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)
The area of a triangle can be calculated as half the product of base length and height. We want the area to be no greater than 10 in².
a) (1/2)(4)(2x-3) ≤ 10
b) 4x -6 ≤ 10 . . . . . simplify
... 4x ≤ 16 . . . . . . . .add 6
... x ≤ 4 . . . . . . . . . divide by the coefficient of x
c) The maximum value of (2x -3) in is (2·4 -3) in = 5 in.
The triangle should be no more than 5 in high to have an area less than 10 in².
y
=
3
x
4
−
3
Use the slope-intercept form to find the slope and y-intercept.
Tap for more steps...
Slope:
3
4
Y-Intercept:
−
3
Answer:
x=−50
Step-by-step explanation:
By=x+2, as this is the only equation that shows the relation between 2 variables
