Answer:
Step-by-step explanation:
Given that,
f(3) = 2
f'(3) = 5.
We want to estimate f(2.85)
The linear approximation of "f" at "a" is one way of writing the equation of the tangent line at "a".
At x = a, y = f(a) and the slope of the tangent line is f'(a).
So, in point slope form, the tangent line has equation
y − f(a) = f'(a)(x − a)
The linearization solves for y by adding f(a) to both sides
f(x) = f(a) + f'(a)(x − a).
Given that,
f(3) = 2,
f'(3) = 5
a = 3, we want to find f(2.85)
x = 2.85
Therefore,
f(x) = f(a) + f'(a)(x − a)
f(2.85) = 2 + 5(2.85 - 3)
f(2.85) = 2 + 5×-0.15
f(2.85) = 2 - 0.75
f(2.85) = 1.25
+3 . I think, hope this helps
Answer:
1. 6
2. -1
3. 0
4. -9
5. 0
6. -21
7. 119
8. -60
9. 64
10. 45 degrees
Step-by-step explanation:
Answer:
- train: 40 kph
- plane: 140 kph
Step-by-step explanation:
Let t represent the speed of the train in km/h. Then 4t-20 is the speed of the plane. Travel times are the same, so we can use the formula ...
time = distance/speed
and equate the travel times.
110/t = 385/(4t-20)
Cross multiplying gives ...
110(4t -20) = 385t
440t -2200 = 385t . . . . . eliminate parentheses
55t -2200 = 0 . . . . . . . . . subtract 385t
t -40 = 0 . . . . . . . . . divide by 55
t = 40 . . . . . . . . . . . add 40; train's speed is 40 kph
4t -20 = 140 . . . . . . find plane's speed; 140 kph
The train's speed is 40 km/h; the plane's speed is 140 km/h.
_____
<em>Check</em>
Train's travel time = 110 km/(40 km/h) = 2.75 h.
Plane's travel time = 385 km/(140 km/h) = 2.75 h.
Answer: The largest you can't make is 7 cents.
Step-by-step explanation:
Once you get 3 consecutive amounts, you can get all larger ones just by adding 3 cent stamps. So let's see what you can make:
1 no
2 no
3 yes
4 no
5 yes
6 yes
7 no
8 yes, and this will give you 11 14 17 etc
9 yes, and this will give you 12 15 18 etc
10 yes, and this will give you 13 16 19 etc.
