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Bond [772]
3 years ago
9

Check the divisibility test of 152875 by 6,5,4, and 8

Mathematics
1 answer:
mixer [17]3 years ago
3 0

Answer:

152875 is onli divide by 5

Step-by-step explanation:

152875÷6=25479.1666

152875÷5=30575

152875÷4=38218.75

152875÷8=19109.375

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Answer:   1. a^2 + 4a + 4

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                3. c^2 -6c + 9



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What are factorials? And how do you do them in fractions like
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4 0
2 years ago
A 5.1-ft-tall person walks away from a 12-ft lamppost at a constant rate of 3.9 ft/sec. What is the rate that the tip of the per
nekit [7.7K]

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

6 0
3 years ago
Determine the values of xfor which the function can be replaced by the Taylor polynomial if the error cannot exceed 0.001.(Enter
MrMuchimi

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)

R_{3} = (x) = (e^z /4!)x^4

Since the aim is R_{3}(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.

3 0
3 years ago
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