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user100 [1]
3 years ago
5

A reality TV show features business owners who try to persuade investors to put money into their companies in exchange for parti

al ownership of the company.
To make a fair deal, the owner estimates the total value of the business and sets the amount of the investment based on the size of the share, as a percentage. For example, an investor could provide $3 million in exchange for 30 percent ownership stake in a company that is worth $10 million.

Abbey owns a company that sells cat toys. She wants an investor to put $100,000 into her company so that she can expand the business.

Which equation correctly relates an investor's share, S (as a percentage), to the total value of Abbey's company, V?

Mathematics
1 answer:
ryzh [129]3 years ago
3 0

C (100% sure now...)

For instance, if Abbey's total value was 1,000,000, then S would be 10% or 0.1. 100,000/1,000,000 is 0.1, which is S.

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A car traveled 133 miles in 4 3/4 hours. What was the cars average speed in miles per hour?
nirvana33 [79]

4 3/4 = 19/4

 

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6 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
Write an equation for the nth term of the arithmetic sequence.<br><br> -2, 1, 4, 7, ...
son4ous [18]

Answer:

Step-by-step explanation:

3x-2   and x≥0  :)

5 0
3 years ago
Read 2 more answers
Can someone help?Im stuck on the attached question
Lana71 [14]
You need to put the parentheses around (1+2)


Your Answer should look like
(1+2)x3=9
4 0
4 years ago
Need the answer ASAP
V125BC [204]

Answer:

Approximately 201 squared inches.

Step-by-step explanation:

So, the composite figure is made up of a square and a semi-circle. The square has side lengths of 12 and the semi-circle has a radius of 6.

The total area of the figure would be the area of the square plus the area of the semi-circle. Thus, find the area of each individual figure.

Square:

The area of a square is given by:

A=l^2

Where l is the side length.

Substitute 12 for l:

A=(12)^2\\A=144\text{ in}^2

So the square is 144 square inches.

Semi-circle:

The area of a semi-circle is given by:

A=\frac{1}{2}\pi r^2

Substitute 6 for r and 3.14 for π:

A=\frac{1}{2}(3.14)(6)^2\\ A=56.52

Therefore, the total area is:

TA=144+56.52\\TA=200.52\text{ in}^2\approx201\text{ in}^2

4 0
4 years ago
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