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Margarita [4]
3 years ago
13

Find the sum of a finite geometric series. The Smiths spent $2,000 on clothes in a particular year. If they increase the amount

they spend on clothes by 10% each year, the Smiths will spend a total of $ on clothing in 4 years.
Mathematics
1 answer:
boyakko [2]3 years ago
7 0

\bf \stackrel{\textit{first year}}{2000}~~,~~\stackrel{\textit{second year}}{2000+\stackrel{\textit{10\% of 2000}}{\frac{2000}{10}}}\implies 2200

now, if we take 2000 to be the 100%, what is 2200? well, 2200 is just 100% + 10%, namely 110%, and if we change that percent format to a decimal, we simply divide it by 100, thus \bf 110\%\implies \cfrac{110}{100}\implies 1.1.

so, 1.1 is the decimal number we multiply a term to get the next term, namely 1.1 is the common ratio.

\bf \qquad \qquad \textit{sum of a finite geometric sequence}\\\\S_n=\sum\limits_{i=1}^{n}\ a_1\cdot r^{i-1}\implies S_n=a_1\left( \cfrac{1-r^n}{1-r} \right)\quad \begin{cases}n=n^{th}\ term\\a_1=\textit{first term's value}\\r=\textit{common ratio}\\----------\\a_1=2000\\r=1.1\\n=4\end{cases}\\\\\\S_4=2000\left[ \cfrac{1-(1.1)^4}{1-1.1} \right]\implies S_4=2000\left(\cfrac{-0.4641}{-0.1}  \right)\\\\\\S_4=2000(4.641)\implies S_4=9282

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2 years ago
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Calculate pt3 such that a line from pt1 to pt3 is perpendicular to the line from pt1 to pt2, and the distance between pt1 and pt
Leni [432]
Let the point_1 = p₁ = (1,4)
and      point_2 = p₂ = (-2,1)
and      Point_3 = p₃ = (x,y)

The line from point_1 to point_2 is L₁ and has slope = m₁
The line from point_1 to point_3 is L₂ and has slope = m₂
m₁ = Δy/Δx = (1-4)/(-2-1) = 1
m₂ = Δy/Δx = (y-4)/(x-1)
L₁⊥L₂ ⇒⇒⇒⇒ m₁ * m₂ = -1
∴ (y-4)/(x-1) = -1 ⇒⇒⇒ (y-4)= -(x-1)
(y-4) = (1-x) ⇒⇒⇒⇒⇒⇒⇒⇒⇒⇒⇒⇒⇒⇒ equation (1)

The distance from point_1 to point_2 is d₁
The distance from point_1 to point_3 is d₂
d = \sqrt{Δx^2+Δy^2}
d₁ = \sqrt{(-2-1)^2+(1-4)^2}
d₂ = \sqrt{(x-1)^2+(y-4)^2}
d₁ = d₂
∴ \sqrt{(-2-1)^2+(1-4)^2} = \sqrt{(x-1)^2+(y-4)^2} ⇒⇒ eliminating the root
∴(-2-1)²+(1-4)² = (x-1)²+(y-4)²
 (x-1)²+(y-4)² = 18
from equatoin (1)  y-4 = 1-x
∴(x-1)²+(1-x)² = 18            ⇒⇒⇒⇒⇒ note: (1-x)² = (x-1)²
2 (x-1)² = 18
(x-1)² = 9
x-1 = \pm \sqrt{9} = \pm 3
∴ x = 4 or x = -2
∴ y = 1 or y = 7

Point_3 = (4,1)  or  (-2,7)












8 0
3 years ago
The length of a rectangle is 5 metres less than twice the breadth. If the perimeter is 50 meters,find the length and breadth
MAXImum [283]
<h3><u>S</u><u> </u><u>O</u><u> </u><u>L</u><u> </u><u>U</u><u> </u><u>T</u><u> </u><u>I</u><u> </u><u>O</u><u> </u><u>N</u><u> </u><u>:</u></h3>

As per the given question, it is stated that the length of a rectangle is 5 m less than twice the breadth.

Assumption : Let us assume the length as "l" and width as "b". So,

\twoheadrightarrow \quad\sf{ Length =2(Width)-5}

\twoheadrightarrow \quad\sf{ \ell=(2b-5) \; m}

Also, we are given that the perimeter of the rectangle is 50 m. Basically, we need to apply here the formula of perimeter of rectangle which will act as a linear equation here.

\\ \twoheadrightarrow \quad\sf{ Perimeter_{(Rectangle)} = 2(\ell +b) } \\

  • <em>l</em> denotes length
  • <em>b</em> denotes breadth

\\ \twoheadrightarrow \quad\sf{50= 2(2b-5+b)} \\

\\ \twoheadrightarrow \quad\sf{50= 2(3b-5)} \\

\\ \twoheadrightarrow \quad\sf{50= 6b - 10} \\

\\ \twoheadrightarrow \quad\sf{50+10= 6b} \\

\\ \twoheadrightarrow \quad\sf{60= 6b} \\

\\ \twoheadrightarrow \quad\sf{\cancel{\dfrac{60}{6}}=b} \\

\\ \twoheadrightarrow \quad\underline{\bf{10\; m = Width }} \\

Now, finding the length. According to the question,

\twoheadrightarrow \quad\sf{ \ell=(2b-5) \; m}

\twoheadrightarrow \quad\sf{ \ell=2(10)-5\; m}

\twoheadrightarrow \quad\sf{ \ell=20-5\; m}

\\ \twoheadrightarrow \quad\underline{\bf{15\; m = Length }} \\

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Whats the formula for interquartile range
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Q3-Q1=IQR is the formula.

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