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Assoli18 [71]
2 years ago
6

If y varies directly as x and given y =9 when x= 5, fInd:

Mathematics
1 answer:
True [87]2 years ago
5 0

Step-by-step explanation:

The statement

y varies directly as x is written as

<h3>y \:  \:  \alpha  \:  \: x</h3>

So we have

<h3>y = kx</h3>

First of all we must find the relationship between them

when

x = 5 and y = 9

We have

9 = 5k

Divide both sides by 5

k =  \frac{9}{5}

So the equation for the variation is

y =  \frac{9}{5} x

<h2>b).</h2>

when x = 15

y =  \frac{9}{5}  \times 15 \\  = 9 \times 3

We have the answer as

<h3>y = 27</h3>

<h2>c).</h2>

when y = 6

6 =  \frac{9}{5} x \\ 30 = 9x \\ x =  \frac{30}{9}

We have the answer as

x =  \frac{10}{3}

Hope this helps you

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What we want is to write \displaystyle{ y=3x^2-18x-6 as y=a(x-h)^2+k

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Finally, distributing 3 over the two terms in the brackets we have:

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Answer: y=3(x-3)^2-33
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