78
Years ago he had a lot to say he had a great day he was the first man I was ever going for the first one I was never gonna get a
Answer:
90t-21t=69t
Step-by-step explanation:
4:00pm to 6:00 am is 14 hours. 14 x 1.5 is in fact 21 which means the temperature dropped 21 degrees in those 14 hours.
First combine like terms ,
5/5x=9/5x
move the variable which will cancel out because it is the same
5=9 is no solution sorry if i’m wrong lol
Ok, I'm going to start off saying there is probably an easier way of doing this that's right in front of my face, but I can't see it so I'm going to use Heron's formula, which is A=√[s(s-a)(s-b)(s-c)] where A is the area, s is the semiperimeter (half of the perimeter), and a, b, and c are the side lengths.
Substitute the known values into the formula:
x√10=√{[(x+x+1+2x-1)/2][({x+x+1+2x-1}/2)-x][({x+x+1+2x-1}/2)-(x+1)][({x+x+1+2x-1}/2)-(2x-1)]}
Simplify:
<span>x√10=√{[4x/2][(4x/2)-x][(4x/2)-(x+1)][(4x/2)-(2x-1)]}</span>
<span>x√10=√[2x(2x-x)(2x-x-1)(2x-2x+1)]</span>
<span>x√10=√[2x(x)(x-1)(1)]</span>
<span>x√10=√[2x²(x-1)]</span>
<span>x√10=√(2x³-2x²)</span>
<span>10x²=2x³-2x²</span>
<span>2x³-12x²=0</span>
<span>2x²(x-6)=0</span>
<span>2x²=0 or x-6=0</span>
<span>x=0 or x=6</span>
<span>Therefore, x=6 (you can't have a length of 0).</span>
This problem is better understood with a given figure. Assuming
that the flight is in a perfect northwest direction such that the angle is 45°,
therefore I believe I have the correct figure to simulate the situation (see
attached).
Now we are asked to find for the value of the hypotenuse
(flight speed) given the angle and the side opposite to the angle. In this
case, we use the sin function:
sin θ = opposite side / hypotenuse
sin 45 = 68 miles per hr / flight
flight = 68 miles per hr / sin 45
<span>flight = 96.17 miles per hr</span>
