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3 years ago

I need help on these questions PLEASE

Mathematics

1 answer:

nordsb [41]3 years ago

6 0

<span>I think the easiest way to answer all of these is to just go ahead and find the equation of the line first.

I'm pretty sure that by the time you reach this problem, you've learned to find
the equation of a line that passes through two given points. You have two points: (2, 100) and (5, 205). The equation of the line through them is [ y = 35x + 30 ].

6). The initial membership fee ... the cost for walking in the door and signing up,
before any months, is $30. (Also the y-intercept.)

7). The cost per month is $35. (The slope of the line.)

8). We already did that, up above. [ y = 35x + 30 ].

9). 'x' in the equation is the number of months.
There are 12 months in one year.

y = 35(12) + 30 = $450</span>

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Subtract and simplify Show work <br> (-5y^2-9y-4)-(-2y^2-y+8)

katrin2010 [14]

Answer:

-3y^2-8y-12

Step-by-step explanation:

3 0

3 years ago

The sum of the series {1(2/3)}²+{2(1)3)}²+3²+{3(2/3)}²+....to 10 term is

ryzh [129]

Step-by-step explanation:

<h3><u>Given Question </u></h3>

The sum of the series is

$\tt{ {\bigg[1\dfrac{2}{3} \bigg]}^{2} + {\bigg[2\dfrac{1}{3} \bigg]}^{2} + {3}^{2} + {\bigg[3\dfrac{2}{3} \bigg]}^{2} + - - 10 \: terms}$

$\green{\begin{gathered}\large{\sf{{\underline{Formula \: Used - }}}} \end{gathered}}$

$\boxed{\tt{ \displaystyle\sum_{k=1}^{n}1 = n \: }}$

$\boxed{\tt{ \displaystyle\sum_{k=1}^{n}k = \frac{n(n + 1)}{2} \: }}$

$\boxed{\tt{ \displaystyle\sum_{k=1}^{n} {k}^{2} = \frac{n(n + 1)(2n + 1)}{6} \: }}$

$\large\underline{\sf{Solution-}}$

Given series is

$\rm :\longmapsto\: {\bigg[1\dfrac{2}{3} \bigg]}^{2} + {\bigg[2\dfrac{1}{3} \bigg]}^{2} + {3}^{2} + {\bigg[3\dfrac{2}{3} \bigg]}^{2} + - - - 10 \: terms$

can be rewritten as

$\rm \: = \: {\bigg[\dfrac{5}{3} \bigg]}^{2} + {\bigg[\dfrac{7}{3} \bigg]}^{2} + {\bigg[\dfrac{9}{3} \bigg]}^{2} + {\bigg[\dfrac{11}{3} \bigg]}^{2} + - - - 10 \: terms$

$\rm \: = \: \dfrac{1}{9}[ {5}^{2} + {7}^{2} + {9}^{2} + - - - 10 \: terms \: ]$

Now, here, 5, 7, 9 forms an AP series with first term 5 and common difference 2.

So, its general term is given by 5 + ( n - 1 )2 = 5 + 2n - 2 = 2n + 3

So, above series can be represented as

$\rm \: = \: \dfrac{1}{9}\displaystyle\sum_{n=1}^{10}(2n + 3) ^{2}$

$\rm \: = \: \dfrac{1}{9}\displaystyle\sum_{n=1}^{10}\bigg[ {4n}^{2} + 9 + 12n\bigg]$

$\rm \: = \: \dfrac{1}{9}\bigg[\displaystyle\sum_{n=1}^{10} {4n}^{2} + \displaystyle\sum_{n=1}^{10}9 + 12\displaystyle\sum_{n=1}^{10}n\bigg]$

$\rm \: = \: \dfrac{1}{9}\bigg[4\displaystyle\sum_{n=1}^{10} {n}^{2} +9 \displaystyle\sum_{n=1}^{10}1 + 12\displaystyle\sum_{n=1}^{10}n\bigg]$

$\rm \: = \: \dfrac{4}{9}\bigg[\dfrac{10(10 + 1)(20 + 1)}{6} \bigg] + 10 + \dfrac{4}{3}\bigg[\dfrac{10(10 + 1)}{2} \bigg]$

$\rm \: = \: \dfrac{4}{9}\bigg[\dfrac{10(11)(21)}{6} \bigg] + 10 + \dfrac{4}{3}\bigg[\dfrac{10(11)}{2} \bigg]$

$\rm \: = \: \dfrac{1540}{9} + 10 + \dfrac{220}{3}$

$\rm \: = \: \dfrac{1540 + 90 + 660}{9}$

$\rm \: = \: \dfrac{2290}{9}$

Hence,

$\boxed{\tt{ {\bigg[1\dfrac{2}{3} \bigg]}^{2} + {\bigg[2\dfrac{1}{3} \bigg]}^{2} + {3}^{2} + {\bigg[3\dfrac{2}{3} \bigg]}^{2} + - - 10 \: terms = \frac{2290}{9}}}$