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Sauron [17]
3 years ago
7

Walter had 2/4 yard of string. He used 3/6 of it on a school project. How much string did he use?

Mathematics
2 answers:
andrezito [222]3 years ago
8 0

1/4 is the correct answer


frozen [14]3 years ago
3 0

2/4 reduces to  1/2

3/6 reduces to 1/2

1/2 * 1/2 = 1/4

 he used 1/4 yard of string

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An artist mixes colors for a painting. She mixes 1.5 ounces of blue with 2.5 ounces of yellow to get the perfect shade of green.
lesya692 [45]

Answer:

4.5 ounces of yellow and 7.5 ounces of blue will create 12 ounces of the perfect shade of green.

Step-by-step explanation:

1.5 + 2.5 = 4 ounces.

Multiply by 3

4.5 + 7.5 = 12 ounces

So, she needs 4.5 ounces of yellow.

4 0
3 years ago
Answer the following questions CORRECTLY I will know if this is wrong. I WILL REPORT ANY INCORRECT ANSWERS!
timurjin [86]

Answer by JKismyhusbandbae: add 5 to 23

Explanation: The numbers are increasing by 5 each term. To double check 13 - 8 equals 5. To make sure it's right add 13 + 5 equals 18 and so you can add 23 plus 5 equals 28. Which means that five is our pattern.

4 0
3 years ago
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What is the solution set of the quadratic inequality 4(x+2)^2<_0?
sattari [20]

The solution is in the picture

8 0
3 years ago
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<img src="https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Csf%5Clim_%7Bx%20%5Cto%200%20%7D%20%5Cfrac%7B1%20-%20%5Cprod%20%5Climits_%
xxTIMURxx [149]

To demonstrate a method for computing the limit itself, let's pick a small value of n. If n = 3, then our limit is

\displaystyle \lim_{x \to 0 } \frac{1 - \prod \limits_{k = 2}^{3} \sqrt[k]{\cos(kx)} }{ {x}^{2} }

Let a = 1 and b the cosine product, and write them as

\dfrac{a - b}{x^2}

with

b = \sqrt{\cos(2x)} \sqrt[3]{\cos(3x)} = \sqrt[6]{\cos^3(2x)} \sqrt[6]{\cos^2(3x)} = \left(\cos^3(2x) \cos^2(3x)\right)^{\frac16}

Now we use the identity

a^n-b^n = (a-b)\left(a^{n-1}+a^{n-2}b+a^{n-3}b^2+\cdots a^2b^{n-3}+ab^{n-2}+b^{n-1}\right)

to rationalize the numerator. This gives

\displaystyle \frac{a^6-b^6}{x^2 \left(a^5+a^4b+a^3b^2+a^2b^3+ab^4+b^5\right)}

As x approaches 0, both a and b approach 1, so the polynomial in a and b in the denominator approaches 6, and our original limit reduces to

\displaystyle \frac16 \lim_{x\to0} \frac{1-\cos^3(2x)\cos^2(3x)}{x^2}

For the remaining limit, use the Taylor expansion for cos(x) :

\cos(x) = 1 - \dfrac{x^2}2 + \mathcal{O}(x^4)

where \mathcal{O}(x^4) essentially means that all the other terms in the expansion grow as quickly as or faster than x⁴; in other words, the expansion behaves asymptotically like x⁴. As x approaches 0, all these terms go to 0 as well.

Then

\displaystyle \cos^3(2x) \cos^2(3x) = \left(1 - 2x^2\right)^3 \left(1 - \frac{9x^2}2\right)^2

\displaystyle \cos^3(2x) \cos^2(3x) = \left(1 - 6x^2 + 12x^4 - 8x^6\right) \left(1 - 9x^2 + \frac{81x^4}4\right)

\displaystyle \cos^3(2x) \cos^2(3x) = 1 - 15x^2 + \mathcal{O}(x^4)

so in our limit, the constant terms cancel, and the asymptotic terms go to 0, and we end up with

\displaystyle \frac16 \lim_{x\to0} \frac{15x^2}{x^2} = \frac{15}6 = \frac52

Unfortunately, this doesn't agree with the limit we want, so n ≠ 3. But you can try applying this method for larger n, or computing a more general result.

Edit: some scratch work suggests the limit is 10 for n = 6.

6 0
2 years ago
What is 11/12times 13/14
AnnyKZ [126]
11/12 is .92 and 13/14 is .93 multiply them together and get .85 roughly
3 0
3 years ago
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