Use the Quadratic Formula to solve 0.09 s 2 − 0.12 s + 0.04 = 0 0.09 s ^ { 2 } - 0.12 s + 0.04 = 0 0.09 s 2 − 0.12 s + 0.04 = 0 . Round your answer(s) to two decimal places.A) s = 0.67 s = 0.67 s = 0.67 B) s = 0.67 , s = − 0.67 s = 0.67 , s = - 0.67 s = 0.67 , s = − 0.67 C) s = 0.89 s = 0.89 s = 0.89 D) s = 0.78 , s = 2.00 s = 0.78 , s = 2.00 s = 0.78 , s = 2.00 E) no solutions

Question

Use the Quadratic Formula to solve   0.09 s 2 − 0.12 s + 0.04 = 0 0.09 s ^ { 2 } - 0.12 s + 0.04 = 0 0.09 s 2 − 0.12 s + 0.04 = 0   . Round your answer(s) to two decimal places.A)    s = 0.67 s = 0.67 s = 0.67 B)    s = 0.67 , s = − 0.67 s = 0.67 , s = - 0.67 s = 0.67 , s = − 0.67 C)    s = 0.89 s = 0.89 s = 0.89 D)    s = 0.78 , s = 2.00 s = 0.78 , s = 2.00 s = 0.78 , s = 2.00 E) no solutions

Jontra Anderson · Accepted Answer

Final Answer : A, Explanation :   The quadratic formula is   s = − b ± b 2 − 4 a c 2 a s = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} s = 2 a − b ± b 2 − 4 a c ​ ​  , where   a = 0.09 a = 0.09 a = 0.09  ,   b = − 0.12 b = -0.12 b = − 0.12  , and   c = 0.04 c = 0.04 c = 0.04  . Plugging these values in gives   s = 0.12 ± ( − 0.12 ) 2 − 4 ( 0.09 ) ( 0.04 ) 2 ( 0.09 ) s = \frac{0.12 \pm \sqrt{(-0.12)^2 - 4(0.09)(0.04)}}{2(0.09)} s = 2 ( 0.09 ) 0.12 ± ( − 0.12 ) 2 − 4 ( 0.09 ) ( 0.04 ) ​ ​  , which simplifies to   s = 0.67 s = 0.67 s = 0.67   after rounding to two decimal places. There is only one solution because the discriminant (  b 2 − 4 a c b^2 - 4ac b 2 − 4 a c  ) equals zero, indicating a perfect square under the square root, leading to one real, repeated solution.

Use the Quadratic Formula to solve $0.09 s ^ { 2 } - 0.12 s + 0.04 = 0$ . Round your answer(s) to two decimal places.

A) $s = 0.67$
B) $s = 0.67, s = - 0.67$
C) $s = 0.89$
D) $s = 0.78, s = 2.00$
E) no solutions

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