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3 years ago

15

The amount of money in Joe’s savings account is increased by 8% each year. Given that he has £500 in 2016. Work out how much Joe

will have in 2021.

1 answer:

ololo11 [35]3 years ago

5 0

Answer:

£700

Step-by-step explanation:

First we find the interest on this money.

Interest = Principal * Rate * Time

= $\frac{500 * 8 * 5}{100}$

=5 * 8 * 5

=£200

Interest = £200

Now you add his actual money to the interest you just found to give you the total amount.

Amount = £500 + £200

= £700

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trapecia [35]

Assuming you want the cosine series expansion over an arbitrary symmetric interval $[-L,L]$ , $L\neq0$ , the cosine series is given by

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You have

$a_0=\displaystyle\frac1L\int_{-L}^Lf(x)\,\mathrm dx$
$a_0=\dfrac1L\left(\dfrac{x^2}2-\dfrac{x^3}3\right)\bigg|_{x=-L}^{x=L}$
$a_0=\dfrac1L\left(\left(\dfrac{L^2}2-\dfrac{L^3}3\right)-\left(\dfrac{(-L)^2}2-\dfrac{(-L)^3}3\right)\right)$
$a_0=-\dfrac{2L^2}3$

$a_n=\displaystyle\frac1L\int_{-L}^Lf(x)\cos nx\,\mathrm dx$

Two successive rounds of integration by parts (I leave the details to you) gives an antiderivative of

$\displaystyle\int(x-x^2)\cos nx\,\mathrm dx=\frac{(1-2x)\cos nx}{n^2}-\dfrac{(2+n^2x-n^2x^2)\sin nx}{n^3}$

and so

$a_n=-\dfrac{4L\cos nL}{n^2}+\dfrac{(4-2n^2L^2)\sin nL}{n^3}$

So the cosine series for $f(x)$ periodic over an interval $[-L,L]$ is

$f_C(x)=-\dfrac{L^2}3+\displaystyle\sum_{n\ge1}\left(-\dfrac{4L\cos nL}{n^2L}+\dfrac{(4-2n^2L^2)\sin nL}{n^3L}\right)\cos nx$

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ANTONII [103]

ANSWER

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EXPLANATION
We were given that
$f(x)$
varies directly as
$x.$

We can write this mathematically as,

$f(x) \propto \: x.$

This implies that,

$f(x) = kx$
where k is the constant of variation.

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This implies that,

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$k = 3$

The equation becomes

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When $x=60$

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Answer:

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