25midowning
Answered

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

Sagot :

Selena328

Answer:

18

Step-by-step explanation:

[tex] \\ \sf16 {x}^{0} + 2 {x}^{2} \times {y}^{ - 1} [/tex]

Now putting the values of x and y.

[tex] \\ \sf16 \times {2}^{0} + 2 \times {2}^{2} \times {4}^{ - 1} [/tex]

[tex] \\ \sf = 16 \times 1 + 2 \times 4 \times \frac{1}{ {4}}[/tex]

[tex] \\ \sf = 16 + 2[/tex]

[tex] \\ \sf = 18[/tex]

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.

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