x2=-0.45.
x1 is between 4 and 5.
x2 is between -1 and 0.
Step-by-step explanation:
We have the function:
As this is a quadratic function, we can calculate the zeros of the function with the quadratic equation:
The zeros are x1=4.45 and x2=-0.45.
x1 is between 4 and 5.
x2 is between -1 and 0.
Answer: C
Step-by-step explanation:
Find the least common denominator of 4, 3, and 12.
4-3-12 | 3
4-1-4 | 4
1-1-1 |-------- 12
The first fractions needs to be multiplied by 3, and the second fraction, by 4
Solve;
Add the fractions with positive signs and subtract the one with negative sign.
Solve;
Simplify by 4;
16/4=4
12/4=3
Let us denote the semi arcs as congruent angles. This means that angles FEJ and EFJ are congruent (That is, they have the same measure). Since angles FEJ and EFJ have the same measure, this implies that sides EJ and FJ are equal. Since angles EJK and FJH are supplementary angles to angle EJF, this implies that EJK and FJH have the same measure.
Using the Angle Side Angle (SAS) criteria, we determine that triangles EKJ and triangle FJH are congruent. This implies that sides EK and FH are equal and that angles EKJ and FHJ are congruent. Note that angle EKJ is the same as EKF and that FHJ is the same as FHE.
Once again, since angles EKF and FHJ are congruent, and angle EKD is supplementary to the angle EKJ when angle FHG is supplementary to angle FHJ, then we have that angles EKD and angle FHG are congruent.
Using again the SAS criteria, we determine that triangles EKD and FHG are congruent.
From this reasoning, we have proved the following facts:
Triangle DEK is congruent to triangl GFH
Angle EKF is congruent to angle FHE
Segment EK is the same as segment FH
