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dexar [7]
4 years ago
6

If point c is between points A and B then AC+CB=___ A.AB B.CA C.BC D.ABC

Mathematics
1 answer:
EastWind [94]4 years ago
3 0

Answer:

A.AB

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
PLEASE HELP ME I AM BEGGING!!!!!!!!!!!!!!!!!!!!!!! I will pick brainliest for the first and the correct answer.
navik [9.2K]

Answer:

x = 5/26

Step-by-step explanation:

3/4 x + 5/8 =4x

3 0
3 years ago
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