64x^2 +y^2=72 and xy=2 find (4x+y/2)^2
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now is correct...........
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farah4:
oh sorry mne wrong cut krdiya ho tm kr loge
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=(4x+y/2)^2
=(4x)^2+2*4x*y/2+(y/2)^2
=16x^2+4xy+y^2/4
=16x^2+8+y^2 /4(xy=2)
=(64x^2+32+y^2)/4
=(64x^2+y^2+32)/4
=(72+32)/4
=26(answer)
=(4x)^2+2*4x*y/2+(y/2)^2
=16x^2+4xy+y^2/4
=16x^2+8+y^2 /4(xy=2)
=(64x^2+32+y^2)/4
=(64x^2+y^2+32)/4
=(72+32)/4
=26(answer)
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