Gym area OK, wall cladding?
A school gymnasium is 24 m long and 12 m wide. It will receive new wall cladding up to a height of 4.20 m.
Calculate the area of ββthis wall covering
A school gymnasium is 24 m long and 12 m wide. It will receive new wall cladding up to a height of 4.20 m.
Calculate the area of ββthis wall covering
If I subtract five times 8 from my number, I get 40 According to the teacher the calculation is: 5×8-0=40 But that's not true, is it?
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If I start from f(x) = – x^2 + c, I have to insert the points, so -2.5 = – (1.5)^2 + c -2.5 = -2.25 + c | +2.25 c = – 0.25 The minus for downward opening is in front of (1.5), otherwise it would be -2.5 = 1.5^2 + c |-2.25 c…
Can someone explain to me how I can solve this task
Hello!
So, you have the two widths and the height for all four sides.
And with A = a Γ b you can calculate and add the areas.
If no windows have to be taken into account and if the doors are “clothed”, no areas have to be drawn off.
Greeting
Martin
Thank you again, “in the picture” π
Hey,
you have four pages, so twice 12*4.20m and twice 24*4.2m, you have to add them.
Greetings
Thank you