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aalyn [17]
3 years ago
8

ASAP NEED HELP PLEASE

Mathematics
1 answer:
Katen [24]3 years ago
7 0

Answer:

B

line n has slope equal to zero

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What is 1/3 x 12 help me quick
MArishka [77]

Answer:

12/3 = 4

Step-by-step explanation:

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The local camera shop contributed 10 antque cameras. The next day, you accepted bibs for 4 of the cameras. What fraction of the
MissTica

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3/5

Step-by-step explanation:

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3 years ago
Simplify each rational expression to lowest terms, specifying the values of xx that must be excluded to avoid division
k0ka [10]

Answer:

(a) \frac{x^2-6x+5}{x^2-3x-10}=\frac{x-1}{x+2}. The domain of this function is all real numbers not equal to -2 or 5.

(b) \frac{x^3+3x^2+3x+1}{x^3+2x^2-x}=1+\frac{x^2+4x+1}{x^3+2x^2-x}. The domain of this function is all real numbers not equal to 0, -1+\sqrt{2} or -1+\sqrt{2}.

(c) \frac{x^2-16}{x^2+2x-8}=\frac{x-4}{x-2}.The domain of this function is all real numbers not equal to 2 or -4.

(d) \frac{x^2-3x-10}{x^3+6x^2+12x+8}=\frac{x-5}{\left(x+2\right)^2}. The domain of this function is all real numbers not equal to -2.

(e) \frac{x^3+1}{x^2+1}=x+\frac{-x+1}{x^2+1}. The domain of this function is all real numbers.

Step-by-step explanation:

To reduce each rational expression to lowest terms you must:

(a) For \frac{x^2-6x+5}{x^2-3x-10}

\mathrm{Factor}\:x^2-6x+5\\\\x^2-6x+5=\left(x^2-x\right)+\left(-5x+5\right)\\x^2-6x+5=x\left(x-1\right)-5\left(x-1\right)\\\\\mathrm{Factor\:out\:common\:term\:}x-1\\x^2-6x+5=\left(x-1\right)\left(x-5\right)

\mathrm{Factor}\:x^2-3x-10\\\\x^2-3x-10=\left(x^2+2x\right)+\left(-5x-10\right)\\x^2-3x-10=x\left(x+2\right)-5\left(x+2\right)\\\\\mathrm{Factor\:out\:common\:term\:}x+2\\x^2-3x-10=\left(x+2\right)\left(x-5\right)

\frac{x^2-6x+5}{x^2-3x-10}=\frac{\left(x-1\right)\left(x-5\right)}{\left(x+2\right)\left(x-5\right)}

\mathrm{Cancel\:the\:common\:factor:}\:x-5\\\\\frac{x^2-6x+5}{x^2-3x-10}=\frac{x-1}{x+2}

The denominator in a fraction cannot be zero because division by zero is undefined. So we need to figure out what values of the variable(s) in the expression would make the denominator equal zero.

To find any values for x that would make the denominator = 0 you need to set the denominator = 0 and solving the equation.

x^2-3x-10=\left(x+2\right)\left(x-5\right)=0

Using the Zero Factor Theorem: = 0 if and only if = 0 or = 0

x+2=0\\x=-2\\\\x-5=0\\x=5

The domain is the set of all possible inputs of a function which allow the function to work. Therefore the domain of this function is all real numbers not equal to -2 or 5.

(b) For \frac{x^3+3x^2+3x+1}{x^3+2x^2-x}

\mathrm{Divide\:the\:leading\:coefficients\:of\:the\:numerator\:}x^3+3x^2+3x+1\mathrm{\:and\:the\:divisor\:}x^3+2x^2-x\mathrm{\::\:}\frac{x^3}{x^3}=1

Quotient = 1

\mathrm{Multiply\:}x^3+2x^2-x\mathrm{\:by\:}1:\:x^3+2x^2-x

\mathrm{Subtract\:}x^3+2x^2-x\mathrm{\:from\:}x^3+3x^2+3x+1\mathrm{\:to\:get\:new\:remainder}

Remainder = x^2+4x+1}

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

  • The domain of this function is all real numbers not equal to 0, -1+\sqrt{2} or -1+\sqrt{2}.

x^3+2x^2-x=0\\\\x^3+2x^2-x=x\left(x^2+2x-1\right)=0\\\\\mathrm{Solve\:}\:x^2+2x-1=0:\quad x=-1+\sqrt{2},\:x=-1-\sqrt{2}

(c) For \frac{x^2-16}{x^2+2x-8}

x^2-16=\left(x+4\right)\left(x-4\right)

x^2+2x-8= \left(x-2\right)\left(x+4\right)

\frac{x^2-16}{x^2+2x-8}=\frac{\left(x+4\right)\left(x-4\right)}{\left(x-2\right)\left(x+4\right)}\\\\\frac{x^2-16}{x^2+2x-8}=\frac{x-4}{x-2}

  • The domain of this function is all real numbers not equal to 2 or -4.

x^2+2x-8=0\\\\x^2+2x-8=\left(x-2\right)\left(x+4\right)=0

(d) For \frac{x^2-3x-10}{x^3+6x^2+12x+8}

\mathrm{Factor}\:x^2-3x-10\\\left(x^2+2x\right)+\left(-5x-10\right)\\x\left(x+2\right)-5\left(x+2\right)

\mathrm{Apply\:cube\:of\:sum\:rule:\:}a^3+3a^2b+3ab^2+b^3=\left(a+b\right)^3\\\\a=x,\:\:b=2\\\\x^3+6x^2+12x+8=\left(x+2\right)^3

\frac{x^2-3x-10}{x^3+6x^2+12x+8}=\frac{\left(x+2\right)\left(x-5\right)}{\left(x+2\right)^3}\\\\\frac{x^2-3x-10}{x^3+6x^2+12x+8}=\frac{x-5}{\left(x+2\right)^2}

  • The domain of this function is all real numbers not equal to -2

x^3+6x^2+12x+8=0\\\\x^3+6x^2+12x+8=\left(x+2\right)^3=0\\x=-2

(e) For \frac{x^3+1}{x^2+1}

\frac{x^3+1}{x^2+1}=x+\frac{-x+1}{x^2+1}

  • The domain of this function is all real numbers.

x^2+1=0\\x^2=-1\\\\\mathrm{For\:}x^2=f\left(a\right)\mathrm{\:the\:solutions\:are\:}x=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}\\\\x=\sqrt{-1},\:x=-\sqrt{-1}

4 0
3 years ago
Find the value of x when 6-2x=5x-9x+6
Ira Lisetskai [31]

Answer:

x = 6

Step-by-step explanation:

Combine like terms here (anything with the same variable)

5x - 9x = -4x

6-2x = -4x+6

Add the lowest term in both sides

-2x + 4x = 2x

2x - 6 = 6

Add 6 to make 2x alone in order to get x

2x = 12

Divide both of the sides by 2 because that is how you get x all by itself

x = 6

4 0
4 years ago
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