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3 years ago

Solve for all values of q that satisfy the equation √(3q+7) = q+3

Mathematics

1 answer:

Salsk061 [2.6K]3 years ago

5 0

Q^2 + 3q + 2 = 0 (q + 2) (q + 1) = 0 q + 2 = 0 q + 1 = 0 q = -2 or q = -1

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Answer:

Step-by-step explanation:

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Answer:

(13, -6.5)

Step-by-step explanation:

Find the distance between points J and the midpoint, which is 8.5 for x-coordinates and 2.5 for y-coordinates.

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we know that

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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

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

$125=5^{3}$

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

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

therefore

the case B) is not a sum of cubes

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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

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we know that

$64=4^{3}$

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

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

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

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

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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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u can you download the gauthmath app i promise it's very helpful :)

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