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4 years ago

What model represents 2x0.9

Mathematics

1 answer:

natima [27]4 years ago

3 0

Answer: the bottom one

Step-by-step explanation:

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You are certain to get a black or a red card when selecting cards from a shuffled deck. Express the indicated degree of likeliho

Vesnalui [34]

Answer:

p=1

Step-by-step explanation:

A deck has a combined 52cards.

Given that you're certain to select a red or black card:-

p(red)=> $p(red)=\frac{26}{52}$

p(black)=> $p(black)=\frac{26}{52}$

Therefore total probability of selection is expressed as:-

$p=p(red)+p(black)\\p=\frac{26}{52}+\frac{26}{52}\\=1$

4 0

3 years ago

Which of the following is equivalent to

IRISSAK [1]

Answer:

$x^{4}$

Step-by-step explanation:

$\frac{x^{5}y^{2} }{xy^{2} }$

= $x^{4}$

7 0

4 years ago

Read 2 more answers

What is the location of the second endpoint if the first endpoint is at (5,2) and

RUDIKE [14]

The second midpoint is (-15,8)

5 0

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 area of the shape shown below.

erastovalidia [21]

Answer:

Step-by-step explanation:

2x2=4

1x2=2

2x2=4

4+2+4=10

write this down if u need to show work

8 0

3 years ago