So you have to add all the numbers to get the perimeter or the outside distance but it looks like t and s is there a few times so that’s were you have to simplify.
So here’s the answer simplified
4t+2s+w+v+x+u
Good luck!
Part A) Find BC, the distance from Tower 2 to the plane, to the nearest foot.
in the triangle ACD
sin16=CD/(7600+BD)--------> CD=sin16*(7600+BD)---------> equation 1
in the triangle BCD
sin24=CD/BD-----------> CD=sin24*BD---------------> equation 2
equation 1=equation 2
sin16*(7600+BD)=sin24*BD-----> sin16*7600+sin16*BD=sin24*BD
sin24*BD-sin16*BD=sin16*7600----> BD=[sin16*7600]/[sin24-sin16]
BD=15979 ft
in the triangle BCD
cos24=BD/BC---------> BC=BD/cos24-------> 15979/cos24-------> 17491
BC=17491 ft
the answer part 1) BC is 17491 ft
Part 2) Find CD, the height of the plane from the ground, to the nearest foot.
CD=sin24*BD ( remember equation 2)
BD=15979 ft
CD=sin24*15979 -----------> CD=6499 ft
the answer part 2) CD is 6499 ft
Answer:
1(x - 2xz - 2xy + 4y²)
Step-by-step explanation:
I notice that no term all of them have. which means itbwill be factorised via 1.
1(x - 2xz - 2xy + 4y²)
Answer: 22.5 minutes
Step-by-step explanation: On Monday Georgia practiced for 3/4 of an hour and on Tuesday she practiced for 3/8 of an hour. The difference between those two is 3/8 of an hour and when you take 3/8 of 60 minutes you get 22.5 minutes
