Check the divisibility test of 152875 by 6,5,4, and 8
1 answer:
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Answer:
6.78ft/sec
Step-by-step explanation:
From the question, dx/dt= 3.9 ft/sec
We are looking for Dy/dt
From geometry,vof this case the relationship between x and y is needed here, there is two similar triangle that exhibited by the man and the lamb
12/y= 5.1/(y-x)
Then ,cross multiply, we have
12(y-x)=5.1y
12y-12x=5.1y
12y-5.1y=12x
6.9y=12x
y=( 12/6.9)x
Differentiating implicitly the bother sides with respect to t, we have
Dy/dt= ( 12/6.9)dx/dt
But dx/dt= 3.9 ft/sec
Then Dy/dt= ( 12/6.9)× 3.9
Dy/dt=6.78ft/sec
Hence, the rate that the tip of the person's shadow moves away from the lamppost when the person is 9 ft away from the lampost is 6.78ft/sec
CHECK THE ATTACHMENT FOR THE FIQURE
Answer:
The values of x for which the function can be replaced by the Taylor polynomial if the error cannot exceed 0.001 is 0 < x < 0.3936.
Step-by-step explanation:
Note: This question is not complete. The complete question is therefore provided before answering the question as follows:
Determine the values of x for which the function can be replaced by the Taylor polynomial if the error cannot exceed 0.001. f(x) = e^x ≈ 1 + x + x²/2! + x³/3!, x < 0
The explanation of the answer is now provided as follows:
Given:
f(x) = e^x ≈ 1 + x + x²/2! + x³/3!, x < 0 …………….. (1)
= (x) = (e^z /4!)x^4
Since the aim is (x) < 0.001, this implies that:
(e^z /4!)x^4 < 0.0001 ………………………………….. (2)
Multiply both sided of equation (2) by (1), we have:
e^4x^4 < 0.024 ……………………….......……………. (4)
Taking 4th root of both sided of equation (4), we have:
|xe^(z/4) < 0.3936 ……………………..........…………(5)
Dividing both sides of equation (5) by e^(z/4) gives us:
|x| < 0.3936 / e^(z/4) ……………….................…… (6)
In equation (6), when z > 0, e^(z/4) > 1. Therefore, we have:
|x| < 0.3936 -----> 0 < x < 0.3936
Therefore, the values of x for which the function can be replaced by the Taylor polynomial if the error cannot exceed 0.001 is 0 < x < 0.3936.