Member: gibsonj338

Becasue a disappears and you get a solution that is true, this means that a can be any number. It can be written as (-∞, ∞) or it can be written as {a|a is all real numbers}.

Solve for b

$\frac{a-1}{ab}=\frac{{1-a}^{-1}}{b}$

$\frac{a-1}{ab}=\frac{1-\frac{1}{{a}^{1}}}{b}$

$\frac{a-1}{ab}=\frac{1-\frac{1}{a}}{b}$

$\frac{a-1}{ab}\times b=\frac{1-\frac{1}{a}}{b}\times b$

$\frac{b(a-1)}{ab}=\frac{1-\frac{1}{a}}{b}\times b$

$\frac{ab-1b}{ab}=\frac{1-\frac{1}{a}}{b}\times b$

$\frac{ab-b}{ab}=\frac{1-\frac{1}{a}}{b}\times b$

$\frac{ab-b}{ab}=\frac{b(1-\frac{1}{a})}{b}$

$\frac{ab-b}{ab}=1(1-\frac{1}{a})$

$\frac{ab-b}{ab}=1-\frac{1}{a}$

$\frac{ab-b}{ab}\times ab=(1-\frac{1}{a})\times ab$

$\frac{ab(ab-b)}{ab}=(1-\frac{1}{a})\times ab$

$1(ab-b)=(1-\frac{1}{a})\times ab$

$ab-b=(1-\frac{1}{a})\times ab$

$ab-b=ab(1-\frac{1}{a})$

$ab-b=1ab-\frac{1}{a}ab$

$ab-b=ab-\frac{1}{a}ab$

$ab-b=ab-\frac{1ab}{a}$

$ab-b=ab-\frac{ab}{a}$

$ab-b=ab-1b$

$ab-b=ab-b$

$ab-b+b=ab-b+b$

$ab+0b=ab-b+b$

$ab+0=ab-b+b$

$ab=ab-b+b$

$ab=ab+0b$

$ab=ab+0$

$ab=ab$

$\frac{ab}{a}=\frac{ab}{a}$

$b=\frac{ab}{b}$

$b=b$

Because b is equal to b, b can be any number. It can be written as (-∞, ∞) or it can be written as {b|b is all real numbers}.

gibsonj338Oct 28, 2016

+1904

$\frac{1}{9}\times45$

$\frac{1}{9}\times\frac{45}{1}$

$\frac{45}{9}$

$\frac{5}{1}$

$5$

gibsonj338Oct 27, 2016

+1904

$nb=p+nb$

$0nb=p+nb-nb$

$0=p+nb-nb$

$0=p+0nb$

$0=p+0$

$0=p$

$p=0$

gibsonj338Oct 27, 2016

+1904

$({x}^{\frac{4}{3}})({}x^{\frac{2}{3}})$

${x}^{\frac{4}{3}+\frac{2}{3}}$

${x}^{\frac{6}{3}}$

${x}^{2}$

gibsonj338Oct 27, 2016

+1904

$4,635.21$

gibsonj338Oct 27, 2016

+1904

All of this arguring and cussing is uncalled for. Enough of that. I will help you wth the math problem.

$6\times(m-3)>5\times(2m+4)$

$6m-18>5\times(2m+4)$

$6m-18>10m+20$

$6m-18-10m>10m+20-10m$

$-4m-18>10m+20-10m$

$-4m-18>20-0m$

$-4m-18>20-0$

$-4m-18>20$

$-4m-18+18>20+18$

$-4m-0>20+18$

$-4m>20+18$

$-4m>38$

$\frac{-4m}{-4}>\frac{38}{-4}$

$1m<\frac{38}{-4}$

$m<\frac{38}{-4}$

$m<-\frac{38}{4}$

$m<-\frac{19}{2}$

$m<-9.5$

gibsonj338Oct 27, 2016

+1904

$f(x)=\sqrt[3]{{x}^{2}+4x}$

${f(x)}^{3}={\sqrt[3]{{x}^{2}+4x}}^{3}$

${f(x)}^{3}={x}^{2}+4x$

${f(x)}^{3}=x(x+4)$

$\frac{{f(x)}^{3}}{x}=\frac{x(x+4)}{x}$

$\frac{{f(x)}^{3}}{x}=x+4$

$\frac{{f(x)}^{3}}{x}-4=x+4-4$

$\frac{{f(x)}^{3}}{x}-4=x+0$

$\frac{{f(x)}^{3}}{x}-4=x$

$x(\frac{{f(x)}^{3}}{x}-4)=x\times x$

${f(x)}^{3}-4x=x\times x$

${f(x)}^{3}-4x={x}^{2}$

${f(x)}^{3}-4x-{x}^{2}={x}^{2}-{x}^{2}$

${f(x)}^{3}-4x-{x}^{2}={0x}^{2}$

${f(x)}^{3}-4x-{x}^{2}=0$

$-{x}^{2}-4x+{f(x)}^{3}=0$

$x = {4 \pm \sqrt{(-4)^2-4(-1)({f(x)}^{3})} \over 2(-1)}$

$x = {4 \pm \sqrt{16-4(-1)({f(x)}^{3})} \over 2(-1)}$

$x = {4 \pm \sqrt{16-(-4)({f(x)}^{3})} \over 2(-1)}$

$x = {4 \pm \sqrt{16+4{f(x)}^{3}} \over 2(-1)}$

$x = {4 \pm \sqrt{4(4+{f(x)}^{3})} \over 2(-1)}$

$x = {4 \pm 2\sqrt{4+{f(x)}^{3}} \over 2(-1)}$

$x = {2(2 \pm 1\sqrt{4+{f(x)}^{3}} )\over 2(-1)}$

$x = {2(2 \pm \sqrt{4+{f(x)}^{3}} )\over 2(-1)}$

$x = {2(2 \pm \sqrt{4+{f(x)}^{3}} )\over -2}$

$x = -1(2 \pm \sqrt{4+{f(x)}^{3}} )$

$x = -(2 \pm \sqrt{4+{f(x)}^{3}} )$

$x = -2 \pm \sqrt{4+{f(x)}^{3}}$

gibsonj338Oct 27, 2016

+1904

${(-1)}^{4/3}$

$\sqrt[3]{{(-1)}^{4}}$

$\sqrt[3]{(-1)(-1)(-1)(-1)}$

$\sqrt[3]{1(-1)(-1)}$

$\sqrt[3]{(-1)(-1)}$

$\sqrt[3]{1}$

$1$

gibsonj338Oct 27, 2016

Username	gibsonj338
Score	1904
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