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3 years ago

9

During February, Kevin will water his ivy every third day, and water his cactus every fifth day. Will Kevin water both plants to

gether again in February? Explain.

2 answers:

777dan777 [17]3 years ago

5 0

Answer: Yes, Kevin can water both plants together again.

Step-by-step explanation:

Given: Kevin waters his ivy every third day, and his cactus every fifth day

such that the number of days after Kevin will water both plants together again=least common multiple of 3 and 5= 15

Thus, after 15 days Kevin can water both plants together again.

As in February, there are 28/29 days , thus if he waters plants before 13/14 february respectively[28-15=13, 19-15=14], then he will sure water both plants in february again.

Sergio039 [100]3 years ago

3 0

No, he will not because there is only 27-29 days in February and 15 is the only common multiple of 3 and 5.

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Find S12 for geometric series: (-7.5) + 15 + (-30) + ...

kolezko [41]

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S12 for geometric series: (-7.5) + 15 + (-30) + ... would be: 10237.5

Step-by-step explanation:

Given the sequence to find the sum up-to 12 terms

$(-7.5) + 15 + (-30) + ...$

As we know that

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

$a_n=a_1\cdot r^{n-1}$

$\mathrm{Compute\:the\:ratios\:of\:all\:the\:adjacent\:terms}:\quad \:r=\frac{a_{n+1}}{a_n}$

$\frac{15}{\left(-7.5\right)}=-2,\:\quad \frac{\left(-30\right)}{15}=-2$

$\mathrm{The\:ratio\:of\:all\:the\:adjacent\:terms\:is\:the\:same\:and\:equal\:to}$

$r=-2$

$\mathrm{The\:first\:element\:of\:the\:sequence\:is}$

$a_1=\left(-7.5\right)$

$a_n=a_1\cdot r^{n-1}$

$\mathrm{Therefore,\:the\:}n\mathrm{th\:term\:is\:computed\:by}\:$

$a_n=\left(-7.5\right)\left(-2\right)^{n-1}$

$a_n=-\left(-2\right)^{n-1}\cdot \:7.5$

$\mathrm{Geometric\:sequence\:sum\:formula:}$

$a_1\frac{1-r^n}{1-r}$

$\mathrm{Plug\:in\:the\:values:}$

$n=12,\:\spacea_1=\left(-7.5\right),\:\spacer=-2$

$=\left(-7.5\right)\frac{1-\left(-2\right)^{12}}{1-\left(-2\right)}$

$=-7.5\cdot \frac{1-\left(-2\right)^{12}}{1+2}$

$\mathrm{Multiply\:fractions}:\quad \:a\cdot \frac{b}{c}=\frac{a\:\cdot \:b}{c}$

$=-\frac{-30712.5}{1+2}$ ∵ $\left(1-\left(-2\right)^{12}\right)\cdot \:7.5=-30712.5$

$=-\frac{-30712.5}{3}$

$=\frac{30712.5}{3}$

$=10237.5$

Thus, S12 for geometric series: (-7.5) + 15 + (-30) + ... would be: 10237.5

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4 years ago