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bogdanovich [222]
3 years ago
13

give an example of two principal amounts and two periods of time for which the simple interest turned at 2.42 percent would be e

qual
Mathematics
1 answer:
diamong [38]3 years ago
6 0
<span>Case 1: Principal: 10,000; Period: 1 year Case 2: Principal: 1,000; Period: 10 years Case 3: Principal: 5,000; Period: 2 years Case 4: Principal: 2,500; Period: 4 years Simple Interest is a product of period, principal and interest rate, for a given interest rate, the interest earned would be equal if the product of principals and period is same for any number of cases. In our examples, we see that the product of principal and period comes out to be 10,000 in each case.</span>
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Please help me finish these answers, I am not finding any thing to really help me out
Mandarinka [93]

Answer:

So

Step-by-step explanation:

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3 years ago
Write 603,478 in word form
Irina18 [472]
Six hundred three thousand four hundred seventy-eight
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Find the Fourier series of f on the given interval. f(x) = 1, ?7 &lt; x &lt; 0 1 + x, 0 ? x &lt; 7
Zolol [24]
f(x)=\begin{cases}1&\text{for }-7

The Fourier series expansion of f(x) is given by

\dfrac{a_0}2+\displaystyle\sum_{n\ge1}a_n\cos\frac{n\pi x}7+\sum_{n\ge1}b_n\sin\frac{n\pi x}7

where we have

a_0=\displaystyle\frac17\int_{-7}^7f(x)\,\mathrm dx
a_0=\displaystyle\frac17\left(\int_{-7}^0\mathrm dx+\int_0^7(1+x)\,\mathrm dx\right)
a_0=\dfrac{7+\frac{63}2}7=\dfrac{11}2

The coefficients of the cosine series are

a_n=\displaystyle\frac17\int_{-7}^7f(x)\cos\dfrac{n\pi x}7\,\mathrm dx
a_n=\displaystyle\frac17\left(\int_{-7}^0\cos\frac{n\pi x}7\,\mathrm dx+\int_0^7(1+x)\cos\frac{n\pi x}7\,\mathrm dx\right)
a_n=\dfrac{9\sin n\pi}{n\pi}+\dfrac{7\cos n\pi-7}{n^2\pi^2}
a_n=\dfrac{7(-1)^n-7}{n^2\pi^2}

When n is even, the numerator vanishes, so we consider odd n, i.e. n=2k-1 for k\in\mathbb N, leaving us with

a_n=a_{2k-1}=\dfrac{7(-1)-7}{(2k-1)^2\pi^2}=-\dfrac{14}{(2k-1)^2\pi^2}

Meanwhile, the coefficients of the sine series are given by

b_n=\displaystyle\frac17\int_{-7}^7f(x)\sin\dfrac{n\pi x}7\,\mathrm dx
b_n=\displaystyle\frac17\left(\int_{-7}^0\sin\dfrac{n\pi x}7\,\mathrm dx+\int_0^7(1+x)\sin\dfrac{n\pi x}7\,\mathrm dx\right)
b_n=-\dfrac{7\cos n\pi}{n\pi}+\dfrac{7\sin n\pi}{n^2\pi^2}
b_n=\dfrac{7(-1)^{n+1}}{n\pi}

So the Fourier series expansion for f(x) is

f(x)\sim\dfrac{11}4-\dfrac{14}{\pi^2}\displaystyle\sum_{n\ge1}\frac1{(2n-1)^2}\cos\frac{(2n-1)\pi x}7+\frac7\pi\sum_{n\ge1}\frac{(-1)^{n+1}}n\sin\frac{n\pi x}7
3 0
2 years ago
Please help! Find the domain of y = 4 square root 4x + 2
olga55 [171]

Answer:

x ≥ -1/2

Step-by-step explanation:

We know that we cannot graph imaginary numbers. Therefore, our <em>x </em>value has to be greater than or equal to 0:

To find our domain, we need to set the square root equal to zero:

√(4x + 2) = 0

4x + 2 = 0

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x = -1/2

We now know that no value below -1/2 can be used or we will get an imaginary number. Therefore, our answer is x ≥ -1/2

Alternatively, we can graph the function and analyze domain:

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alexgriva [62]
It should include zero
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