Question

Solve:
$\sf \sqrt{ \dfrac{44 \times 4}{ \sin = \theta} }$
Or
$\sf \dfrac{ \sqrt{44} }{ \sqrt{8888} } \times 3449$
Solve any one of these​

Answer 1

Answer:

Hi young brother!

$\huge \dashrightarrow \sf {\blue {\underline \red{question}}}$

$\sf \dfrac{ \sqrt{44} }{ \sqrt{8888} } \times 3449$

$\green{\boxed{ \huge\sf \red{ anwser}}} \huge\downarrow$

$\sf \: cancel \: \frac{ \sqrt{44} }{ \sqrt{8888} } \\ \\ = \frac{1}{ \sqrt{202} }$

$= \frac{1}{ \sqrt{202} } \times 3449$

Now convert 3449 to fraction

$= \frac{3449}{1}$

$= \frac{1}{ \sqrt{202} } \times \frac{3449}{1} \\ \\ = \frac{1 \times 3449}{ \sqrt{202} \times 1 } \\ \\ = \frac{3449}{ \sqrt{202} }$

Hope it helps you from my side

:)

Answer 2

Answer:

REFER TO THE ATTACHMENT PLEASE

a²+b²+c²=10

a²+b²+c²=10squaring both sides

a²+b²+c²=10squaring both sides=> a+b+c+ + 2(a²b² + b²c² + a²c²) = 100

a²+b²+c²=10squaring both sides=> a+b+c+ + 2(a²b² + b²c² + a²c²) = 100=> a+b+c² + 2(25) = 100

a²+b²+c²=10squaring both sides=> a+b+c+ + 2(a²b² + b²c² + a²c²) = 100=> a+b+c² + 2(25) = 100=> a+b+c4 +50= 100

a²+b²+c²=10squaring both sides=> a+b+c+ + 2(a²b² + b²c² + a²c²) = 100=> a+b+c² + 2(25) = 100=> a+b+c4 +50= 100=> a + b + c = 50

a²+b²+c²=10squaring both sides=> a+b+c+ + 2(a²b² + b²c² + a²c²) = 100=> a+b+c² + 2(25) = 100=> a+b+c4 +50= 100=> a + b + c = 50Learn More:

a²+b²+c²=10squaring both sides=> a+b+c+ + 2(a²b² + b²c² + a²c²) = 100=> a+b+c² + 2(25) = 100=> a+b+c4 +50= 100=> a + b + c = 50Learn More:a³ + b³ + c³-3abc = (a + b + c)(a + b² + c²-ab- bc - ca).

a²+b²+c²=10squaring both sides=> a+b+c+ + 2(a²b² + b²c² + a²c²) = 100=> a+b+c² + 2(25) = 100=> a+b+c4 +50= 100=> a + b + c = 50Learn More:a³ + b³ + c³-3abc = (a + b + c)(a + b² + c²-ab- bc - ca).brainly.in/question/1195178

a²+b²+c²=10squaring both sides=> a+b+c+ + 2(a²b² + b²c² + a²c²) = 100=> a+b+c² + 2(25) = 100=> a+b+c4 +50= 100=> a + b + c = 50Learn More:a³ + b³ + c³-3abc = (a + b + c)(a + b² + c²-ab- bc - ca).brainly.in/question/1195178If a+b+c=6 and a2+b2+c2=14 and a3+b3+c3-36 find value of abc ...