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garri49 [273]
3 years ago
11

How many pairs of whole numbers have the sum of 99

Mathematics
1 answer:
valentina_108 [34]3 years ago
7 0
50 pairs of whole numbers
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60.3 60.44 witch one is greater
Sindrei [870]
60.44 is greater than 60.3
-you can see this when you subtract 60.3 from 60.44 this would leave a 0.11 which is a positive number.
-thus meaning 60.44 is larger/greater.
4 0
3 years ago
Help pls !! how would I write this equation
aev [14]

2x = Twice your age

200 = Two hundred less

> -1 = more than negative


With the information given, your equation would be:

200-2x>-1

7 0
3 years ago
What is the speed that a car reaches in m/s, if it travels 956km in 56sec?Help me!! MRU (Uniform Rectilinear Movement)
In-s [12.5K]

Answer:

17.07m/s

Step-by-step explanation:

v=d/t

d=956km

t=56sec

v=956/56

v=17.07m/s

8 0
3 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
What is the run of a 50% slope with a rise of 50 feet?
Alinara [238K]
25 is the answer because 50% of 50 is 25
5 0
3 years ago
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