I believe it is 1, but i'm not completely sure.
Shorter side: 130ft
longer side: 260ft
greatest possible area: 33800ft^2
basically get the equations for area and perimeter which will be
A: lw (length x width)
P: l + 2w
Substitute them into eachother and you get the equation:
-2w^2+520w = A
find the vertex of this parabola which will give you the greatest width which is 130m
Then u can find length and area from this width
4,56 км / ч просто разделят ускорението и точката на среща.
Add 2p² to each side of the equation. Then you have
2p² + 16p + 24 = 0 .
Before you roll up your sleeves and start working on it, you can make it
even more convenient if you divide each side by 2 . Then you have:
p² + 8p + 12 = 0 .
Now you have a nice, comfortable, familiar-looking quadratic equation.
You can either factor the left side into (p + 6) (p + 2), or, if you can't find
the factors, you can apply the quadratic formula to it.
That's how to solve it, and find its two solutions.
If tan theta is -1, we know immediately that theta is in either Quadrant II or Q IV. We need to focus on Q IV due to the restrictions on theta.
Because tan theta is -1, the ray representing theta makes a 45 degree angle with the horiz axis, and a 45 degree angle with the negative vert. axis. Thus the hypotenuse, by the Pythagorean Theorem, tells us that the hyp is sqrt(2).
Thus, the cosine of theta is adj / hyp, or +1 / sqrt(2), or [sqrt(2)]/2
The secant of theta is the reciprocal of that, and thus is
2 sqrt(2)
---------- * ------------ = sqrt(2) (answer)
sqrt(2) sqrt(2)
