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makkiz [27]
2 years ago
7

Tony writes an equation to represent the distance he walks in meters, d, in relation to the number of minutes he walks, m. d = 1

08m Mark has created a table to record the number of minutes it took him to walk different distances in meters. Time (minutes) Distance (meters) 10 1,100 20 2,200 40 4,400 Assuming that Tony and Mark both walk at a constant rate, the unit rate for Tony is meters per minute and the unit rate for Mark is meters per minute. On a graph, the line representing Tony’s distance covered with respect to time would be Mark’s line. So, Tony’s walking speed is Mark’s.
Mathematics
2 answers:
anastassius [24]2 years ago
7 0

Answer:

Tony : d = 108m.....so Tony walks 108 meters per minute

Mark :

(10,1100),(20,2200)

slope = (2200 - 1100) / (20 - 10) = 1100/10 = 110 meters per minute

So Tony's walk speed is 2 meters per minute slower then Tony's walk speed

Step-by-step explanation:  i just did it and got it right

liraira [26]2 years ago
6 0
Tony : d = 108m.....so Tony walks 108 meters per minute

Mark :
(10,1100),(20,2200)
slope = (2200 - 1100) / (20 - 10) = 1100/10 = 110 meters per minute

So Tony's walk speed is 2 meters per minute slower then Tony's walk speed
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What is a3 in an arithmetic sequence in which a10=41 and a15=61
USPshnik [31]
\bf \begin{array}{llll}
term&value\\
-----&-----\\
a_{10}&41\\
a_{11}&41+d\\
a_{12}&(41+d)+d\\
&41+2d\\
a_{13}&(41+2d)+d\\
&41+3d\\
a_{14}&(41+3d)+d\\
&41+4d\\
a_{15}&(41+4d)+d\\
&41+5d=61
\end{array}
\\\\\\
41+5d=61\implies 5d=20\implies d=\cfrac{20}{5}\implies \boxed{d=4}\\\\
-------------------------------\\\\

\bf n^{th}\textit{ term of an arithmetic sequence}\\\\
a_n=a_1+(n-1)d\qquad 
\begin{cases}
n=n^{th}\ term\\
a_1=\textit{first term's value}\\
d=\textit{common difference}\\
----------\\
d=4\\
n=10\\
a_{10}=41
\end{cases}
\\\\\\
41=a_1+(10-1)4\implies 41=a_1+36\implies \boxed{5=a_1}

thus

\bf n^{th}\textit{ term of an arithmetic sequence}\\\\
a_n=a_1+(n-1)d\qquad 
\begin{cases}
n=n^{th}\ term\\
a_1=\textit{first term's value}\\
d=\textit{common difference}\\
----------\\
d=4\\
n=3\\
a_{1}=5
\end{cases}
\\\\\\
a_3=a_1+(3-1)4\implies a_3=5+(3-1)4

and surely you know how much that is.
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2 years ago
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Answer:

The width is: w(x)=4x^2

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Step-by-step explanation:

The given rectangle has area given algebraically by the function:

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The width of the rectangle is the greatest common factor of 8x^5,  12x^3 and 20x^2

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We now divide the area by the width to obtain the length of the rectangle:

l(x)=\frac{8x^5+12x^3+20x^2}{4x^2}

This simplifies to:

l(x)=\frac{8x^5}{4x^2}+\frac{12x^3}{4x^2}+\frac{20x^2}{4x^2}

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f(x) = -5x^2 - x + 20;

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