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4 years ago

14

Can someone please help, Im doing a practice test and cant move on wihtout answering this. The $1000 prize for a lottery is to b

e divided evenly among the winners. Initially there are x winners, but then one more winner comes forward, causing each winner to receive $50 less. Create an equation that represents the situation and can be used to solve for x , the initial number of winners.

2 answers:

Verizon [17]4 years ago

8 0

1000=x+1 then it works out

Mila [183]4 years ago

3 0

950÷(x+1)

1p0p-50=950 dollars

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Which expression is a sum of cubes?

erik [133]

we know that

A polynomial in the form $a^{3} +b^{3}$ is called a sum of cubes

so

Let's verify each case to determine the solution

<u>case A)</u> $-64x^{6} y^{12} +125x^{16} y^{3}$

we know that

$-64=-4^{3}$

$x^{6}= (x^{2})^{3}$

$y^{12}= (y^{4})^{3}$

$125=5^{3}$

$x^{16}=x^{15} *x=x*(x^{5})^{3}$ -------> is not a perfect cube

$y^{3}= (y)^{3}$

therefore

the case A) is not a sum of cubes

<u>case B)</u> $-32x^{6} y^{12} +125x^{16} y^{3}$

we know that

$-32=-2^{5}$ -------> is not a perfect cube

$x^{6}= (x^{2})^{3}$

$y^{12}= (y^{4})^{3}$

$125=5^{3}$

$x^{16}=x^{15} *x=x*(x^{5})^{3}$ -------> is not a perfect cube

$y^{3}= (y)^{3}$

therefore

the case B) is not a sum of cubes

<u>case C)</u> $32x^{6} y^{12} +125x^{9} y^{3}$

we know that

$32=2^{5}$ -------> is not a perfect cube

$x^{6}= (x^{2})^{3}$

$y^{12}= (y^{4})^{3}$

$125=5^{3}$

$x^{9}=(x^{3})^{3}$

$y^{3}= (y)^{3}$

therefore

the case C) is not a sum of cubes

<u>case A)</u> $64x^{6} y^{12} +125x^{9} y^{3}$

we know that

$64=4^{3}$

$x^{6}= (x^{2})^{3}$

$y^{12}= (y^{4})^{3}$

$125=5^{3}$

$x^{9}=(x^{3})^{3}$

$y^{3}= (y)^{3}$

Substitute

$4^{3}(x^{2})^{3}(y^{4})^{3} +5^{3}(x^{3})^{3}(y)^{3}$

$(4x^{2}y^{4})^{3} +(5x^{3}y)^{3}$

therefore

<u>the answer is</u>

$64x^{6} y^{12} +125x^{9} y^{3}$ is a sum of cubes

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Anuta_ua [19.1K]

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