2.5 years = 30 months
7500$ in 30 months
x$ in 1 month
___________________
x = (7500*1)/30 = 250
The answer is 250$/month.
Answer: I would say Side-Angle-Side
Step-by-step explanation:
Answer:
x<1, C
Step-by-step explanation:
ily ! have a good day, i hope i'm right lol
6. Take your compass and place the pointed edge on point B. Place one point on each side of B, each the same distance away from B. Next, place your compass on one of the two new points and extend your compass to draw a circle. Repeat with the SAME radian from the other point. Find where the two circles intersect with each other and draw a line from the points of intersection to point B. Place point A anywhere on that line that you just created and then you're done!
7. Select any place along either line and place point S on it. Next, using the same method as above, draw two circles with the same radius around both points S and R. Draw a line through the intersection points. Locate the intersection where your new line connects with the line across from the shared line of RS. Place a point at the intersection, for your reference, then connect that point to point S. Now you have completed this problem as well.
8. Use a straight edge to draw one line. Place points A and B on each end. Use the circle method yet again to find a line perpendicular to line AB. Next, take your compass and set it to the distance from point A to B. Use that same distance to make a point on the perpendicular line. This creates point C. The final step is to connect A with C and B with C.
Answer:
Yes, the function satisfies the hypothesis of the Mean Value Theorem on the interval [1,5]
Step-by-step explanation:
We are given that a function
Interval [1,5]
The given function is defined on this interval.
Hypothesis of Mean Value Theorem:
(1) Function is continuous on interval [a,b]
(2)Function is defined on interval (a,b)
From the graph we can see that
The function is continuous on [1,5] and differentiable at(1,5).
Hence, the function satisfies the hypothesis of the Mean Value Theorem.
