John got 20% discount at a sale .He got a new shirt for $12 .What is the original price of the shirt ?

Mathematics

1 answer:

Arte-miy333 [17]2 years ago

3 0

The original price of the shirt must be $15 if a 20% discount resulted in the final price becoming $12

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Using a table of values, determine the solution to the equation below to the nearest fourth of a unit.

Leto [7]

2x-4=3-x +1
<span>Simplifying 2x + -4 = 3 + -1x + 1
Reorder the terms: -4 + 2x = 3 + -1x + 1
Reorder the terms: -4 + 2x = 3 + 1 + -1x
Combine like terms: 3 + 1 = 4 -4 + 2x = 4 + -1x
Solving -4 + 2x = 4 + -1x
Solving for variable 'x'.
Move all terms containing x to the left, all other terms to the right.
Add 'x' to each side of the equation. -4 + 2x + x = 4 + -1x + x
Combine like terms: 2x + x = 3x -4 + 3x = 4 + -1x + x
Combine like terms: -1x + x = 0 -4 + 3x = 4 + 0 -4 + 3x = 4
Add '4' to each side of the equation. -4 + 4 + 3x = 4 + 4
Combine like terms: -4 + 4 = 0 0 + 3x = 4 + 4 3x = 4 + 4
Combine like terms: 4 + 4 = 8 3x = 8
Divide each side by '3'. x = 2.666666667
Simplifying x = 2.666666667</span>

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

5. h(x)=0.5x - 1 with domain (-♾, 4) what is the range

Rudiy27

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