Answer:
yes she is wrong
Step-by-step explanation:
becuse you add all of them up and get higher.
Answer:
The value of the printer on the first year was $ 23,750.00. On the second year it was $ 22,562.5. On the third year it was $ 21,434.38.
Step-by-step explanation:
Since the printer depreciates at a rate of 5% per year, I believe the stated equation is miss typed. Therefore I'll answer this with the correct equation that would represent that setting:
![y(x) = 25,000*0.95^x](https://tex.z-dn.net/?f=y%28x%29%20%3D%2025%2C000%2A0.95%5Ex)
In the first year the value of the printer is:
![y(1) = 25,000*0.95^1 = 23,750](https://tex.z-dn.net/?f=y%281%29%20%3D%2025%2C000%2A0.95%5E1%20%3D%2023%2C750)
On the second year the value of the printer is:
![y(2) = 25,000*0.95^2 = 22,562.5\\](https://tex.z-dn.net/?f=y%282%29%20%3D%2025%2C000%2A0.95%5E2%20%3D%2022%2C562.5%5C%5C)
On the third year the value of the printer is:
![y(3) = 25,000*0.95^3 = 21,434.38\\](https://tex.z-dn.net/?f=y%283%29%20%3D%2025%2C000%2A0.95%5E3%20%3D%2021%2C434.38%5C%5C)
The value of the printer on the first year was $ 23,750.00. On the second year it was $ 22,562.5. On the third year it was $ 21,434.38.
Answer:
y = 1/e
y = 0.37
Step-by-step explanation:
y = (ln x)/x; (e, 1/e)
Step 1
Find the point of tangency.
It's given as (e,1/e) or (2.72,0.37)
Step 2
Find the first derivative, and evaluate it at x=e
Using product rule.
d(uv) = udv + vdu
Where u = lnx and v = 1/x
du = 1/x , dv = -1/x^2
f'(x) = -lnx/x^2 +1/x^2
f'(x) = (1-lnx)/x^2
At x = e
f'(e) = (1-lne)/e^2
f'(e) = 0
The slope of the tangent line at this point is m= 0
Step 3
Find the equation of the tangent line at (e,1/e) with a slope of m=0
y−y1 = m(x - x1)
y-(1/e) = 0(x-1)
y = 1/e
y = 0.37
Answer:
Hope they get the points!