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artcher [175]
4 years ago
5

You have a jug of milk from which you have to remove one cup milk aside. but you only have a three-cup measuring container and a

five-cup measuring container with you. so how will you remove exactly one cup milk aside?
Mathematics
1 answer:
cluponka [151]4 years ago
8 0
Assuming the jug has an infinite amount of milk:

Pour into the five cup
Pour the five cups into the three cup, there is now two cups inside of the five cup container and three cups inside of the three cup container
Return the three cups into the jug, so now there is only two cups inside of the five cup container.
Move the two cups into the three cup container. 
Fill the five cup container again, so now there are two cups inside of the three cup container and five cups inside the five cup container.
Pour one cup from the five cup container into the three cup container, now there are three cups inside of the three cup container and four cups inside of the five cup container
Empty out the three cup container, so now there are only four cups inside of the five cup container.
Pour the four cups into the three cups, until you can't anymore, so there are three cups in the three cup container, and one cup in the five cup container.
Pour the three cups back into the container, and you're left with one cup of milk.

I hope this helped, there was probably a simpler way to do this, but whatever.
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Step-by-step explanation:

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Find the Maclaurin polynomials of orders n = 0, 1, 2, 3, and 4, and then find the nth Maclaurin polynomials for the function in
Zielflug [23.3K]

Answer:

Σ(-1)^kx^k for k = 0 to n

Step-by-step explanation:

The nth Maclaurin polynomials for f to be

Pn(x) = f(0) + f'(0)x + f''(0)x²/2! + f"'(0)x³/3! +. ......

The given function is.

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

Differentiate four times with respect to x

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

f'(x) = -1/(1+x)²

f''(x) = 2/(1+x)³

f'''(x) = -6/(1+x)⁴

f''''(x) = 24/(1+x)^5

To calculate with a coefficient of 1

f(0) = 1

f'(0) = -1

f''(0) = 2

f'''(0) = -6

f''''(0) = 24

Findinf Pn(x) for n = 0 to 4.

Po(x) = 1

P1(x) = 1 - x

P2(x) = 1 - x + x²

P3(x) = 1 - x+ x² - x³

P4(x) = 1 - x+ x² - x³+ x⁴

Hence, the nth Maclaurin polynomials is

1 - x+ x² - x³+ x⁴ +.......+(-1)^nx^n

= Σ(-1)^kx^k for k = 0 to n

6 0
3 years ago
30 POINTSSS!!!!
NARA [144]
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3 0
3 years ago
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Factor completely x^8- 16.
Nataly_w [17]

Answer:

Step-by-step explanation:


So in this example we'll be using the difference of squares which essentially states that: (a-b)(a+b)=a^2-b^2 or another way to think of it would be: a-b=(\sqrt{a}-\sqrt{b})(\sqrt{a}+\sqrt{b}). So in this example you'll notice both terms are perfect squares. in fact x^n is a perfect square as long as n is even. This is because if it's even it can be split into two groups evenly for example, in this case we have x^8. so the square root is x^4 because you can split this up into (x * x * x * x) * (x * x * x * x) = x^8. Two groups with equal value multiplying to get x^8, that's what the square root is. So using these we can rewrite the equation as:

x^8-16 = (x^4-4)(x^4+4)

Now in this case you'll notice the degree is still even (it's 4) and the 4 is also a perfect square, and it's a difference of squares in one of the factors, so it can further be rewritten:

x^4-4 = (x^2-2)(x^2+2)

So completely factored form is: (x^2-2)(x^2+4)(x^4+4)

I'm assuming that's considered completely factored but you can technically factor it further. While the identity difference of squares technically only applies to difference of squares, it can also be used on the sum of squares, but you need to use imaginary numbers. Because x^2+4 = x^2-(-4). and in this case a=x^2 and b=-4. So rewriting it as the difference of squares becomes: x^4+4 = x^4 - (-4) = (x^2-\sqrt{-4})(x^2+\sqrt{-4}) = (x^2-2i)(x^2+2i) just something that might be useful in some cases.

7 0
2 years ago
Read 2 more answers
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