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Category: Algebra

and-are-2-roots-of-eq-ax-2-bx-c-0-with-conditions-2-b-2-a-find-c-in-terms-of-a-and-b-

Question Number 55039 by behi83417@gmail.com last updated on 16/Feb/19 $$\alpha\:{and}\:\beta,{are}\:\mathrm{2}\:{roots}\:{of}\:{eq}: \\ $$$$\:\:\:\:{ax}^{\mathrm{2}} +{bx}+{c}=\mathrm{0}\:{with}\:{conditions}: \\ $$$$\:\:\:\:\begin{cases}{\alpha^{\mathrm{2}} =\beta+{b}}\\{\beta^{\mathrm{2}} =\alpha+{a}}\end{cases} \\ $$$${find}:\:\:\boldsymbol{{c}}\:{in}\:{terms}\:{of}:\:\boldsymbol{{a}}\:\:{and}\:\:\boldsymbol{{b}}. \\ $$ Commented by mr W…

1-Solve-for-the-linear-system-x-2y-3z-5-2x-y-4z-0-3x-4y-11z-5-2-Ghana-railways-co-operation-has-20-trains-for-it-s-operations-It-is-observed-that-x-trains-can-accommodate-2-passengers-y-trains-

Question Number 186099 by akuba last updated on 01/Feb/23 $$\mathrm{1}.\:{Solve}\:{for}\:{the}\:{linear}\:{system}: \\ $$$${x}−\mathrm{2}{y}+\mathrm{3}{z}=\mathrm{5} \\ $$$$\mathrm{2}{x}+{y}−\mathrm{4}{z}=\mathrm{0} \\ $$$$\mathrm{3}{x}+\mathrm{4}{y}−\mathrm{11}{z}=−\mathrm{5} \\ $$$$\mathrm{2}.\:{Ghana}\:{railways}\:{co}−{operation}\:{has}\:\mathrm{20}\: \\ $$$${trains}\:{for}\:{it}'{s}\:{operations}.\:{It}\:{is}\:{observed}\:{that} \\ $$$${x}\:{trains}\:{can}\:{accommodate}\:\mathrm{2}\:{passengers}, \\ $$$${y}\:{trains}\:\mathrm{3}\:{passengers}\:{and}\:{z}\:{trains}\:\mathrm{5}\: \\…

Question-120556

Question Number 120556 by bobhans last updated on 01/Nov/20 Commented by bobhans last updated on 01/Nov/20 $$\mathrm{The}\:\mathrm{graph}\:\mathrm{y}=\sqrt{\mathrm{x}}\:\mathrm{and}\:\mathrm{y}=\mathrm{5}−\mathrm{x}^{\mathrm{2}} \:\mathrm{intersect}\:\mathrm{at} \\ $$$$\mathrm{one}\:\mathrm{point}\:\mathrm{x}=\mathrm{r}\:.\:\mathrm{use}\:\mathrm{newton}\:\mathrm{method} \\ $$$$\mathrm{to}\:\mathrm{estimate}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{r}\:\mathrm{to} \\ $$$$\mathrm{four}\:\mathrm{decimal}\:\mathrm{places} \\…

x-x-x-1-2-x-

Question Number 120545 by pooooop last updated on 01/Nov/20 $$\left(\sqrt{\boldsymbol{{x}}}\right)^{\frac{\boldsymbol{{x}}}{\:\sqrt{\boldsymbol{{x}}}}} \:=\:\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:\:\:\:\:\boldsymbol{{x}}=? \\ $$ Commented by pooooop last updated on 01/Nov/20 $$ \\ $$$$\:\:\:\left(\sqrt{{x}}\right)^{\frac{{x}\sqrt{{x}}}{{x}}} =\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} \\…