Solve 7^2+7/4^3*64/63

$$\frac { 7 ^ { 2 } + 7 } { 4 ^ { 3 } } \cdot \frac { 64 } { 63 }$$

Evaluate

$$\frac{8}{9}\approx 0.888888889$$

Solution Steps

Calculate 7 to the power of 2 and get 49.

$$\frac{49+7}{4^{3}}\times \frac{64}{63}$$

Add 49 and 7 to get 56.

$$\frac{56}{4^{3}}\times \frac{64}{63}$$

Calculate 4 to the power of 3 and get 64.

$$\frac{56}{64}\times \frac{64}{63}$$

Reduce the fraction \frac{56}{64} to lowest terms by extracting and canceling out 8.

$$\frac{7}{8}\times \frac{64}{63}$$

Multiply \frac{7}{8} times \frac{64}{63} by multiplying numerator times numerator and denominator times denominator.

$$\frac{7\times 64}{8\times 63}$$

Do the multiplications in the fraction \frac{7\times 64}{8\times 63}.

$$\frac{448}{504}$$

Reduce the fraction \frac{448}{504} to lowest terms by extracting and canceling out 56.

$$\frac{8}{9}$$

Factor

$$\frac{2 ^ {3}}{3 ^ {2}} = 0.8888888888888888$$