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MrRissso [65]
3 years ago
5

Eduardo spent 35% of his time at the theme park on a roller coaster if he was there for eight hours how much time did you spend

on roller coasters
Mathematics
1 answer:
MakcuM [25]3 years ago
3 0

Answer:

2.8

Step-by-step explanation:

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What are the intercepts of the equation <br> 2x+3/2y+3z=6
Ne4ueva [31]
X-intercept => 2x = 6 => x = 3
x-intercept is 3.

y-intercept => 3/2y = 6 => y = 4
y-intercept is 4.

z-intercept => 3z = 6 => z = 2
z-intercept is 2.
3 0
3 years ago
In 2000, U.S. Airways burned about 115 litres of fuel per passenger. In 2014 that dropped
ankoles [38]

Answer:

22% reduction

Step-by-step explanation:

Find the percent decrease by dividing the difference in amounts by the original amount.

Find the difference:

115 - 90 = 25

Divide this by the original amount:

25/115

= 0.217

So, the percent decrease is 21.7%. Round this to the nearest whole percentage:

= 22

The percent reduction in fuel used was approx. 22%

6 0
3 years ago
Jeremy wanted to buy a used car for $4000. He had to pay 8% in sales tax. How much is he paying for the car?
lord [1]

Answer:

he pays 4320 :)

Step-by-step explanation:

5 0
3 years ago
Account A and Account B both have a principal of $1,000 and an annual interest rate of 4%. No additional deposits or withdrawals
postnew [5]
Account A
Total-$1800
Interest-$800

Account B-
Total-$2191.12
Interest-$1191.12

Account B by $391.12
4 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}&#10;\textit{circumference of a circle}\\\\ &#10;2\pi r&#10;\end{array}\qquad \qquad \qquad \qquad &#10;\begin{array}{cllll}&#10;\textit{arc's length}\\\\&#10;s=\cfrac{\theta r\pi }{180}&#10;\end{array}\\\\&#10;-------------------------------

\bf \stackrel{\stackrel{green~circle}{perimeter}}{2\pi(7.5) }~-~\stackrel{\stackrel{green~circle}{arc}}{\cfrac{(135)(7.5)\pi }{180}}~+&#10;\stackrel{\stackrel{red~section}{perimeter}}{2\pi(9.5) }~-~\stackrel{\stackrel{red~section}{arc}}{\cfrac{(135)(9.5)\pi }{180}}+\stackrel{endpoints}{2+2}&#10;\\\\\\&#10;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}&#10;\textit{area of a circle}\\\\ &#10;\pi r^2&#10;\end{array}\qquad \qquad \qquad \qquad &#10;\begin{array}{cllll}&#10;\textit{area of a sector of a circle}\\\\&#10;s=\cfrac{\theta r^2\pi }{360}&#10;\end{array}\\\\&#10;-------------------------------

\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}}&#10;\\\\\\&#10;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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