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finlep [7]
3 years ago
5

What are the next three terms in the sequence 100, 20, 4, 0.8, 0.16​

Mathematics
1 answer:
Yuri [45]3 years ago
4 0

Answer:0.032, 0.0064, 0.00128. PLEASE GIVE BRAINLIEST

Step-by-step explanation:

You can see that the sequence is /5. 100/5 Is 20, 20/5 is 4, 4/5 is 0.8 and so on.

The next 3 terms should Be 0.16/5= 0.032, 0.032/5=0.0064, 0.0064/5 is 0.00128

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Answer:

A

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Unlike fractions find the sum 2/8+1/3=
chubhunter [2.5K]

Answer:

7/12

Step-by-step explanation:

Find the common denominator 24 so fraction would change to 6/24 + 8/24

that equals 14/24 and simplified that 7/12

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Read 2 more answers
1/2 cm x 3/2cm x 7/4cm... Find the volume
forsale [732]
Remember, to the find the volume, you have to do length*width*height.

So it would be 7/4 * 3/2 * 1/2

To answer this, multiply all the numerators together, and multiply all the denominator together.

This would be:

21/16



So the answer is 21/16 cm^3 or 1.3125 cm^3.


Hope this helps. If you have any questions, place it in the comment section below.


6 0
3 years ago
consider the function and then use calculus to answer the questions that follow 1 1/x 5/x^2 1/x^3 (a) Find the interval(s) where
boyakko [2]

Answer:

a)X=((-15-\sqrt{201},(-15+\sqrt{201}),(0,\infty)

b)Y=(\infty,\frac{1}{2}(-15-\sqrt{201} ) ),(\frac{1}{2}()-15+\sqrt{201)},0  )

Step-by-step explanation:

From the question we are told that

The Function

f(x)=1+\frac{1}{x}  +\frac{5}{x^2} +\frac{1}{x^3}

Generally the differentiation of function f(x) is mathematically solved as

f(x)=1+\frac{1}{x}  +\frac{5}{x^2} +\frac{1}{x^3}

f(x)=\frac{x^3+x^2+5x+1}{x^2}

Therefore

f'(x)=\frac{x^2+10x+3}{x^4}

Generally critical point is given as

f'(x)=0

\frac{x^2+10x+3}{x^4}=0

x=-5 \pm\sqrt{22}

Generally the maximum and minimum x value for critical point is mathematically solved as

f'(-5 \pm\sqrt{22})

Where

Maximum value of x

f'(-5 +\sqrt{22})

Minimum value of x

f'(-5 +\sqrt{22})

Therefore interval of increase is mathematically given by

f'(-5 -\sqrt{22}),f'(-5 +\sqrt{22})

f(x)

Therefore interval of decrease is mathematically given by

(-\infty,-5 -\sqrt{22}),f'(-5 +\sqrt{22},0),(0,\infty)

Generally the second differentiation of function f(x) is mathematically solved as

f''(x)=\frac{2(x^2+15x+6)}{x^5}

Generally the point of inflection is mathematically solved as

f''(x)=0

x^2+15x+6=0

Therefore inflection points is given as

x=\frac{1}{2} (-15 \pm \sqrt{201}

f''(x)>0,\frac{1}{2}(-15-\sqrt{201})

a)Generally the concave upward interval X is mathematically given as

X=((-15-\sqrt{201},(-15+\sqrt{201}),(0,\infty)

f''(x)

b)Generally the concave downward interval Y is mathematically given as

Y=(\infty,\frac{1}{2}(-15-\sqrt{201} ) ),(\frac{1}{2}()-15+\sqrt{201)},0  )

5 0
2 years ago
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