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Korvikt [17]
3 years ago
10

Which table of values is correct for the equation y = 5(2)x

Mathematics
1 answer:
Tems11 [23]3 years ago
5 0

Answer:

Option D is correct.

x             y

0             5

1              10

2              20

Step-by-step explanation:

Given the equation:  y = 5(2)^x           .....[1]

Here, x is the input variable and y is the output variable.

For x =0

Substitute in equation [1]; we have;

y = 5(2)^{0} = 5 \cdot 1= 5

For x = 1

Substitute in equation [1]; we have;

y = 5(2)^{1} = 5 \cdot 2= 10

For x =2

Substitute in equation [1]; we have;

y = 5(2)^{2} = 5 \cdot 4= 20

Therefore, the table values which is correct for the equation y = 5(2)^x  is;

x             y

0             5

1              10

2              20

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ANTONII [103]
Megan:
x to the one third power =  x ^{1/3}
<span>x to the one twelfth power = </span>x ^{1/12}

<span>The quantity of x to the one third power, over x to the one twelfth power is:
</span>\frac{x ^{1/3}}{x ^{1/12}}
<span>
Since </span>\frac{ x^{a} }{ x^{b} } = x ^{a-b}
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Julie:
x times x to the second times x to the fifth = x * x² * x⁵

<span>The thirty second root of the quantity of x times x to the second times x to the fifth is
</span>\sqrt[32]{x* x^{2} * x^{5} }
<span>
Since </span>x^{a}* x^{b}= x^{a+b}
Then \sqrt[32]{x* x^{2} * x^{5} }= \sqrt[32]{ x^{1+2+5} } =\sqrt[32]{ x^{8} }

Since \sqrt[n]{x^{m}} = x^{m/n} }
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Since both Megan and Julie got the same result, it can be concluded that their expressions are equivalent.
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