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jeka57 [31]
3 years ago
6

Type a digit that makes this statement true. 17,421,55 is divisible by 6.

Mathematics
2 answers:
Anettt [7]3 years ago
8 0

Answer:

<h2>290359.166667</h2>

Step-by-step explanation:

17,421,55/6=290359.166667

kompoz [17]3 years ago
5 0

Answer:

290359.166667

Step-by-step explanation:

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Factories fully 8x^2+6x
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8x^2+6x

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2x(4x+3)

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The half-life of the radioactive element unobtanium-43 is 10 seconds. If 48 grams of unobtanium-43 are initially present, how ma
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Answer:

At time, 10 seconds, the mass remaining is 24 grams

At time, 20 seconds, the mass remaining is 12 grams

At time, 30 seconds, the mass remaining is 6 grams

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At time, 50 seconds, the mass remaining is 1.5 grams

Step-by-step explanation:

Given;

initial mass of the radioactive element  = 48 grams

half life, t = 10 seconds

time of decay (s)                          remaining mass of the  radioactive element

0 ------------------------------------------------- 48 grams

10------------------------------------------------- 24 grams

20 ------------------------------------------------- 12 grams

30 -------------------------------------------------- 6 grams

40 --------------------------------------------------- 3 grams

50 ----------------------------------------------------- 1.5 grams

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Thus;

At time, 10 seconds, the mass remaining is 24 grams

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7 0
3 years ago
1) Let f(x)=6x+6/x. Find the open intervals on which f is increasing (decreasing). Then determine the x-coordinates of all relat
brilliants [131]

Answer:

1) increasing on (-∞,-1] ∪ [1,∞), decreasing on [-1,0) ∪ (0,1]

x = -1 is local maximum, x = 1 is local minimum

2) increasing on [1,∞), decreasing on (-∞,0) ∪ (0,1]

x = 1 is absolute minimum

3) increasing on (-∞,0] ∪ [8,∞), decreasing on [0,4) ∪ (4,8]

x = 0 is local maximum, x = 8 is local minimum

4) increasing on [2,∞), decreasing on (-∞,2]

x = 2 is absolute minimum

5) increasing on the interval (0,4/9], decreasing on the interval [4/9,∞)

x = 0 is local minimum, x = 4/9 is absolute maximum

Step-by-step explanation:

To find minima and maxima the of the function, we must take the derivative and equalize it to zero to find the roots.

1) f(x) = 6x + 6/x

f\prime(x) = 6 - 6/x^2 = 0 and x \neq 0

So, the roots are x = -1 and x = 1

The function is increasing on the interval (-∞,-1] ∪ [1,∞)

The function is decreasing on the interval [-1,0) ∪ (0,1]

x = -1 is local maximum, x = 1 is local minimum.

2) f(x)=6-4/x+2/x^2

f\prime(x)=4/x^2-4/x^3=0 and x \neq 0

So the root is x = 1

The function is increasing on the interval [1,∞)

The function is decreasing on the interval (-∞,0) ∪ (0,1]

x = 1 is absolute minimum.

3) f(x) = 8x^2/(x-4)

f\prime(x) = (8x^2-64x)/(x-4)^2=0 and x \neq 4

So the roots are x = 0 and x = 8

The function is increasing on the interval (-∞,0] ∪ [8,∞)

The function is decreasing on the interval [0,4) ∪ (4,8]

x = 0 is local maximum, x = 8 is local minimum.

4) f(x)=6(x-2)^{2/3} +4=0

f\prime(x) = 4/(x-2)^{1/3} has no solution and x = 2 is crtitical point.

The function is increasing on the interval [2,∞)

The function is decreasing on the interval (-∞,2]

x = 2 is absolute minimum.

5) f(x)=8\sqrt x - 6x for x>0

f\prime(x) = (4/\sqrt x)-6 = 0

So the root is x = 4/9

The function is increasing on the interval (0,4/9]

The function is decreasing on the interval [4/9,∞)

x = 0 is local minimum, x = 4/9 is absolute maximum.

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3 years ago
The purpose of a battery in a circuit is to _____.
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Answer:

Increase the potential of charges in the circuit

Step-by-step explanation:

Hope this is correct

HAVE A GOOD DAY!

7 0
3 years ago
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