Draw or sketch out any problems like this, otherwise they appear abstract.
A circle’s area can be calculated by (pi d^2)/4 We have an area of 56 cm (^2?), so
pi d^2 = 56 x 4 (or 224) d^2 = 224/pi, d = √(224/pi)
A circle circumscribed around a square has a diameter equivalent to the length of the square’s diagonal, so the square’s diagonal is √(224/pi) (same as the circle diameter…)
A square’s side can be calculated, knowing its diagonal length, by use of Pythagoras’ theorem… The diagonal √(224/pi) is squared, divided by two, since the square’s sides are all equal, and the resulting number’s square root is calculated.
Squaring √(224/pi), we get 224/pi, and dividing by two, we get 112/pi, which is 35.6507 (cm^2), and the square root is 5.9708 cm, the side of the square.
I cannot emphasize enough that a drawing or sketch is an invaluable tool for these tasks, it saves having to retain a “picture” in your head. Note that a calculator was not required up until the last moment, dividing 112 by pi, and finding the square root of that answer. Picking up the calculator too early obliges you to transcribe numbers from the calculator to paper, and that can lead to issues. Try to enjoy maths, see it as a challenge not a chore. (and use correct units!)
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Related Questions (More Answers Below)
Answer:
Check below
Step-by-step explanation:
Hi,
We're dealing with linear functions.
We have here the first function.
That's a linear function, with slope, and no linear coefficient since
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Even though each one has the same slope value, they have different non zero linear coefficients, {-6,-18,6,18}. Unlike, the first one
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Answer: Option (3)
Step-by-step explanation:
The slope of the given line is
Thus, the answer must be Option (3)
Answer: It is only one can that can be filled up.
Step-by-step explanation: If 1 gas can can hold 10 L of gas and you only have 7 L then how can you fill up more than 1 gas can with only 7 L? You don't have enough gas to fill up more than 1 gas can. So you are left with only 1 gas can filled but only with 7 L.
The distance between any two points is:
d^2=(x2-x1)^2+(y2-y1)^2
d^2=(6--2)^2+(4--4)^2
d^2=8^2+8^2
d^2=64+64
d^2=128
d=√128 units