Substitute the value of the variable into the expression and simplify, its theta arrow-to-the-right, pi/4 times 1-tan theta/sin theta - cos theta (i cant type the signs on computer, sorry
I mean I think so I’m not completely sure it’s a really hard question
My triangle solver says EF ≈ 7.5 ft.
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The law of cosines is usually used for this.
EF^2 = DE^2 +DF^2 -2*DE*DF*cos(D)
.. = 6^2 +11^2 -2*6*11*cos(40°)
.. ≈ 55.88
EF ≈ √55.88 ≈ 7.5 . . . ft
Answer:
its 68+1x200000000000
Step-by-step explanation:
When you read this, the first thing that should jump out at you is:
What does "largest" mean ?
Does it mean the longest possible playground ? The widest possible ?
The playground with the most possible area ?
Well, we can narrow it down right away. If you try and find the longest
or the widest possible playground, then what you get is: The longest or
the widest possible playground is 250 feet by zero. It has a perimeter
of 500 ft, and nobody can play in it. That's silly.
It makes a lot more sense if we look for the playground that has
the greatest AREA.
I happen to remember that if you have a certain fixed amount of
fence and you want to use it to enclose the most possible area,
then you should form it into a circle. And if it has to be a rectangle,
then the next most area will be enclosed when you form it into a square.
So you want to take your 500 feet of fence and make a playground
that's 125-ft long and 125-ft wide.
Its area is (125-ft x 125-ft) = 15,625 square feet.
Just to make sure that a square is the right answer, let's test
what we would have if we made it not quite square ... let's say
1 foot longer and 1 foot narrower:
Length = 126 feet
Width = 124 feet
Perimeter = 2 (126 + 124) = 500-ft good
Area = (126-ft x 124-ft) = 15,624 square feet.
Do you see what happened ? We kept the same perimeter, but
as soon as we started to make it not-square, the area started to
decrease.
The square is the rectangle with the most possible area.
