Sorry I forgot to take the picture.please solve 16-17

Mathematics

1 answer:

kaheart [24]3 years ago

4 0

Answer:

Can't see the picture because old computer but isn't 16 minus 17 = -1 (negative 1)

Step-by-step explanation:

You might be interested in

1 The 14 members of the baseball team are going to stop at an ice cream shop after the game. The ice cream shop has a special tr

sergeinik [125]

Answer:

###########################################################

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

Can someone help me understand this?

ASHA 777 [7]

< means 'less than'. 3.5 is less than 4.5, so A is correct. Since -4.5 is NEGATIVE it is less than 0. As for C, since both numbers are negative, it would be the seemingly 'smaller' number that is actually bigger. It may help to think of the numbers on a number line. In the negatives, the numbers start to go in reverse, so the further from 0 a negative number is, the smaller it is. -4.5 IS LESS than -3.5. All answers are correct.

4 0

3 years ago

Read 2 more answers

Find the perimeter of a square that is 3 3/4 inches on a side.

Dahasolnce [82]

perimeter would be s x 4

so 3 3/4 x 4 = 15

B is the correct answer

6 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.

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

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