Please help! I will award brainliest!! Find x and y. Lines AB , CD , and EF are straight lines.

Mathematics

1 answer:

NikAS [45]3 years ago

4 0

y = 14 & x = 15.6

Hope that helps!!

Let me know if its right or not plz :)

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Rewrite the function by completing the square. 4x^2-4x+1=0

andrey2020 [161]

Do in completing the square method.

${4x}^{2} - 4x + 1 = 0 \\ 4x^{2}-4x=-1 \\ \frac{4x^{2}-4x}{4}=\frac{-1}{4} \\ x^{2}+\frac{-4}{4}x=\frac{-1}{4} \\ x^{2}-x=\frac{-1}{4} \\ x^{2}-x+\left(-\frac{1}{2}\right)^{2}=-\frac{1}{4}+\left(-\frac{1}{2}\right)^{2} \\ x^{2}-x+\frac{1}{4}=\frac{-1+1}{4} \\ x^{2}-x+\frac{1}{4}=0 \\ \left(x-\frac{1}{2}\right)^{2}=0 \\ \sqrt{\left(x-\frac{1}{2}\right)^{2}}=\sqrt{0}$

Now simplify.

$x-\frac{1}{2}=0 \\ x-\frac{1}{2}=0 \\ \\ \rightarrow \: x=\frac{1}{2} \\ \rightarrow \: x=\frac{1}{2}$

So, answer is..

$\boxed{x=\frac{1}{2} }$

Now rewrite the equation.

$4 x ^ { 2 } - 4 x + 1 = 0 \\ \boxed{4\times \left(\frac{1}{4}\right)-4\times \left(\frac{1}{2}\right)+1=0 }$

Verification :-

$4 x ^ { 2 } - 4 x + 1 = 0 \\ 4\times \left(\frac{1}{4}\right)-4\times \left(\frac{1}{2}\right)+1=0 \\ 1-4\times \left(\frac{1}{2}\right)+1=0 \\ 1-\frac{4}{2}+1=0 \\ 1-2+1=0 \\ -1+1=0 \\ \underline{ 0 = 0} \: \\ \dashrightarrow \text{true}$