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erma4kov [3.2K]
3 years ago
10

Divide £16 in the ratio 3 : 1

Mathematics
2 answers:
lana66690 [7]3 years ago
7 0

Answer:

12 : 4

Step-by-step explanation:

3:1

3+1 = 4

16 / 4 = 4

3 x 4 =12

1 x 4 =4

12 : 4

seraphim [82]3 years ago
4 0
£12:£4
Add the ratio together to get 4 and then divide 16 by 4
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Step-by-step explanation:

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value if x is -21.25.

...............

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We have

$\sum_{x=8}^{4}x^2 + 9\left(\dfrac{\cos^2(\infty)}{\sin^2(\infty)} \right)$

For the sum it is not correct to assume

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Note that for

$\sum_{x=a}^b f(x)$

it is assumed a\leq x \leq b and in your case \nexists x\in\mathbb{Z}: a\leq x\leq b for a>b

In fact, considering a set S we have

$\sum_{x=a}^b (S \cup \varnothing) = \sum_{x=a}^b S + \sum_{x=a}^b \varnothing $ that satisfy S = S \cup \varnothing

This means that, by definition \sum_{x=a}^b \varnothing = 0

Therefore,

$\sum_{x=8}^{4}x^2 = 0$

because the sum is empty.

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9\left(\dfrac{\cos^2(\infty)}{\sin^2(\infty)} \right)

we have other problems. Actually, this case is really bad.

Note that \cos^2(\infty) has no value. In fact, if we consider for the case

$\lim_{x \to \infty} \cos^2(x)$, the cosine function oscillates between [-1, 1] , and therefore it is undefined. Thus, we cannot evaluate

9\left(\dfrac{\cos^2(\infty)}{\sin^2(\infty)} \right)

and then

$\sum_{x=8}^{4}x^2 + 9\left(\dfrac{\cos^2(\infty)}{\sin^2(\infty)} \right)$

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