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3 years ago

What is the greatest common factor of 15n^5, 30n^3, and 45n^2?

Mathematics

2 answers:

bixtya [17]3 years ago

8 0

Answer:

the greatest common factor is 45n^2

Step-by-step explanation:

Mama L [17]3 years ago

4 0

Answer:15n^2

Step-by-step explanation: The greatest common factor of 15,30, and 45 is 15. The greatest common factor of n^5, n^3, and n^2 is n^2.

so together is 15n^2

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Find the sum of the first 25 terms in this geometric series:<br> 8 + 6 + 4.5...

Ksivusya [100]

Step-by-step explanation:

Given the geometric sequence

8 + 6 + 4.5...

A geometric sequence has a constant ratio and is defined by

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

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

$\frac{6}{8}=\frac{3}{4},\:\quad \frac{4.5}{6}=\frac{3}{4}$

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

$r=\frac{3}{4}$

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

$a_1=8$

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

$a_n=8\left(\frac{3}{4}\right)^{n-1}$

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

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

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

$n=25,\:\spacea_1=8,\:\spacer=\frac{3}{4}$

$=8\cdot \frac{1-\left(\frac{3}{4}\right)^{25}}{1-\frac{3}{4}}$

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

$=\frac{\left(1-\left(\frac{3}{4}\right)^{25}\right)\cdot \:8}{1-\frac{3}{4}}$

$=\frac{8\left(-\left(\frac{3}{4}\right)^{25}+1\right)}{\frac{1}{4}}$

$\mathrm{Apply\:exponent\:rule}:\quad \left(\frac{a}{b}\right)^c=\frac{a^c}{b^c}$

$=\frac{8\left(-\frac{3^{25}}{4^{25}}+1\right)}{\frac{1}{4}}$

$\mathrm{Apply\:the\:fraction\:rule}:\quad \frac{a}{\frac{b}{c}}=\frac{a\cdot \:c}{b}$

$=\frac{\left(1-\frac{3^{25}}{4^{25}}\right)\cdot \:8\cdot \:4}{1}$

$\mathrm{Multiply\:the\:numbers:}\:8\cdot \:4=32$

$=\frac{32\left(-\frac{3^{25}}{4^{25}}+1\right)}{1}$

$=\frac{32\cdot \frac{4^{25}-3^{25}}{4^{25}}}{1}$ ∵ $\mathrm{Join}\:1-\frac{3^{25}}{4^{25}}:\quad \frac{4^{25}-3^{25}}{4^{25}}$

$=32\cdot \frac{4^{25}-3^{25}}{4^{25}}$

$=\frac{\left(4^{25}-3^{25}\right)\cdot \:32}{4^{25}}$

$=\frac{2^5\left(4^{25}-3^{25}\right)}{2^{50}}$ ∵ $\mathrm{Factor}\:32:\ 2^5$ , $\mathrm{Factor}\:4^{25}:\ 2^{50}$

$=\frac{4^{25}-3^{25}}{2^{45}}$ ∵ $\mathrm{Cancel\:}\frac{\left(4^{25}-3^{25}\right)\cdot \:2^5}{2^{50}}:\quad \frac{4^{25}-3^{25}}{2^{45}}$

$\mathrm{Apply\:the\:fraction\:rule}:\quad \frac{a\pm \:b}{c}=\frac{a}{c}\pm \frac{b}{c}$

$=\frac{4^{25}}{2^{45}}-\frac{3^{25}}{2^{45}}$

$=32-\frac{3^{25}}{2^{45}}$ ∵ $\frac{4^{25}}{2^{45}}=32$

$=32-0.024$ ∵ $\frac{3^{25}}{2^{45}}=0.024$

$=31.98$

Therefore, the sum of the first 25 terms in this geometric series: 31.98

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How do I solve equations with a variable to the power of something (give steps please)

lozanna [386]

Answer:

Step-by-step explanation:

Factoring will reveal the solution. So we divide the equation by the greatest common factor of the terms and use that factor as the coefficient. In this case the greatest common factor is just x.

2x^2+5x

x(2x+5) so the equation will equal zero when either of those expressions is zero because zero times anything is zero. x=0 and x=-5/2

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