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lorasvet [3.4K]
3 years ago
5

What is the common ratio of 2, 10/3, 50/9

Mathematics
1 answer:
mojhsa [17]3 years ago
6 0

The common ratio of 2, 10/3, 50/9 is \frac{10}{6}

<u>Solution:</u>

Given, series of elements are 2, \frac{10}{3}, \frac{50}{9}

We have to find the common ratio of the above given series.

We know that, common ratio of an G.P is division of any number in that series with the previous number of the series.

So, now take \frac{10}{3} and 2

\text { Then, common ratio }=\frac{\frac{10}{3}}{2}

\begin{array}{l}{\text { Common ratio }=\frac{10}{3} \times \frac{1}{2}} \\\\ {\text { Common ratio }=\frac{10}{3 \times 2}=\frac{10}{6}}\end{array}

Hence, the common ratio of the given series is \frac{10}{6}

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iogann1982 [59]
-1/3(x - 15) = -4

step 1 : multiply both sides by -3, step 2 : add 15 to both sides

because when u multiply both sides by -3, this cancels out the -1/3 on the left side leaving u with : x - 15 = -4 * -3.....x - 15 = 12.....and then u would add 15 to both sides....giving u : x = 12 + 15.....x = 27
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2 years ago
In the equation (x^2+y)^5, what is the coefficient of the term x^4y^3? what is the coefficient of the same term in the expansion
lys-0071 [83]

\displaystyle&#10;(x+y)^n=\sum_{k=0}^n\binom{n}{k}x^{n-k}y^k

<em>-------------------------------------------------------------</em>


\displaystyle&#10;(x^2+y)^n=\sum_{k=0}^n\binom{n}{k}x^{2n-2k}y^k\\&#10;n=5\\&#10;k=3\\\\\binom{5}{3}=\dfrac{5!}{3!2!}=\dfrac{4\cdor5}{2}=10

<u>It's 10.</u>

----------------------------------------------------

\displaystyle&#10;(3x^2+y)^n=\sum_{k=0}^n\binom{n}{k}(3x)^{2n-2k}y^k=\sum_{k=0}^n\binom{n}{k}\cdot 3^{2n-2k}\cdot x^{2n-2k}y^k\\\\&#10;n=5\\&#10;k=4\\\\&#10;\binom{5}{3}\cdot3^{2\cdot5-2\cdot4}=10\cdot3^{2}=10\cdot9=90

<u>It's 90</u>

6 0
3 years ago
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