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Illusion [34]
2 years ago
10

40 pounds of sugar are repackaged of 1/16 pounds each. how many packet of sugar are there?

Mathematics
1 answer:
jarptica [38.1K]2 years ago
8 0
We know to each pound of such, there is 16 packets produced per?

So, 16 x 40 = 640.
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How do change 2/5 into a decimal​
kolbaska11 [484]

Answer: 0.4

Step-by-step explanation:

Divide the 2 by 5 and get 0.4, which is the decimal form.

7 0
3 years ago
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SOMEONE PLEASE HELP ALL THE QUESTIONS R ATTACHED AS WELL AS THE ANSWERS!!!!!!!!!!!!!!
Reil [10]

The error is in <em>Step-4</em>.

A negative exponent does NOT mean that the number turns negative. A negative exponent means the number is in the denominator.

(4)⁻⁴ means (1/4⁴) .  That's (1/256) .  All positive numbers.


8 0
3 years ago
What values would make a ploynimal equal to zero if the factors are (x+3) and (x+12)
Anna71 [15]
The values 35 and 25 would make the answer to this 0
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7. Solve the formula M=2P+3Q for the variable Q.
vichka [17]
The answer is <span>Q=(M−2P)/3
</span>
<span>M = 2P + 3Q

Subtract 2P from both sides:
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M - 2P = 3Q

Divide both sides by 3:
(M - 2P)/3 = 3Q/3
(M - 2P)/3 = Q</span>
7 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. 
6 0
3 years ago
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