At the farmer's market, there is a scale that measures the weight of fruit and vegetables. The needle is 16 cm long, and it rota
2 answers:
It may help to draw a picture as you start planning this problem and draw a circle, showing a circle which represents the scale needle's range of motion 360 degrees around. The needle length will then be the radius of that circle.
We know that the needle rotated 68 degrees and are trying to find the distance it traveled in cm. To find out how far the needle traveled, we need to find the arc length - in this case, the distance between the needle's starting point and where it ended up after the oranges were placed on the scale on the circle.
There is a formula you can easily use to find the arc length
arc length = 2*
*r ( 
)
We need to input the radius, r into this formula, as well as the central angle measure c, which in this case is 68 degrees.
2(16) ( 
)
If you multiply the above formula components all together using a calculator, you will get ~18.99. (When rounded to the second decimal place.)
So, the answer is around 19 cm. (Your answer may vary depending on how your teacher usually wants you to round answers.)
We know that the needle rotated 68 degrees and are trying to find the distance it traveled in cm. To find out how far the needle traveled, we need to find the arc length - in this case, the distance between the needle's starting point and where it ended up after the oranges were placed on the scale on the circle.
There is a formula you can easily use to find the arc length
arc length = 2*
We need to input the radius, r into this formula, as well as the central angle measure c, which in this case is 68 degrees.
2
If you multiply the above formula components all together using a calculator, you will get ~18.99. (When rounded to the second decimal place.)
So, the answer is around 19 cm. (Your answer may vary depending on how your teacher usually wants you to round answers.)
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Answer:
(x, y) = (2, 5)
Step-by-step explanation:
I find it easier to solve equations like this by solving for x' = 1/x and y' = 1/y. The equations then become ...
3x' -y' = 13/10
x' +2y' = 9/10
Adding twice the first equation to the second, we get ...
2(3x' -y') +(x' +2y') = 2(13/10) +(9/10)
7x' = 35/10 . . . . . . simplify
x' = 5/10 = 1/2 . . . . divide by 7
Using the first equation to find y', we have ...
y' = 3x' -13/10 = 3(5/10) -13/10 = 2/10 = 1/5
So, the solution is ...
x = 1/x' = 1/(1/2) = 2
y = 1/y' = 1/(1/5) = 5
(x, y) = (2, 5)
_____
The attached graph shows the original equations. There are two points of intersection of the curves, one at (0, 0). Of course, both equations are undefined at that point, so each graph will have a "hole" there.
