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Papessa [141]
3 years ago
13

-x<-29−x<−29 pls answer pls

Mathematics
1 answer:
Tasya [4]3 years ago
3 0

Solve for x.

x − 29 > x > 0

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H(x)= 9/x( x^2-49)<br> Determine the domain of the function
Norma-Jean [14]

Answer: The domain of the function h(x) = \frac{9}{x(x^2-49)} is:

Interval Notation: (-∞ , -7) ∪ (-7 , 0) ∪ (0 , 7) ∪ (7, ∞)

Set-Builder Notation: { x | x ≠ 0 , 7 , -7 }

All real numbers besides 0, 7, and -7.

Step-by-step explanation:

In order to find the domain of your rational function, we need to simplify it:

h(x) = \frac{9}{x(x^2-49)} = \frac{9}{(x)(x+7)(x-7)}

Remember, most of the time, the domain of a rational function consists of all real numbers besides zero.

To find the domain, we equal the equations in the denominator to zero.

x=0

x+7=0 --> x=-7

x-7=0 --> x=7

So all real numbers except for 0, -7, and 7 are in the domain of this rational function.

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3 years ago
Place the indicated product in the proper location on the grid. (x + y)(x2 - xy + y2)
mihalych1998 [28]

Answer:

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Step-by-step explanation:

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3 years ago
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If a triangle has a side length of 10 <br> 12 and 15 what type of triangle is it
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(5x^3-8x^2+9x+12)/(x-3)<br><br> please answer with sentences on how to do it step by step. thanks!
Anastaziya [24]

Answer:

5x^2+7x+30+\frac{102}{x-3}

Step-by-step explanation:

\mathrm{Divide\:the\:leading\:coefficients\:of\:the\:numerator\:}5x^3-8x^2+9x+12\\\mathrm{and\:the\:divisor\:}x-3\mathrm{\::\:}\frac{5x^3}{x}=5x^2\\\mathrm{Quotient}=5x^2\\\mathrm{Multiply\:}x-3\mathrm{\:by\:}5x^2:\:5x^3-15x^2\\\mathrm{Subtract\:}5x^3-15x^2\mathrm{\:from\:}5x^3-8x^2+9x+12\mathrm{\:to\:get\:new\:remainder}\\\mathrm{Remainder}=7x^2+9x+12\\\mathrm{Therefore}\\=5x^2+\frac{7x^2+9x+12}{x-3}

\mathrm{Divide\:the\:leading\:coefficients\:of\:the\:numerator\:}7x^2+9x+12\\\mathrm{and\:the\:divisor\:}x-3\mathrm{\::\:}\frac{7x^2}{x}=7x\\\mathrm{Quotient}=7x\\\mathrm{Multiply\:}x-3\mathrm{\:by\:}7x:\:7x^2-21x\\\mathrm{Subtract\:}7x^2-21x\mathrm{\:from\:}7x^2+9x+12\mathrm{\:to\:get\:new\:remainder}\\\mathrm{Remainder}=30x+12\\\mathrm{Therefore}\\=5x^2+7x+\frac{30x+12}{x-3}\\

\mathrm{Divide\:the\:leading\:coefficients\:of\:the\:numerator\:}30x+12\\\mathrm{and\:the\:divisor\:}x-3\mathrm{\::\:}\frac{30x}{x}=30\\\mathrm{Quotient}=30\\\mathrm{Multiply\:}x-3\mathrm{\:by\:}30:\:30x-90\\\mathrm{Subtract\:}30x-90\mathrm{\:from\:}30x+12\mathrm{\:to\:get\:new\:remainder}\\\mathrm{Remainder}=102\\\mathrm{Therefore}\\=5x^2+7x+30+\frac{102}{x-3}

6 0
3 years ago
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Adam bought a new flat screen tv with an area of 21 ft the screen is 3 feet wide how tall is it​
Tema [17]

Answer:

7 ft bRo

Step-by-step explanation:

wow i took 21 and divided it by three and i got SEVEN

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