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dimulka [17.4K]
3 years ago
9

What is 6 / 247 and what is the remainder

Mathematics
1 answer:
OLEGan [10]3 years ago
6 0

Answer:

Set up the long division.


2 4 7 | 6



2 Calculate 600 ÷ 247, which is 2 with a remainder of 106.


0 . 0 2

2 4 7 | 6 .  

4 . 9 4

1 . 0 6


3 Bring down 0, so that 1060 is large enough to be divided by 247.


0 . 0 2

2 4 7 | 6 .  

4 . 9 4

1 . 0 6 0


4 Calculate 1060 ÷ 247, which is 4 with a remainder of 72.


0 . 0 2 4

2 4 7 | 6 .  

4 . 9 4

1 . 0 6 0

9 8 8

7 2


5 Bring down 0, so that 720 is large enough to be divided by 247.


0 . 0 2 4

2 4 7 | 6 .    

4 . 9 4  

1 . 0 6 0

9 8 8

7 2 0


6 Calculate 720 ÷ 247, which is 2 with a remainder of 226.


0 . 0 2 4 2

2 4 7 | 6 .    

4 . 9 4  

1 . 0 6 0

9 8 8

7 2 0

4 9 4

2 2 6


7 Therefore, 6 ÷ 247 ≈ 0.0242.

0

.

0

2

4

2

0.0242




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klio [65]
C bc 294x*10-3 x +10 = -27
7 0
3 years ago
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Mashcka [7]

For this case we have that by definition of power properties it is fulfilled that:

(a ^ n) ^ m = a ^ {n * m}

We must rewrite the following function:

f (x) = 7 (\frac {1} {2})^{3x}}

Using the mentioned property we have:

f (x) = 7 ((\frac {1} {2}) ^ 3) ^ x

Solving the operation within the parenthesis we have:

f (x) = 7 (\frac {1 ^ 3} {2 ^ 3}) ^{ x}\\f (x) = 7 (\frac {1} {8})^{x}

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Option B

5 0
3 years ago
I think I know this much so far for A (but I could be wrong):
Eduardwww [97]
Part A. You have the correct first and second derivative.

---------------------------------------------------------------------

Part B. You'll need to be more specific. What I would do is show how the quantity (-2x+1)^4 is always nonnegative. This is because x^4 = (x^2)^2 is always nonnegative. So (-2x+1)^4 >= 0. The coefficient -10a is either positive or negative depending on the value of 'a'. If a > 0, then -10a is negative. Making h ' (x) negative. So in this case, h(x) is monotonically decreasing always. On the flip side, if a < 0, then h ' (x) is monotonically increasing as h ' (x) is positive.

-------------------------------------------------------------

Part C. What this is saying is basically "if we change 'a' and/or 'b', then the extrema will NOT change". So is that the case? Let's find out

To find the relative extrema, aka local extrema, we plug in h ' (x) = 0
h  ' (x) = -10a(-2x+1)^4
0 = -10a(-2x+1)^4
so either
-10a = 0 or (-2x+1)^4 = 0
The first part is all we care about. Solving for 'a' gets us a = 0. 
But there's a problem. It's clearly stated that 'a' is nonzero. So in any other case, the value of 'a' doesn't lead to altering the path in terms of finding the extrema. We'll focus on solving (-2x+1)^4 = 0 for x. Also, the parameter b is nowhere to be found in h ' (x) so that's out as well. 
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Alexxandr [17]
Hello,

Your answer would be:

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4 0
4 years ago
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