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2 years ago

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Solve for x -

exFormula1" title=" x^{2} + 5x + 6 = 0 \\ \\ " alt=" x^{2} + 5x + 6 = 0 \\ \\ " align="absmiddle" class="latex-formula">

ty! :)

1 answer:

Volgvan2 years ago

3 0

Hello,

x² + 5x + 6 = 0

a = 1 ; b = 5 ; c = 6

∆ = b² - 4ac = 5² - 4 × 1 × 6 = 25 - 24 = 1 > 0

2 solutions :

$x_{1} = \frac{ - b - \sqrt{\Delta} }{2a} = \frac{ - 5 - 1}{2 \times 1} = \frac{ - 6}{2} = - 3$

$x_{2} = \frac{ - b + \sqrt{\Delta} }{2a} = \frac{ - 5 + 1}{2 \times 1} = \frac{ - 4}{2} = - 2$

S = { -3 ; -2}

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For the following telescoping series, find a formula for the nth term of the sequence of partial sums {Sn}. Then evaluate limn→[

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Answer:

The following are the solution to the given points:

Step-by-step explanation:

Given value:

$1) \sum ^{\infty}_{k = 1} \frac{1}{k+1} - \frac{1}{k+2}\\\\2) \sum ^{\infty}_{k = 1} \frac{1}{(k+6)(k+7)}$

Solve point 1 that is $\sum ^{\infty}_{k = 1} \frac{1}{k+1} - \frac{1}{k+2}\\\\$ :

when,

$k= 1 \to s_1 = \frac{1}{1+1} - \frac{1}{1+2}\\\\$

$= \frac{1}{2} - \frac{1}{3}\\\\$

$k= 2 \to s_2 = \frac{1}{2+1} - \frac{1}{2+2}\\\\$

$= \frac{1}{3} - \frac{1}{4}\\\\$

$k= 3 \to s_3 = \frac{1}{3+1} - \frac{1}{3+2}\\\\$

$= \frac{1}{4} - \frac{1}{5}\\\\$

$k= n^ \to s_n = \frac{1}{n+1} - \frac{1}{n+2}\\\\$

Calculate the sum $(S=s_1+s_2+s_3+......+s_n)$

$S=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+.....\frac{1}{n+1}-\frac{1}{n+2}\\\\$

$=\frac{1}{2}-\frac{1}{5}+\frac{1}{n+1}-\frac{1}{n+2}\\\\$

When $s_n \ \ dt_{n \to 0}$

$=\frac{1}{2}-\frac{1}{5}+\frac{1}{0+1}-\frac{1}{0+2}\\\\=\frac{1}{2}-\frac{1}{5}+\frac{1}{1}-\frac{1}{2}\\\\= 1 -\frac{1}{5}\\\\= \frac{5-1}{5}\\\\= \frac{4}{5}\\\\$

$\boxed{\text{In point 1:} \sum ^{\infty}_{k = 1} \frac{1}{k+1} - \frac{1}{k+2} =\frac{4}{5}}$

In point 2: $\sum ^{\infty}_{k = 1} \frac{1}{(k+6)(k+7)}$

when,

$k= 1 \to s_1 = \frac{1}{(1+6)(1+7)}\\\\$

$= \frac{1}{7 \times 8}\\\\= \frac{1}{56}$

$k= 2 \to s_1 = \frac{1}{(2+6)(2+7)}\\\\$

$= \frac{1}{8 \times 9}\\\\= \frac{1}{72}$

$k= 3 \to s_1 = \frac{1}{(3+6)(3+7)}\\\\$

$= \frac{1}{9 \times 10} \\\\ = \frac{1}{90}\\\\$

$k= n^ \to s_n = \frac{1}{(n+6)(n+7)}\\\\$

calculate the sum: $S= s_1+s_2+s_3+s_n\\$

$S= \frac{1}{56}+\frac{1}{72}+\frac{1}{90}....+\frac{1}{(n+6)(n+7)}\\\\$

when $s_n \ \ dt_{n \to 0}$

$S= \frac{1}{56}+\frac{1}{72}+\frac{1}{90}....+\frac{1}{(0+6)(0+7)}\\\\= \frac{1}{56}+\frac{1}{72}+\frac{1}{90}....+\frac{1}{6 \times 7}\\\\= \frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{42}\\\\=\frac{45+35+28+60}{2520}\\\\=\frac{168}{2520}\\\\=0.066$

$\boxed{\text{In point 2:} \sum ^{\infty}_{k = 1} \frac{1}{(n+6)(n+7)} = 0.066}$

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3 years ago

How much money has to be invested at 5.9% interest compounded continuously to have $15,000 after 12 years?

scoray [572]

The amount of $7389.43 has to be invested at 5.9% interested continuously to have $15,000 after 12 years.

Step-by-step explanation:

The given is,

Future value, F = $15,000

Interest, i = 5.9%

( compounded continuously )

Period, t = 12 years

Step:1

Formula to calculate the present with compounded continuously,

$F=Pe^{(i)(t)}$ ...............(1)

Substitute the values in equation (1) to find the P value,

$15000=Pe^{(0.059)(12)}$ ( ∵ $i = \frac{5.9}{100}=0.059$ )

$15000=Pe^{0.708}$

$15000=P(2.0299)$ ( ∵ $e^{o.708} =2.0299$ )

We change the P (Present value) into the left side,

$P=\frac{15000}{2.0299}$

$=7389.427$

≅ 7389.43

P = $ 7389.43

Result:

The amount of $7389.43 has to be invested at 5.9% interested continuously to have $15,000 after 12 years.

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3 years ago

1.5x - 3y = 6 (solve for y)

d1i1m1o1n [39]

Hello :
<span>1.5x - 3y = 6
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4 years ago

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