Question

all quetion sloved please fast

Answer 1

$<b >Hey here is your answer$

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$\color{blue} {\boxed{\mathbb{\underline{QUESTION \: NO \: :- \: 6}}}}$

Given :-

<B = 80°
AB = BC

To find :- all the angles of ∆.

In ∆ ABC

AB = BC
<A = <C (angles opposite to equal side of ∆). -(1)

Now ,

In ∆ ABC
<A + <B + <C = 180°. (angle sum property of ∆)
<A + 80° + <A = 180°
2<A = 180° - 80°
2<A = 100°
<A = 100°/2
<A = 50°

so all the angles of ∆ are :-

<A = 50°
<B = 80°
<C = 50°

$\color{blue} {\boxed{\mathbb{\underline{QUESTION \: NO \: :- \: 7}}}}$

$= \frac{6 - 4\sqrt{3} }{6 + 4 \sqrt{3} } \\ \\ = \frac{6 - 4 \sqrt{3} }{6 + 4 \sqrt{3} } \times\frac{6 - 4 \sqrt{3} }{6 - 4 \sqrt{3} } \\ \\ = \frac{ {(6 - 4 \sqrt{3}) }^{2} }{ {(6)}^{2} - {(4 \sqrt{3}) }^{2} } \\ \\ = \frac{ {(6)}^{2} + {(4 \sqrt{3} )}^{2} - 2 \times 6 \times 4 \sqrt{3} }{36 - 48} \\ \\ = \frac{36 + 48 - 48 \sqrt{3} }{ - 12} \\ \\ = \frac{84 - 48 \sqrt{3} }{ - 12} \\ \\ = \frac{7 - 4 \sqrt{3} }{ - 1} \\ \\ = - 7 + 4 \sqrt{3}$

answer

$\color{blue} {\boxed{\mathbb{\underline{QUESTION \: NO \: :- \: 8}}}}$

$evaluate \: \: {( - 12)}^{3} + {7}^{3} + {5}^{3}$
using identity when a + b + c = 0

${a}^{3} + {b}^{3} + {c}^{3} = 3abc$
so,

$= {( - 12)}^{3} + {7}^{3} + {5}^{3} \\ \\ = 3 \times - 12 \times 7 \times 5 \\ \\ = - 1260$
answer

$\color{blue} {\boxed{\mathbb{\underline{QUESTION \: NO \: :- \: 10}}}}$

see in the above attachment pls .

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Hope this helps you

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