If $$x=2t+1 $$ and $$y=7t^2 $$, then what is the value of $$y$$ in terms of $$x$$?

Correct.
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__Plugging In__ $$x=5$$ into this answer choice results in
>$$\displaystyle \frac{7(x-1{)}^{2}}{4} = \frac{7\cdot 4^2}{ 4} = 7\cdot 4 = 28$$.
All other answer choices are eliminated for this Plug-In, so this answer choice has to be correct.

Incorrect.
[[snippet]]
__Plugging In__ $$x=5$$ into this answer choice results in
>$$\displaystyle (x+2)^{2} = (5+2)^{2} = 7^2 = 49$$.

Incorrect.
[[snippet]]
__Plugging In__ $$x=5$$ into this answer choice results in
>$$\displaystyle \frac{4(x+1)^{2}}{7} = \frac{4(5+1)^{2}}{7} = \frac{4 \cdot 36}{7}$$.
This is a fraction so it is not equal to 28.

Incorrect.
[[snippet]]
__Plugging In__ $$x=5$$ into this answer choice results in
>$$\displaystyle \frac{7(x+2)^{2}}{4} = \frac{7(5+2)^{2}}{4} = \frac{7 \cdot 49}{7}$$.
This is a fraction so it is not equal to 28.

Incorrect.
[[snippet]]
__Plugging In__ $$x=5$$ into this answer choice results in
>$$\displaystyle (x-1)^{2} = (5-1)^{2} = 4^2 = 16$$.

$$(x-1)^{2}$$

$$(x+2)^{2}$$

$$\displaystyle \frac{4(x+1)^{2}}{7}$$

$$\displaystyle \frac{7(x-1)^{2}}{4}$$

$$\displaystyle \frac{7(x+2)^{2}}{4}$$