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KengaRu [80]
3 years ago
12

Make me laugh a little. Free Brainliest.

Mathematics
2 answers:
Yuri [45]3 years ago
7 0

Answer:

HELLOOOOOO

Step-by-step explanation:

ima crazy person super outgoing :,) ive tried to eat cake up my nose before ... ANYWAYSSSSS hiiii

Arturiano [62]3 years ago
7 0

Answer:

How bout a good Old Joke :)

Step-by-step explanation:

What did the little mountain say to the big mountain?

HIII CLIFF!!!!!!!!!!!!!!

Hahaha, Well. I hope that was funny. :p

Life is just wayyyy to 2020 right now :p

~PumpkinSpice1

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Sever21 [200]
Lisa has 21 books because you add 1 to Felix because she Lisa had 1 more than him & Joe has 10 than Lisa .
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The St. Patrick's Day Parade has 214 floats in it. If each float is carrying
Leona [35]

Step-by-step explanation:

214

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--------

428

+2140

————

2568

6 0
3 years ago
Find the volume of air in three different basketballs. Use π = 3.14. Round your answers to the nearest whole number
KiRa [710]
Volume of a sphere = (4/3)*pi*radius^3; so

Childs = (4/3)*3.14*4^3 = 267.9 = 268

Womens = (4/3)*3.14*4.5^3 = 381.5 = 382

Mens = (4/3)*3.14*4.8^3 = 463.0 = 463

3 0
4 years ago
Read 2 more answers
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
HELP ME PLEASE <br> PLEASE PLEASE
nika2105 [10]

Ans

Step-by-step explanation:

4 0
3 years ago
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