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xenn [34]
3 years ago
7

Are 6r +3 and 9r equivalent ​

Mathematics
1 answer:
Vaselesa [24]3 years ago
6 0

Answer:

No

Step-by-step explanation:

u cant add 6r+3 because only one number has a variable. 6r+3r would equal 9r

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Westkost [7]

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the smallest number is 7

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Answer ALL the questions, SHOW ALL YOUR WORK. Explain your answers when necessary. Answer ALL questions
Rufina [12.5K]

Answer:

  d. Anthony rode 14 more miles per week than Malique.

Step-by-step explanation:

Anthony rode (141 mi)/(3 wk) = 47 mi/wk.

Malique rode (165 mi)/(5 wk) = 33 mi/wk.

Anthony rode more miles per week by ...

  (47 -33) mi/wk = 14 mi/wk

Anthony rode 14 more miles per week than Malique.

5 0
3 years ago
What is the nth term of this sequence?
lions [1.4K]

Answer:

D, you simply just add 1.3 each time.

4 0
2 years ago
How do you solve his with working
AlexFokin [52]
Check the picture below.

a)

so the perimeter will include "part" of the circumference of the green circle, and it will include "part" of the red encircled section, plus the endpoints where the pathway ends.

the endpoints, are just 2 meters long, as you can see 2+15+2 is 19, or the radius of the "outer radius".

let's find the circumference of the green circle, and then subtract the arc of that sector that's not part of the perimeter.

and then let's get the circumference of the red encircled section, and also subtract the arc of that sector, and then we add the endpoints and that's the perimeter.

\bf \begin{array}{cllll}
\textit{circumference of a circle}\\\\ 
2\pi r
\end{array}\qquad \qquad \qquad \qquad 
\begin{array}{cllll}
\textit{arc's length}\\\\
s=\cfrac{\theta r\pi }{180}
\end{array}\\\\
-------------------------------

\bf \stackrel{\stackrel{green~circle}{perimeter}}{2\pi(7.5) }~-~\stackrel{\stackrel{green~circle}{arc}}{\cfrac{(135)(7.5)\pi }{180}}~+
\stackrel{\stackrel{red~section}{perimeter}}{2\pi(9.5) }~-~\stackrel{\stackrel{red~section}{arc}}{\cfrac{(135)(9.5)\pi }{180}}+\stackrel{endpoints}{2+2}
\\\\\\
15\pi -\cfrac{45\pi }{8}+19\pi -\cfrac{57\pi }{8}+4\implies \cfrac{85\pi }{4}+4\quad \approx \quad 70.7588438888



b)

we do about the same here as well, we get the full area of the red encircled area, and then subtract the sector with 135°, and then subtract the sector of the green circle that is 360° - 135°, or 225°, the part that wasn't included in the previous subtraction.


\bf \begin{array}{cllll}
\textit{area of a circle}\\\\ 
\pi r^2
\end{array}\qquad \qquad \qquad \qquad 
\begin{array}{cllll}
\textit{area of a sector of a circle}\\\\
s=\cfrac{\theta r^2\pi }{360}
\end{array}\\\\
-------------------------------

\bf \stackrel{\stackrel{red~section}{area}}{\pi(9.5^2) }~-~\stackrel{\stackrel{red~section}{sector}}{\cfrac{(135)(9.5^2)\pi }{360}}-\stackrel{\stackrel{green~circle}{sector}}{\cfrac{(225)(7.5^2)\pi }{360}}
\\\\\\
90.25\pi -\cfrac{1083\pi }{32}-\cfrac{1125\pi }{32}\implies \cfrac{85\pi }{4}\quad \approx\quad 66.75884

7 0
3 years ago
Elephant<br> A. 30 kilograms<br> B. 90 grams<br> C. 5,000 grams<br> D. 5,000 kilograms
Mariulka [41]
Pretty sure it’s d. 5,000 kilograms
7 0
3 years ago
Read 2 more answers
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