$, and the $BE is intercepted on $AB $ = bc$, which proves that the target is $AE = ad$.by $A, B, C, d$ four points round and tangent properties of $$\angle{bec} = \angle{bce} = {1\over2} (180^\circ-\angle{ebc}) = {1\OVER2}\ANGLE{ADC} = \angl e{odc},$$ therefore $O, E, C, d$ Four points are all round. This can be $$\angle{aed} = \angle{ocd} = {1\over 2}\angle{bcd} = {1\over2} (180^\circ-\angle{a}), $$ is proof $ae = ad$.Q.E.D.7, the two circles with each other in the $D $, the line cut a round
. $$ (+ x) \cdot \frac{7}{10} = \times \frac{3}{5} + x$$ $$\rightarrow x = \frac{200}{3}.$$ so should be added copper $\d frac{200}{3}$ G.4. Hard aluminum is $4\%$ copper, $1\%$ manganese, $0.5\%$ magnesium and $94.5\%$ aluminum preparation, melting loss rate according to $1.5\%$ calculation. Manufacturing $250$ kg hard aluminum, how many kilograms of the above materials? (Keep one decimal place)Answer:The total amount of raw materials required for the establishment is $x $. $$\rightarrow X\cdot
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