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2 years ago

Find a polynomial function that has the given zeros. (there are many correct answers.) 0, 1, 6

1 answer:

7 0

Since those are the zeros, we can say that the equation would have to be
x(x-1)(x-6), then we just expand to get x^3 - 7x^2 + 6x

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Revenue is 3,000,cost of goods is 1,500, selling expense is 500 what is the prophit?

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In this situation the revenue is the total sales you make before all expenses.
To find the profit: 3000 - 1500 - 500 = 1000.
Your profit is $1,000.

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2 years ago

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If a insect is 0.095 long how long is it if looking through a microscope 100 times how long would the insect be

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3 years ago

-1<9+n<17<br><br><br>i don't understand how to solve it

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3 years ago

Nick had 3 1/4 bottles of water for his wrestling practice. When he finished he had 1 2/4 bottles of water left. He said he used

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3 years ago

A student solves the following equation and

Phoenix [80]

Answer:

Since the equation is undefined for -2

Therefore, NO SOLUTION for the given equation.

Step-by-step explanation:

Considering the expression

$\frac{3}{a+2}-6\cdot \frac{a}{-4+a^2}=\frac{1}{a-2}$

$\frac{3}{a+2}-\frac{6a}{-4+a^2}=\frac{1}{a-2}$

$\mathrm{Find\:Least\:Common\:Multiplier\:of\:}a+2,\:-4+a^2,\:a-2:\quad \left(a+2\right)\left(a-2\right)$

$\mathrm{Multiply\:by\:LCM=}\left(a+2\right)\left(a-2\right)$

$\frac{3}{a+2}\left(a+2\right)\left(a-2\right)-\frac{6a}{-4+a^2}\left(a+2\right)\left(a-2\right)=\frac{1}{a-2}\left(a+2\right)\left(a-2\right)$

$\frac{3}{a+2}\left(a+2\right)\left(a-2\right):\quad 3\left(a-2\right)$

$-\frac{6a}{-4+a^2}\left(a+2\right)\left(a-2\right):\quad -6a$

$\frac{1}{a-2}\left(a+2\right)\left(a-2\right):\quad a+2$

so equation becomes

$3\left(a-2\right)-6a=a+2$

$-3a-6=a+2$

$-3a-6+6=a+2+6$

$-4a=8$

$\mathrm{Divide\:both\:sides\:by\:}-4$

$\frac{-4a}{-4}=\frac{8}{-4}$

$a=-2$

$\mathrm{Verify\:Solutions}$

$\mathrm{Take\:the\:denominator\left(s\right)\:of\:}\frac{3}{a+2}-6\frac{a}{-4+a^2}-\frac{1}{a-2}\mathrm{\:and\:compare\:to\:zero}$

$\mathrm{Solve\:}\:a+2=0:\quad a=-2$

$\mathrm{Solve\:}\:-4+a^2=0:\quad a=2,\:a=-2$

$\mathrm{Solve\:}\:a-2=0:\quad a=2$

So the following points are undefined

$a=-2,\:a=2$

Since the equation is undefined for -2

Therefore, NO SOLUTION for the given equation.

4 0

3 years ago