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Misha Larkins [42]
2 years ago
9

A camera priced at ​$220 is marked down 30​%. Find the price of the camera after the markdown.

Mathematics
1 answer:
Usimov [2.4K]2 years ago
6 0

Answer:

154

Step-by-step explanation:

220-(220*30%)=154

Use this on google next time so you do not waste points on brainly

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T(z)= sin^2(z/4), {-3 ≤ z ≤ 8}
Usimov [2.4K]

Answer:

T´(z)=n^2s/4*i

Step-by-step explanation:

T(z)=d/dz(sin^2*z/4)

T´(z)=d/dz(sn^2zi/4)

T´(z)=d/dz(sn^2z/4*i)

T´(z)i*d/dz(sn^2z/4)

T´(z)=n^2s/4*i

Simplify

3 0
2 years ago
Find the perimeter of this problem
GrogVix [38]

Answer: 119 4/7

Step-by-step explanation:

Add all the sides together.

4 0
3 years ago
You roll a die. What is the probability that 2 will not appear on the die
Shalnov [3]

Answer: 5/6

Step-by-step explanation: its because it has a 5/6 chance to not land on 2 if i were to roll a die and ExPeCt to land on a 2 it would be 1/6

so the answer is "the probability is 5/6 chance"

3 0
3 years ago
Read 2 more answers
Question in the picture below
tekilochka [14]
The answer is C . cube 2
Explanation:
Using PEMDAS its Parentheses, exponents ,multiple then divide and add , subtract
8 0
2 years ago
Read 2 more answers
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
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