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3 years ago

Solve for k 10/3= k/9

Mathematics

2 answers:

MArishka [77]3 years ago

8 0

Answer:

k = 30

Step-by-step explanation:

$\frac{10}{3}=\frac{k}{9}\\\\\frac{k}{9}=\frac{10}{3}\\\\k=\frac{10}{3}*9\\\\k=10*3=30$

agasfer [191]3 years ago

6 0

Answer:

k = 30

Step-by-step explanation:

$\frac{10}{3}$ = $\frac{k}{9}$

We can easily solve this by cross multiplying. Multiply 10 by 9, and multiply the variable (k) by 3.

10 x 9 = 90

3 x k = 3k

These two (90 and 3k) are equal to one another and thus can both be set up into an equation.

3k = 90

Solve for k.

3k/3 = 90/3

k = 30

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

Let's verify each case to determine the solution

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we know that

$-64=-4^{3}$

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therefore

the case A) is not a sum of cubes

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

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

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

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

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

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

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

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therefore

the answer is

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

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