Which expression is equivalent to one over fivem − 20?

Mathematics

2 answers:

Len [333]2 years ago

6 0

Answer:

$\large\boxed{\dfrac{1}{5(m-4)}}$

Step-by-step explanation:

One (1) over (fraction) five m (5m) - 20:

$\dfrac{1}{5m-20}$ <em>distributive</em>

$\dfrac{1}{(5)(m)-(5)(4)}=\dfrac{1}{5(m-4)}$

weeeeeb [17]2 years ago

4 0

Answer:

I'd say 1/5- 4m

Step-by-step explanation:

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$x=-\frac{-20+\sqrt{-20w+3600}}{10},\:x=-\frac{-20-\sqrt{-20w+3600}}{10}$

Step-by-step explanation:

$w=-5\left(x-8\right)\left(x+4\right)\\\mathrm{Expand\:}-5\left(x-8\right)\left(x+4\right):\quad -5x^2+20x+160\\w=-5x^2+20x+160\\Switch\:sides\\-5x^2+20x+160=w\\\mathrm{Subtract\:}w\mathrm{\:from\:both\:sides}\\-5x^2+20x+160-w=w-w\\Simplify\\-5x^2+20x+160-w=0\\Solve\:with\:the\:quadratic\:formula\\\mathrm{Quadratic\:Equation\:Formula:}\\\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}\\x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$\mathrm{For\:}\quad a=-5,\:b=20,\:c=160-w:\quad x_{1,\:2}=\frac{-20\pm \sqrt{20^2-4\left(-5\right)\left(160-w\right)}}{2\left(-5\right)}\\x=\frac{-20+\sqrt{20^2-4\left(-5\right)\left(160-w\right)}}{2\left(-5\right)}:\quad -\frac{-20+\sqrt{-20w+3600}}{10}\\x=\frac{-20-\sqrt{20^2-4\left(-5\right)\left(160-w\right)}}{2\left(-5\right)}:\quad -\frac{-20-\sqrt{-20w+3600}}{10}\\The\:solutions\:to\:the\:quadratic\:equation\:are\\x=-\frac{-20+\sqrt{-20w+3600}}{10},\:x=-\frac{-20-\sqrt{-20w+3600}}{10}$