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Leviafan [203]
2 years ago
14

Pls HELPPP !!!!!!!!Variables y and x have a proportional relationship, and y = 21 when x = 14. What is the value of x when y = 1

2?
Mathematics
1 answer:
Drupady [299]2 years ago
8 0

Answer:

8

Step-by-step explanation:

y=21 divided by x=14 = 1.5 so Y is 1.5 times larger than X so you should divide y=12 by 1.5 to get 8

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4 years ago
Can someone PLZZZZ HELP!!!!
swat32

Answer:

The solution is:

Part A. \sqrt{5}^{\frac{7k}{3}}) which is sqrt(5)^7k/3[/tex]

Part B. k = 18/7

Step-by-step explanation:

Part A.

To solve this part, we're going two use THREE important properties of exponents:

1. (x^{n})^{m} = x^{nm}

2. \frac{x^{n}}{x^{m}} = x^{n-m}

3. \sqrt[n]{x^{m}} = x^{\frac{m}{n} }

Let's work the numerator using the properties 1, 2 and 3:

(\sqrt{5}^{3} )^{\frac{k}{9} } }  = (\sqrt{5}^{3\frac{k}{9}}) = (\sqrt{5}^{\frac{k}{3}})

Let's work the denominator using the properties 1, 2 and 3:

(\sqrt{5}^{6} )^{-\frac{k}{3} } }  = (\sqrt{5}^{6\frac{k}{3}}) = (\sqrt{5}^(2k))

Now dividing the numerator by the denominator:

\sqrt{5}^{\frac{k}{3}-(-2k)})=\sqrt{5}^{\frac{7k}{3}})

Part B

if 5^{\frac{3}{2} } 5^{\frac{3}{2}} = \sqrt{5}^{\frac{7k}{3}})

Then:

5^{3} = 5^{\frac{7k}{6}})

So \frac{7k}{6}} = 3

Solving for k, we have:

k = 18/7

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