Which segment is congruent to segment GJ

1 answer:

7 0

<span>well congruent means they must be equal so since from point G to point J is -3 spaces you would just need to find another segment that is also -3 spaces, JM. For the second question i think you mean find the midpoint between NH, and if so you would need to figure out 0 - 6, which gives you -3 which is located at point K. Hope this is the right one.</span>

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Reika [66]

Answer: The correct option is (c) $\dfrac{49}{100}.$

Step-by-step explanation: We are given to solve the following quadratic equation by the method of completing the square:

$5x^2-7x+2=0~~~~~~~~~~~~~~~~~~~(i)$

Also, we are to find the constant added on both sides to form the perfect square trinomial.

We have from equation (i) that

$5x^2-7x+2=0\\\\\Rightarrow x^2-\dfrac{7}{5}x+\dfrac{2}{5}=0\\\\\\\Rightarrow x^2-2\times x\times \dfrac{7}{10}+\left(\dfrac{7}{10}\right)^2+\dfrac{2}{5}=\left(\dfrac{7}{10}\right)^2\\\\\\\Rightarrow \left(x-\dfrac{7}{10}\right)^2=\dfrac{49}{100}-\dfrac{2}{5}\\\\\\\Rightarrow \left(x-\dfrac{7}{10}\right)^2=\dfrac{49-40}{100}\\\\\\\Rightarrow \left(x-\dfrac{7}{10}\right)^2=\dfrac{9}{100}\\\\\\\Rightarrow x-\dfrac{7}{10}=\pm\dfrac{3}{10}\\\\\\\Rightarrow x=\pm\dfrac{3}{10}+\dfrac{7}{10}.$

So,

$x=\dfrac{3}{10}+\dfrac{7}{10},~~~~~~~~x=-\dfrac{3}{10}+\dfrac{7}{10}\\\\\\\Rightarrow x=\dfrac{10}{10},~~~~~~~~\Rightarrow x=\dfrac{-3+7}{10}\\\\\\\Rightarrow x=1,~-\dfrac{2}{5}.$

Thus, the required solution is $x=1,~-\dfrac{2}{5}.$ and the value of the constant added is $\dfrac{49}{100}.$

Option (c) is correct.

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barxatty [35]

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(x-3) ^2 =11

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