PLZZZZ NEED HELP!!!!!!

Mathematics

1 answer:

jeka942 years ago

7 0

Notice the 7) 1st sequence, and the 8) second one

$\bf \begin{array}{llll}
term&value\\
-----&-----\\
1&t_1=-2(1)+5\\
2&t_2=-2(2)+5\\
3&t_3=-2(3)+5\\
4&t_4=-2(4)+5\\
5&t_5=-2(5)+5\\
6&t_6=-2(6)+5\\
7&t_7=-2(7)+5\\
8&t_8=-2(8)+5\\
9&t_9=-2(9)+5\\
\end{array}\qquad \qquad 
\begin{array}{llll}
term&value\\
-----&-----\\1&t_1=2(3)^{1-1}\\2&t_2=2(3)^{2-1}\\3&t_3=2(3)^{3-1}\\\end{array}$

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Last year , there were 30 male members and 20 female members in a Chess Club . This year , the number of male members is decreas

LekaFEV [45]

The r% of a quantity x is computed by dividing x in 100 parts, and considering r of such parts. So, the r% of the male is

$30\times\cfrac{r}{100} = 0.3 r$

and similarly, the r% of female is

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and the number of female increased by this quantity, so now it is

$20+0.2r$

we know that these two new counts are the same number, so we can build and solve the equality

$30 - 0.3r = 20+0.2r$

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Divide both sides by 0.5 to solve for r:

$r = 20$

Let's check the answer

The 20% of 30 is $30 \times \frac{20}{100} = 6$ , while the 20% of 20 is 4. So, we are stating that $30-6 = 20+4$ which is true because both expressions evaluate to 24.