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2 years ago

What is the inverse to "if i listen to this song, then it will get stuck in my head"

Mathematics

1 answer:

VikaD [51]2 years ago

3 0

Answer:

The answer is: If I do not listen to this song, then it will not get stuck in my head.

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Which is a function and which one isn’t. please help me i need to get this done please help meh.

Anton [14]

Answer:

1) not a function

2) Function

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Step-by-step explanation:

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3 years ago

What are the solutions of the quadratic equation (x + 3)2 = 49

Korvikt [17]

the right answer is

x = 4 and x = –10

I just take the test and I got it right I hope yall get it right to

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3 years ago

Read 2 more answers

1) Marcos deseja comprar uma guitarra. Seu amigo possui um e quer vender a vista por R$850,00. Caso Marcos prefira, pode parcela

e-lub [12.9K]

Answer:

1)C

2)C

Step-by-step explanation:

Fiz no Classroom!!!

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2 years ago

EASY MATH!!! GIVING BRAINLIEST!!!

Harman [31]

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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

2 years ago