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GrogVix [38]
3 years ago
8

Suppose y varies directly with x. Write a direct variation equation that relates x and y. Then find the value of y when x = 7.

Mathematics
2 answers:
horsena [70]3 years ago
5 0
From the statement, y varies directly with x. It can be expressed as:

y α x

To change it to an equality, we insert a proportionality constant k.

y = kx

We first solve for k,

8 = k(-4)
k = -2

Therefore, when x = 7,

y = -2(7)
y = -14
Sonja [21]3 years ago
4 0
For direct variation, y = kx
8 = -4k
k = 8/-4 = -2

Therefore, required equation is y = -2x

when x = 7, y = -2(7) = -14
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BRAINLIEST AWARD NO.2
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Answer:

A whole number first term to render as fifth term a value larger than 10000, should be at least 121

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The formula is given as recursive since it involves the previous number of the sequence, and defined as:

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Then in this case, the first five terms are:

a_1=4\\a_2=4*3+6=18\\a_3=18*3+6=60\\a_4=60*3+6=186\\a_5=186*3+6=564\\

So if we want to find the first term in the case that the fifth one is greater than 10,000 using this recursive formula, now we have to start backwards, and say that the fifth term is "> 10000" and what the fourth one is.

Notice that if you have this definition for the nth term, we can obtain from it, what the previous term is to find the general rule:

a_n=a_{n-1}*3+6\\a_n-6=a_{n-1}*3\\\frac{a_n-6}{3} = a_{n-1}\\a_{n-1}=\frac{a_n}{3} -2

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a_5>10000\\\frac{a_5}{3} -2>\frac{10000}{3} -2\\a_4>=\frac{10000}{3} -2\\\frac{a_4}{3} -2>\frac{\frac{10000}{3}-2}{3}-2 =\frac{10000}{9}-\frac{8}{3} \\a_3>\frac{10000}{9}-\frac{8}{3} \\\frac{a_3}{3} -2>\frac{\frac{10000}{9}-\frac{8}{3} }{3} -2=\frac{10000}{27} -\frac{8}{9} -2=\frac{10000}{27} -\frac{26}{9}\\a_2=\frac{10000}{27} -\frac{26}{9}\\\frac{a_2}{3} -2>\frac{\frac{10000}{27} -\frac{26}{9}}{3} -2=\frac{10000}{81} -\frac{80}{27} \\a_1>\frac{10000}{81} -\frac{80}{27}\approx 120.49

therefore, the starting first term should be at least about 121 to give a fifth term larger than 10,000

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