D. C. Solve for the indicated variable in each of the following. (3 pts each show your
solution)
1. If y varies directly as x and y=3 when x=4, find y when x6
2. If r varies inversely as s and r=100 when s=27, find the values of r when s=45
3. If varies inversely as the square of q and p=3 when q4, find p when q =16.



Sagot :

Answer:

1. y = 9/2 or lowest term of 4 1/2

2. r = 60

3. 3/16

Step-by-step explanation:

1. y = kx

3 = 4k

¾ = k

y = ¾x

y = ¾(6)

Multiply 3 to 6

y = 18/4

Divide the numerator and denominator by 2

[tex]y = \frac{18 \div 2}{14 \div 2} [/tex]

Divide the numbers

[tex]y = \frac{9}{2} \: or \: lowest \: term \: of \: 4 \frac{1}{2} [/tex]

[tex]2. \: r = \frac{k}{s} [/tex]

[tex]100 = \frac{k}{27} [/tex]

k = 27(100)

k = 2,700

[tex]r = \frac{2700}{45} [/tex]

Divide 2,700 to 45

r = 60

[tex]3. \: p = \frac{k}{ {q}^{2} } [/tex]

[tex]3 = \frac{k}{ {4}^{2} } [/tex]

[tex]3 = \frac{k}{16} [/tex]

k = 16(3)

k = 48

[tex]p = \frac{48}{ {16}^{2} } [/tex]

[tex]p = \frac{48}{256} [/tex]

Divide the 48 to 256

[tex]p = \frac{48}{256} [/tex]

Divide the numerator and denominator by 16

[tex]p = \frac{48 \div 16}{256 \div 16} [/tex]

Divide the number

[tex]p = \frac{3}{16} [/tex]

I hope it's help

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