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2 years ago

Is this right? isn't the Directly proportional equation this. x=(1/k)y

1 answer:

7 0

It's y=kx. But 'k' is any constant. So my 'k' can easily be 1/(your k). The key is that when 'x' or 'y' increases by some factor, the other one also increases, by some constant multiple of the same factor. If so, then they're proportional.

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3 years ago

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3 years ago

Directions:Simplify the following monomials.SHOW ALL STEPS!

Mandarinka [93]

Answer:

7. Ans;

$\frac{6 {a}^{5} {b}^{7} }{ - 2 {a}^{3} {b}^{7} } = \frac{2 {a}^{3} {b}^{7}(3 {a}^{2} ) }{ - 2 {a}^{3} {b}^{7} } = - 3 {a}^{2}$

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8. Ans;

$\frac{ - 20 {x}^{3} {y}^{2} }{ - 5 {x}^{3}y } = \frac{ - 5 {x}^{3}y(4y) }{ - 5 {x}^{3}y } = 4y$

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9. Ans;

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10. Ans;

$\frac{21 {m}^{8} {n}^{5} }{27 {m}^{5} {n}^{4} } = \frac{3 {m}^{5} {n}^{4}(7 {m}^{3} n) }{3 {m}^{5} {n}^{4}(9) } = \frac{7 {m}^{3}n }{9}$

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$\frac{ - 15 {x}^{5} {y}^{4} }{45x {y}^{3} } = \frac{ 15x {y}^{3} ( - {x}^{4} y)}{15x {y}^{3}(3) } = - \frac{ {x}^{4}y }{3}$

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12.Ans;

$\frac{7 {p}^{2} {q}^{2} }{14 {p}^{2} {q}^{2} } = \frac{7 {p}^{2} {q}^{2} }{7 {p}^{2} {q}^{2} (2) } = \frac{1}{2} = 0.5$

I hope I helped you^_^

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2 years ago

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Find the missing side, please

AVprozaik [17]

Answer:

a. √442, b. √540, c. 8

Step-by-step explanation:

remember that a^2 + b^2 = c^2

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2 years ago