Pretty sure it’s irrational
Answer:
1.) Exponential Growth
2.) Exponential Decay
3.) Exponential Growth
4.) Exponential Decay
Step-by-step explanation:
<u>1.) </u><u><em>f (x) </em></u><u>= 0.5 (7/3)^</u><u><em>x</em></u>
↓
always increasing
<u>2.) </u><u><em>f (x) </em></u><u>= 0.9 (0.5)^</u><u><em>x</em></u>
<em> </em>↓
always decreasing
<u>3.) </u><u><em>f (x) </em></u><u>= 21 (1/6)^</u><u><em>x</em></u>
↓
always increasing
<u>4.) </u><u><em>f (x) </em></u><u>= 320 (1/6)^</u><u><em>x</em></u>
<em> </em> ↓
always decreasing
<u><em>EXPLANATION:</em></u>
It's exponential growth when the base of our exponential is bigger than 1, which means those numbers get bigger. It's exponential decay when the base of our exponential is in between 1 and 0 and those numbers get smaller.
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
Answer:
perpendicular
Step-by-step explanation:
When 2 lines intersect at any point, not just in the middle of each (that has nothing to do with it), and they meet to form right angles, the lines are perpendicular to one another and their slopes are opposite reciprocals.
CD is called the middle base and is equal to :(AB + EF)/2
CD = <span>(AB + EF)/2
4x = [(x</span>²+10) + (8-x)]/2
8x = x²+10+8-x
8x= x²-x+18
x²-9x+18 = 0
x' = [-b+√(b²-4.a.c)]/2a & x" = [-b-√(b<span>²-4.a.c)]/2a
x'= 6 and x" = 3, So we have2 values of x</span>
