Question

16÷√41-5 give me solutions ​

Answer 1

$\bigstar$$\boxed{\huge{\red{\mathfrak{Answer}}}}$$\bigstar$

_________________________

→ $\frac{16}{\sqrt{41}-5}$

$\mathtt{Rationalising\:The\:denominator}$

→ $\frac{16}{\sqrt{41}-5}\times\frac{\sqrt{41}+5}{\sqrt{41}+5}$

$\mathtt{Using\:identity ,}$

$\mathtt{(a+b)(a-b)={a}^{2}-{b}^{2}}$

→ $\frac{16(\sqrt{41}+5)}{41-25}$

→ $\frac{16(\sqrt{41}+5)}{16}$

→ $\mathtt{\huge{\sqrt{41}+5}}$

Answer 2

Answer:

Two point reoccurring seven.

Step-by-step explanation:

41-5 = 36

= /~36

= 6

16÷6

= Two point reoccurring seven.

Hope this helps.

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