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Ostrovityanka [42]
3 years ago
12

Simplify the following cos(-x)tanx

Mathematics
1 answer:
RSB [31]3 years ago
8 0
\bf \textit{symmetry identites}\\\\
cos(-\theta)=cos(\theta)\\\\
-----------------------------\\\\
cos(-x)tan(x)\implies cos(-x)\cdot \cfrac{sin(x)}{cos(x)}\\\\\\ cos(x)\cdot \cfrac{sin(x)}{cos(x)}\implies sin(x)
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How to simplify fractions
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\frac{2}{3} + \frac{2}{9} \\\\\frac{6+2}{9}\\\\\frac{8}{9}

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Ella used 37.2 pounds of apples to make applesauce . She used one -tenth as a many pounds of sugar as pounds of apples . How man
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Robert gets a loan from his bank.
Goryan [66]

Answer:

£10,200

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5 0
3 years ago
What is the completely factored form of f(x)=x^3-7x^2+2x+4
xxMikexx [17]

Solution, \mathrm{Factor}\:x^3-7x^2+2x+4:\quad \left(x-1\right)\left(x^2-6x-4\right)

Steps:

x^3-7x^2+2x+4

Use\:the\:rational\:root\:theorem,\\a_0=4,\:\quad a_n=1,\\\mathrm{The\:dividers\:of\:}a_0:\quad 1,\:2,\:4,\:\quad \mathrm{The\:dividers\:of\:}a_n:\quad 1,\\\mathrm{Therefore,\:check\:the\:following\:rational\:numbers:\quad }\pm \frac{1,\:2,\:4}{1},\\\frac{1}{1}\mathrm{\:is\:a\:root\:of\:the\:expression,\:so\:factor\:out\:}x-1,\\\left(x-1\right)\frac{x^3-7x^2+2x+4}{x-1}

\frac{x^3-7x^2+2x+4}{x-1}

\mathrm{Divide}\:\frac{x^3-7x^2+2x+4}{x-1},\\\mathrm{Divide\:the\:leading\:coefficients\:of\:the\:numerator\:}x^3-7x^2+2x+4\mathrm{\:and\:the\:divisor\:}x-1\mathrm{\::\:}\frac{x^3}{x},\\\mathrm{Quotient}=x^2,\\\mathrm{Multiply\:}x-1\mathrm{\:by\:}x^2:\:x^3-x^2,\\\mathrm{Subtract\:}x^3-x^2\mathrm{\:from\:}x^3-7x^2+2x+4\mathrm{\:to\:get\:new\:remainder},\\\mathrm{Remainder}=-6x^2+2x+4,\\Therefore,\\\frac{x^3-7x^2+2x+4}{x-1}=x^2+\frac{-6x^2+2x+4}{x-1}

\mathrm{Divide}\:\frac{-6x^2+2x+4}{x-1},\\\mathrm{Divide\:the\:leading\:coefficients\:of\:the\:numerator\:}-6x^2+2x+4\mathrm{\:and\:the\:divisor\:}x-1\mathrm{\::\:}\frac{-6x^2}{x}=-6x,\\\mathrm{Quotient}=-6x,\\\mathrm{Multiply\:}x-1\mathrm{\:by\:}-6x:\:-6x^2+6x,\\\mathrm{Subtract\:}-6x^2+6x\mathrm{\:from\:}-6x^2+2x+4\mathrm{\:to\:get\:new\:remainder},\\\mathrm{Remainder}=-4x+4,\\\mathrm{Therefore},\\\frac{-6x^2+2x+4}{x-1}=-6x+\frac{-4x+4}{x-1}

\mathrm{Divide}\:\frac{-4x+4}{x-1},\\\mathrm{Divide\:the\:leading\:coefficients\:of\:the\:numerator\:}-4x+4\mathrm{\:and\:the\:divisor\:}x-1\mathrm{\::\:}\frac{-4x}{x}=-4,\\\mathrm{Quotient}=-4,\\\mathrm{Multiply\:}x-1\mathrm{\:by\:}-4:\:-4x+4,\\\mathrm{Subtract\:}-4x+4\mathrm{\:from\:}-4x+4\mathrm{\:to\:get\:new\:remainder},\\\mathrm{Remainder}=0,\\\mathrm{Therefore},\\\frac{-4x+4}{x-1}=-4

\mathrm{The\:Correct\:Answer\:is\:\left(x-1\right)\left(x^2-6x-4\right)}

\mathrm{Hope\:This\:Helps!!!}

\mathrm{-Austint1414}

8 0
3 years ago
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