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3 years ago

Y varies directly as x and y = 9 when x = 72

Mathematics

1 answer:

PIT_PIT [208]3 years ago

6 0

Answer:

y = $\frac{x}{8}$

Step-by-step explanation:

since y varies directly as x, then the equation relating them is

y = kx ← k is the constant of variation

to find k use the given condition y = 9 when x = 72

k = $\frac{9}{x}$ = $\frac{9}{72}$ = $\frac{1}{8}$ , hence

y = $\frac{1}{8}$ × x = $\frac{x}{8}$

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40, 10, 5/2 5/8... (fractions) is it arithmetic or geometric and what are the next two terms PLZ HELP DUE IN 10 MINS

patriot [66]

Answer:

Please check the explanation.

Step-by-step explanation:

Given the sequence

$40,\:10,\:\frac{5}{2},\:\frac{5}{8}$

A geometric sequence has a constant ratio 'r' and is defined by

$\:a_n=a_0\cdot r^{n-1}$

Computing the ratios of all the adjacent terms

$\frac{10}{40}=\frac{1}{4},\:\quad \frac{\frac{5}{2}}{10}=\frac{1}{4},\:\quad \frac{\frac{5}{8}}{\frac{5}{2}}=\frac{1}{4}$

The ratio of all the adjacent terms is the same and equal to

$r=\frac{1}{4}$

Thus, the given sequence is a geometric sequence.

As the first element of the sequence is

$a_1=40$

Therefore, the nth term is calculated as

$\:a_n=a_0\cdot r^{n-1}$

$a_n=40\left(\frac{1}{4}\right)^{n-1}$

Put n = 5 to find the next term

$a_5=40\left(\frac{1}{4}\right)^{5-1}$

$a_5=40\cdot \frac{1}{4^4}$

$a_5=\frac{40}{4^4}$

$=\frac{2^3\cdot \:5}{2^8}$

$a_5=\frac{5}{2^5}$

$a_5=\frac{5}{32}$

now, Put n = 6 to find the 6th term

$a_6=40\left(\frac{1}{4}\right)^{6-1}$

$a_6=40\cdot \frac{1}{4^5}$

$a_6=\frac{40}{4^5}$

$=\frac{2^3\cdot \:5}{2^{10}}$

$a_6=\frac{5}{2^7}$

$a_6=\frac{5}{128}$

Thus, the next two terms of the sequence 40, 10, 5/2, 5/8... is:

$a_5=\frac{5}{32}$

$a_6=\frac{5}{128}$

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Answer:

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Answer:

Step-by-step explanation:

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