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bearhunter [10]

2 years ago

14

Compute the amount of interest earned in the following simple interest problem. A deposit of $1,295 at 7% for 180 days = _____.

(Note: Use 365 days in a year)

1 answer:

givi [52]2 years ago

7 0

I = Prt, where P = $1,295; r = 7/100 = 0.07, t = 180/365

I = 1,295 x 0.07 x 180/365 = $44.70

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andrey2020 [161]

First observe that if $a+b>0$ ,

$(a + b)^2 = a^2 + 2ab + b^2 \\\\ \implies a + b = \sqrt{a^2 + 2ab + b^2} = \sqrt{a^2 + ab + b(a + b)} \\\\ \implies a + b = \sqrt{a^2 + ab + b \sqrt{a^2 + ab + b(a+b)}} \\\\ \implies a + b = \sqrt{a^2 + ab + b \sqrt{a^2 + ab + b \sqrt{a^2 + ab + b(a+b)}}} \\\\ \implies a + b = \sqrt{a^2 + ab + b \sqrt{a^2 + ab + b \sqrt{a^2 + ab + b \sqrt{\cdots}}}}$

Let $a=0$ and $b=x$ . It follows that

$a+b = x = \sqrt{x \sqrt{x \sqrt{x \sqrt{\cdots}}}}$

Now let $b=1$ , so $a^2+a=4x$ . Solving for $a$ ,

$a^2 + a - 4x = 0 \implies a = \dfrac{-1 + \sqrt{1+16x}}2$

which means

$a+b = \dfrac{1 + \sqrt{1+16x}}2 = \sqrt{4x + \sqrt{4x + \sqrt{4x + \sqrt{\cdots}}}}$

Now solve for $x$ .

$x = \dfrac{1 + \sqrt{1 + 16x}}2 \\\\ 2x = 1 + \sqrt{1 + 16x} \\\\ 2x - 1 = \sqrt{1 + 16x} \\\\ (2x-1)^2 = \left(\sqrt{1 + 16x}\right)^2$

(note that we assume $2x-1\ge0$ )

$4x^2 - 4x + 1 = 1 + 16x \\\\ 4x^2 - 20x = 0 \\\\ 4x (x - 5) = 0 \\\\ 4x = 0 \text{ or } x - 5 = 0 \\\\ \implies x = 0 \text{ or } \boxed{x = 5}$

(we omit $x=0$ since $2\cdot0-1=-1\ge0$ is not true)

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1 year ago

Helppp me get this this right

irakobra [83]

Answer:

$4^{16}$

Step-by-step explanation:

$16^{8}=\\16*16*16*16*16*16*16*16=\\4*4*4*4*4*4*4*4*4*4*4*4*4*4*4*4=\\4^{16}$

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Juli2301 [7.4K]

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