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3 years ago

7+3√5/3+√5 - 7-3√5/3-√5 = a + b√5

Mathematics

2 answers:

Olenka [21]3 years ago

8 0

$\begin{gathered} \\\frac{7 + 3 \sqrt{5} }{3 + \sqrt{5} } - \frac{7 - 3 \sqrt{5} }{3 - \sqrt{5} } = a + \sqrt{5} b \\ \end{gathered}$

$\begin{gathered} \\ \frac{( \: 7 + 3 \sqrt{5} \: \: ( 3 - \sqrt{5}) \: \: - 7 - 3 \sqrt{5} \: \: ( 3 + \sqrt{5}) \: }{(3 + \sqrt{5}) \: \:(3 + \sqrt{5}) } = a + \sqrt{5} \: b\\ \end{gathered}$

$\begin{gathered} \\ \frac{( \: 21 - 7 \sqrt{5} \: + 9 \sqrt{5} - 15) \: \: - ( \: 21 + 7 \sqrt{5} \: - 9 \sqrt{5} + 15)\: }{(3 + \sqrt{5}) \: \:(3 + \sqrt{5}) } = a + \sqrt{5} \: b\\ \end{gathered}$

$\begin{gathered} \\ \frac{( \: 6 + 2 \sqrt{5} ) \: \: - ( \: 6 - 2 \sqrt{5} )\: }{(3 + \sqrt{5}) \: \:(3 + \sqrt{5}) } = a + \sqrt{5} \: b\\ \end{gathered}$

$\begin{gathered} \\ \frac{\: 6 + 2 \sqrt{5} \: \: - \: \: 6 - 2 \sqrt{5} \: }{(3 + \sqrt{5}) \: \:(3 + \sqrt{5}) } = a + \sqrt{5} \: b\\ \end{gathered}$

$\begin{gathered} \\ \frac{\: 4 \sqrt{5} \: \: }{3 {}^{2} - {\sqrt{5} }^{2} } = a + \sqrt{5} \: b\\ \end{gathered}$

$\begin{gathered} \\ \frac{\: 4 \sqrt{5} \: \: }{ \: \: \: \: 9 - 5 \: \: \: } = a + \sqrt{5} \: b\\ \end{gathered}$

$\begin{gathered} \\ \frac{\: 4 \sqrt{5} \: \: }{ \: \: \: \: 4 \: \: \: } = a + \sqrt{5} \: b\\ \end{gathered}$

$\begin{gathered} \\ \: \sqrt{5} = a + \sqrt{5} \: b\\ \end{gathered}$

we can also write it as ;

$\begin{gathered} \\ \: 0 + \sqrt{5} = a + \sqrt{5} \: b\\ \end{gathered}$

★ Henceforth, the value of a and b are :

→ a = 0

→ b = 1

daser333 [38]3 years ago

6 0

<h3>Given:-</h3>

$\\ \sf \implies\frac{7 + 3 \sqrt{5} }{3 + \sqrt{5} } - \frac{7 - 3 \sqrt{5} }{3 - \sqrt{5} } = a + \sqrt{5} b \\$

The value of a and b

<h3>Solution:-</h3>

$\\ \sf \implies\frac{7 + 3 \sqrt{5} }{3 + \sqrt{5} } - \frac{7 - 3 \sqrt{5} }{3 - \sqrt{5} } = a + \sqrt{5} b \\$

$\\ \sf \implies\frac{( \: 7 + 3 \sqrt{5} \: \: ( 3 - \sqrt{5}) \: \: - 7 - 3 \sqrt{5} \: \: ( 3 + \sqrt{5}) \: }{(3 + \sqrt{5}) \: \:(3 + \sqrt{5}) } = a + \sqrt{5} \: b\\$

$\\ \sf \implies\frac{( \: 21 - 7 \sqrt{5} \: + 9 \sqrt{5} - 15) \: \: - ( \: 21 + 7 \sqrt{5} \: - 9 \sqrt{5} + 15)\: }{(3 + \sqrt{5}) \: \:(3 + \sqrt{5}) } = a + \sqrt{5} \: b\\$

$\\ \sf \implies\frac{( \: 6 + 2 \sqrt{5} ) \: \: - ( \: 6 - 2 \sqrt{5} )\: }{(3 + \sqrt{5}) \: \:(3 + \sqrt{5}) } = a + \sqrt{5} \: b\\$

$\\ \sf \implies\frac{\: 6 + 2 \sqrt{5} \: \: - \: \: 6 - 2 \sqrt{5} \: }{(3 + \sqrt{5}) \: \:(3 + \sqrt{5}) } = a + \sqrt{5} \: b\\$

$\\ \sf \implies\frac{\: 4 \sqrt{5} \: \: }{3 {}^{2} - {\sqrt{5} }^{2} } = a + \sqrt{5} \: b\\$

$\\ \sf \implies\frac{\: 4 \sqrt{5} \: \: }{ \: \: \: \: 9 - 5 \: \: \: } = a + \sqrt{5} \: b\\$

$\\ \sf \implies\frac{\: 4 \sqrt{5} \: \: }{ \: \: \: \: 4 \: \: \: } = a + \sqrt{5} \: b\\$

$\\ \sf \implies\frac{\: \cancel{4 } \sqrt{5} \: \: }{ \: \: \: \: \cancel{4 }\: \: \: } = a + \sqrt{5} \: b\\$

$\\ \sf \implies \: \sqrt{5} = a + \sqrt{5} \: b\\$

we can also write it as ;

$\\ \sf \implies \: 0 + \sqrt{5} = a + \sqrt{5} \: b\\$

★<u> </u><u>Henceforth, the value of a and b are</u> :

→ a = 0

→ b = 1

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Find the area that the curve encloses and then sketch it.<br> r = 3 + 8 sin(6)

Rudiy27

Answer:

$A=41\pi\: \text{units}^2\approxA\approx128.8053\:\text{units}^2$

Step-by-step explanation:

I assume you mean $r=3+8\sin\theta$ :

Use the formula $\displaystyle A=\int\limits^a_b \frac{1}{2} {r(\theta)^2} \, d\theta$ where $a$ and $b$ are the lower and upper bounds and $r(\theta)$ is the equation of the polar curve.

Since the graph is symmetrical about the line $\displaystyle \theta=\frac{\pi}{2}$ , let the bounds of integration be $\displaystyle \biggr(-\frac{\pi}{2},\frac{\pi}{2}\biggr)$ to find half the area of the curve, and then find twice of that area:

$\displaystyle A=\int\limits^a_b \frac{1}{2} {r(\theta)^2} \, d\theta\\\\A=2\int\limits^{\frac{\pi}{2}}_{-\frac{\pi}{2}} \frac{1}{2} {(3+8\sin\theta)^2} \, d\theta\\\\A=\int\limits^{\frac{\pi}{2}}_{-\frac{\pi}{2}} 9+48\sin\theta+64\sin^2\theta \, d\theta\\\\A=\int\limits^{\frac{\pi}{2}}_{-\frac{\pi}{2}} 9+48\sin\theta+64\biggr(\frac{1-\cos2\theta}{2} \biggr) \, d\theta\\\\\\A=\int\limits^{\frac{\pi}{2}}_{-\frac{\pi}{2}} (9+48\sin\theta+32-32\cos2\theta) \, d\theta$

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Thus, the area of the curve is 41π square units. See below for a graph of the curve and its shaded area.

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nevsk [136]

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I may be wrong, ask for more opinions
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