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100-121k^(2)=0
What are the solutions to the given equation?
Choose 1 answer:
(A)
k=(100)/(121)
(B)
k=-(100)/(121) and
k=(100)/(121)
(c)
k=(10)/(11)
(D)
k=-(10)/(11) and
k=(10)/(11)

$100-121 k^{2}=0$ $\newline$ What are the solutions to the given equation? $\newline$ Choose $1$ answer: $\newline$ (A) $k=\frac{100}{121}$ $\newline$ (B) $k=-\frac{100}{121}$ and $k=\frac{100}{121}$ $\newline$ (c) $k=\frac{10}{11}$ $\newline$ (D) $k=-\frac{10}{11}$ and $k=\frac{10}{11}$

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View step-by-step help

Solve a quadratic equation using the zero product property

Full solution

Q.

100-121 k^{2}=0

\newline

What are the solutions to the given equation?

\newline

Choose

1

answer:

\newline

(A)

k=\frac{100}{121}

\newline

(B)

k=-\frac{100}{121}

and

k=\frac{100}{121}

\newline

(c)

k=\frac{10}{11}

\newline

(D)

k=-\frac{10}{11}

and

k=\frac{10}{11}

Question Prompt: question_prompt: ＂What are the solutions to the given equation $100-121k^{2}=0$ ?＂
Addition Step: Step $1$ : Add $121k^2$ to both sides to isolate the $k^2$ term, so it becomes $121k^2 = 100$ .
Division Step: Step $2$ : Divide both sides by $121$ to solve for $k^2$ , getting $k^2 = \frac{100}{121}$ .
Square Root Step: Step $3$ : Take the square root of both sides to solve for $k$ , remembering that there are two solutions: $k = \pm\sqrt{\frac{100}{121}}$ .
Final Solution: Step $4$ : Simplify the square root to get $k = \pm\frac{10}{11}$ , which means $k = \frac{10}{11}$ or $k = -\frac{10}{11}$ .

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