Solve frac{sqrt{6,7*10^-113,3*10^23}}{2440}

$$\frac { \sqrt { 6,7 \cdot 10 ^ { - 11 } 3,3 \cdot 10 ^ { 23 } } } { 2440 }$$

Evaluate

$$\frac{2500\sqrt{2211}}{61}\approx 1927.101302542$$

Solution Steps

To multiply powers of the same base, add their exponents. Add -11 and 23 to get 12.

$$\frac{\sqrt{6,7\times 10^{12}\times 3,3}}{2440}$$

Calculate 10 to the power of 12 and get 1000000000000.

$$\frac{\sqrt{6,7\times 1000000000000\times 3,3}}{2440}$$

Multiply 6,7 and 1000000000000 to get 6700000000000.

$$\frac{\sqrt{6700000000000\times 3,3}}{2440}$$

Multiply 6700000000000 and 3,3 to get 22110000000000.

$$\frac{\sqrt{22110000000000}}{2440}$$

Factor 22110000000000=100000^{2}\times 2211. Rewrite the square root of the product \sqrt{100000^{2}\times 2211} as the product of square roots \sqrt{100000^{2}}\sqrt{2211}. Take the square root of 100000^{2}.

$$\frac{100000\sqrt{2211}}{2440}$$

Divide 100000\sqrt{2211} by 2440 to get \frac{2500}{61}\sqrt{2211}.

$$\frac{2500}{61}\sqrt{2211}$$