we'll start off by grouping some
so we have a missing guy at the end in order to get the a perfect square trinomial from that group, hmmm, what is it anyway?
well, let's recall that a perfect square trinomial is
so we know that the middle term in the trinomial, is really 2 times the other two without the exponent, well, in our case, the middle term is just "x", well is really -x, but we'll add the minus later, we only use the positive coefficient and variable, so we'll use "x" to find the last term.
so, there's our fellow, however, let's recall that all we're doing is borrowing from our very good friend Mr Zero, 0, so if we add (1/2)², we also have to subtract (1/2)²
![\bf \left( x^2 -x +\left[ \cfrac{1}{2} \right]^2-\left[ \cfrac{1}{2} \right]^2 \right)=6\implies \left( x^2 -x +\left[ \cfrac{1}{2} \right]^2 \right)-\left[ \cfrac{1}{2} \right]^2=6 \\\\\\ \left(x-\cfrac{1}{2} \right)^2=6+\cfrac{1}{4}\implies \left(x-\cfrac{1}{2} \right)^2=\cfrac{25}{4}\implies x-\cfrac{1}{2}=\sqrt{\cfrac{25}{4}} \\\\\\ x-\cfrac{1}{2}=\cfrac{\sqrt{25}}{\sqrt{4}}\implies x-\cfrac{1}{2}=\cfrac{5}{2}\implies x=\cfrac{5}{2}+\cfrac{1}{2}\implies x=\cfrac{6}{2}\implies \boxed{x=3}](https://tex.z-dn.net/?f=%5Cbf%20%5Cleft%28%20x%5E2%20-x%20%2B%5Cleft%5B%20%5Ccfrac%7B1%7D%7B2%7D%20%5Cright%5D%5E2-%5Cleft%5B%20%5Ccfrac%7B1%7D%7B2%7D%20%5Cright%5D%5E2%20%5Cright%29%3D6%5Cimplies%20%5Cleft%28%20x%5E2%20-x%20%2B%5Cleft%5B%20%5Ccfrac%7B1%7D%7B2%7D%20%5Cright%5D%5E2%20%5Cright%29-%5Cleft%5B%20%5Ccfrac%7B1%7D%7B2%7D%20%5Cright%5D%5E2%3D6%20%5C%5C%5C%5C%5C%5C%20%5Cleft%28x-%5Ccfrac%7B1%7D%7B2%7D%20%5Cright%29%5E2%3D6%2B%5Ccfrac%7B1%7D%7B4%7D%5Cimplies%20%5Cleft%28x-%5Ccfrac%7B1%7D%7B2%7D%20%5Cright%29%5E2%3D%5Ccfrac%7B25%7D%7B4%7D%5Cimplies%20x-%5Ccfrac%7B1%7D%7B2%7D%3D%5Csqrt%7B%5Ccfrac%7B25%7D%7B4%7D%7D%20%5C%5C%5C%5C%5C%5C%20x-%5Ccfrac%7B1%7D%7B2%7D%3D%5Ccfrac%7B%5Csqrt%7B25%7D%7D%7B%5Csqrt%7B4%7D%7D%5Cimplies%20x-%5Ccfrac%7B1%7D%7B2%7D%3D%5Ccfrac%7B5%7D%7B2%7D%5Cimplies%20x%3D%5Ccfrac%7B5%7D%7B2%7D%2B%5Ccfrac%7B1%7D%7B2%7D%5Cimplies%20x%3D%5Ccfrac%7B6%7D%7B2%7D%5Cimplies%20%5Cboxed%7Bx%3D3%7D)
Answer:
Part 1) The value of x is 50°
Part 2) The measure of angle LOK is 114°
Option C
Step-by-step explanation:
we know that
if two angles are supplementary, then their sum is equal to 180 degrees
so
(x+64)°+66°=180°
solve for x
x+130°=180°
x=180°-130°=50°
Find the measure of angle LOK
∠LOK=(x+64)°
∠LOK=(50+64)°=114°
Answer:
The answer is B!!
Step-by-step explanation:
What do we know given from the statement above the photo? We are told <Q is congruent to <S, and QR is congruent to SR. So by given that, we have an angle (angle Q or angle S), a side (side QR or SR), and another angle (angle R), so we have ASA. I hope this helps!!
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