Question

I have the equation. 16x^0+2x^2*y^-1 How do I solve the equation if x = 2 and y = 4?

Answer 1

Answer:

18

Step-by-step explanation:

$\\ \sf16 {x}^{0} + 2 {x}^{2} \times {y}^{ - 1}$

Now putting the values of x and y.

$\\ \sf16 \times {2}^{0} + 2 \times {2}^{2} \times {4}^{ - 1}$

$\\ \sf = 16 \times 1 + 2 \times 4 \times \frac{1}{ {4}}$

$\\ \sf = 16 + 2$

$\\ \sf = 18$

Note:

•If the power of a number is 0 then it's value will be 1.

•If the power of a number is negative then reciprocal the number to make the power positive.