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If a, b, c, d are in proportion, prove that: (ma? + nb) : (mc + nd?) = (ma? - nb) : (mc? : (me² - nd²)give correct answer or I will report u |
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Answer» Answer: If (ma + nb) : B : : (MC + nd) : d, prove that a, b, C, d are in proportion. Step-by-step explanation: Consider, (ma+nb):b :: (mc+nd):d Consider, (ma+nb):b :: (mc+nd):d⟹ Consider, (ma+nb):b :: (mc+nd):d⟹ b Consider, (ma+nb):b :: (mc+nd):d⟹ b(ma+nb) Consider, (ma+nb):b :: (mc+nd):d⟹ b(ma+nb) Consider, (ma+nb):b :: (mc+nd):d⟹ b(ma+nb) = Consider, (ma+nb):b :: (mc+nd):d⟹ b(ma+nb) = d Consider, (ma+nb):b :: (mc+nd):d⟹ b(ma+nb) = d(mc+nd) Consider, (ma+nb):b :: (mc+nd):d⟹ b(ma+nb) = d(mc+nd) Consider, (ma+nb):b :: (mc+nd):d⟹ b(ma+nb) = d(mc+nd) Consider, (ma+nb):b :: (mc+nd):d⟹ b(ma+nb) = d(mc+nd) d(ma+nb)=b(mc+nd) Consider, (ma+nb):b :: (mc+nd):d⟹ b(ma+nb) = d(mc+nd) d(ma+nb)=b(mc+nd)dma+dnb=bmc+bnd Consider, (ma+nb):b :: (mc+nd):d⟹ b(ma+nb) = d(mc+nd) d(ma+nb)=b(mc+nd)dma+dnb=bmc+bnddma−bmc=bnd−dnb Consider, (ma+nb):b :: (mc+nd):d⟹ b(ma+nb) = d(mc+nd) d(ma+nb)=b(mc+nd)dma+dnb=bmc+bnddma−bmc=bnd−dnbm(da−bc)=n(bd−db) Consider, (ma+nb):b :: (mc+nd):d⟹ b(ma+nb) = d(mc+nd) d(ma+nb)=b(mc+nd)dma+dnb=bmc+bnddma−bmc=bnd−dnbm(da−bc)=n(bd−db)m(da−bc)=n(0) Consider, (ma+nb):b :: (mc+nd):d⟹ b(ma+nb) = d(mc+nd) d(ma+nb)=b(mc+nd)dma+dnb=bmc+bnddma−bmc=bnd−dnbm(da−bc)=n(bd−db)m(da−bc)=n(0)m(da−bc)=0 Consider, (ma+nb):b :: (mc+nd):d⟹ b(ma+nb) = d(mc+nd) d(ma+nb)=b(mc+nd)dma+dnb=bmc+bnddma−bmc=bnd−dnbm(da−bc)=n(bd−db)m(da−bc)=n(0)m(da−bc)=0da−bc=0 ---- As m cannot be equal to 0 Consider, (ma+nb):b :: (mc+nd):d⟹ b(ma+nb) = d(mc+nd) d(ma+nb)=b(mc+nd)dma+dnb=bmc+bnddma−bmc=bnd−dnbm(da−bc)=n(bd−db)m(da−bc)=n(0)m(da−bc)=0da−bc=0 ---- As m cannot be equal to 0∴da=bc Consider, (ma+nb):b :: (mc+nd):d⟹ b(ma+nb) = d(mc+nd) d(ma+nb)=b(mc+nd)dma+dnb=bmc+bnddma−bmc=bnd−dnbm(da−bc)=n(bd−db)m(da−bc)=n(0)m(da−bc)=0da−bc=0 ---- As m cannot be equal to 0∴da=bcb Consider, (ma+nb):b :: (mc+nd):d⟹ b(ma+nb) = d(mc+nd) d(ma+nb)=b(mc+nd)dma+dnb=bmc+bnddma−bmc=bnd−dnbm(da−bc)=n(bd−db)m(da−bc)=n(0)m(da−bc)=0da−bc=0 ---- As m cannot be equal to 0∴da=bcba Consider, (ma+nb):b :: (mc+nd):d⟹ b(ma+nb) = d(mc+nd) d(ma+nb)=b(mc+nd)dma+dnb=bmc+bnddma−bmc=bnd−dnbm(da−bc)=n(bd−db)m(da−bc)=n(0)m(da−bc)=0da−bc=0 ---- As m cannot be equal to 0∴da=bcba Consider, (ma+nb):b :: (mc+nd):d⟹ b(ma+nb) = d(mc+nd) d(ma+nb)=b(mc+nd)dma+dnb=bmc+bnddma−bmc=bnd−dnbm(da−bc)=n(bd−db)m(da−bc)=n(0)m(da−bc)=0da−bc=0 ---- As m cannot be equal to 0∴da=bcba = Consider, (ma+nb):b :: (mc+nd):d⟹ b(ma+nb) = d(mc+nd) d(ma+nb)=b(mc+nd)dma+dnb=bmc+bnddma−bmc=bnd−dnbm(da−bc)=n(bd−db)m(da−bc)=n(0)m(da−bc)=0da−bc=0 ---- As m cannot be equal to 0∴da=bcba = d Consider, (ma+nb):b :: (mc+nd):d⟹ b(ma+nb) = d(mc+nd) d(ma+nb)=b(mc+nd)dma+dnb=bmc+bnddma−bmc=bnd−dnbm(da−bc)=n(bd−db)m(da−bc)=n(0)m(da−bc)=0da−bc=0 ---- As m cannot be equal to 0∴da=bcba = DC Consider, (ma+nb):b :: (mc+nd):d⟹ b(ma+nb) = d(mc+nd) d(ma+nb)=b(mc+nd)dma+dnb=bmc+bnddma−bmc=bnd−dnbm(da−bc)=n(bd−db)m(da−bc)=n(0)m(da−bc)=0da−bc=0 ---- As m cannot be equal to 0∴da=bcba = dc |
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