Respuesta :

Answer in fraction form = 33/2

Answer in decimal form = 16.5

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Work Shown:

n = some number

2n = twice that number

2n+3 = "sum of twice a number and 3"

[tex]\sqrt{2n+3}[/tex] = square root of the previous expression

The equation we need to solve is [tex]\sqrt{2n+3} = 6[/tex]

We'll follow PEMDAS in reverse, undoing each operation, to isolate n

[tex]\sqrt{2n+3} = 6\\\\\left(\sqrt{2n+3}\right)^2 = 6^2 \ \ \text{ ... square both sides}\\\\2n+3 = 36\\\\2n+3-3 = 36-3 \ \ \text{ ... subtract 3 from both sides}\\\\2n = 33\\\\\frac{2n}{2}=\frac{33}{2} \ \ \text{ ... divide both sides by 2}\\\\n = \frac{33}{2}\\\\n = 16.5[/tex]

harrypeart

Answer:

[tex]\frac{33}{2}[/tex]

Step-by-step explanation:

 In the problem, "The square root of the sum of twice a number and 3 is 6" can be converted to an equation that looks like this:

Let x be the number:

[tex]\sqrt{2x+3 = 6 }[/tex]

In order to solve the problem:

[tex]\sqrt{2x+3}^2=6^2[/tex]

By squaring both sides to remove the exponent:

[tex]{2x + 3 = 36}[/tex]

Simplify:

[tex]{2x+3-3=36-3}[/tex]

Subtract 3 from both sides:

[tex]{2x = 33 }[/tex]

Then Simplifying further:

[tex]\frac{2x}{2} = \frac{33}{2}[/tex]

Divide both sides by 2 to remove 2 from the x.

[tex]{x = \frac{33}{2} }[/tex]

Thus, meaning the answer is [tex]\frac{33}{2}[/tex]

Hope this helps you.

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