Solve frac{sqrt{3452}}{260}*7^3/55+1/55

$$\frac { \sqrt { 3452 } } { 260 } \times \frac { 7 ^ { 3 } } { 55 } + \frac { 1 } { 55 }$$

Evaluate

$$\frac{343\sqrt{863}}{7150}+\frac{1}{55}\approx 1.427449447$$

Solution Steps

Factor 3452=2^{2}\times 863. Rewrite the square root of the product \sqrt{2^{2}\times 863} as the product of square roots \sqrt{2^{2}}\sqrt{863}. Take the square root of 2^{2}.

$$\frac{2\sqrt{863}}{260}\times \frac{7^{3}}{55}+\frac{1}{55}$$

Divide 2\sqrt{863} by 260 to get \frac{1}{130}\sqrt{863}.

$$\frac{1}{130}\sqrt{863}\times \frac{7^{3}}{55}+\frac{1}{55}$$

Calculate 7 to the power of 3 and get 343.

$$\frac{1}{130}\sqrt{863}\times \frac{343}{55}+\frac{1}{55}$$

Multiply \frac{1}{130} times \frac{343}{55} by multiplying numerator times numerator and denominator times denominator.

$$\frac{1\times 343}{130\times 55}\sqrt{863}+\frac{1}{55}$$

Do the multiplications in the fraction \frac{1\times 343}{130\times 55}.

$$\frac{343}{7150}\sqrt{863}+\frac{1}{55}$$

Factor

$$\frac{343 \sqrt{863} + 130}{7150} = 1.4274494466567704$$