I'm not sure if this is the easiest way of doing this, but it surely work.
Let the base of the triangle be AB, and let CH be the height. Just for reference, we have

Moreover, let CH=y and BC=z
Now, AHC, CHB and ABC are all right triangles. If we write the pythagorean theorem for each of them, we have the following system

If we solve the first two equations for y squared, we have

And we can deduce

So that the third equation becomes

(we can't accept the negative root because negative lengths make no sense)
1) the polynomial representing the area is
2x^4 + 6x^3 - 5x^2 -7x + 24
2) the constant term is
24
3) the polynomial is a 4th degree
4) if my guess is ryt... the LEADING coefficient is the coefficient of the highest degree of X ...i.e 2
Answer:
The answer to your question is below
Step-by-step explanation:
Formula
Triangle's Area = 1/2 bh/2
Trapezoid's area = 1/2 (b1 + b2)h
Parallelogram's area = bh
Rectangle's area = bh
Considering the formulas, we can conclude that:
a) The first choice is true, both formulas have 1/2 in.
b) The second choice is also true, both equations are the same
c) The third choice is incorrect
d) This choice is correct, the bases are added,
e) This choice is incorrect, the sides are not added.
It is -9. Rise over run. 9 up, 1 to the left, 9/1 = 9. Because the line goes down to the right, it is negative.
Please put me as the brainiest.
Answer:
Non-linear
Step-by-step explanation:
Here, we want to determine the function type that would model the height of a football from the time it was kicked to the time it lands on the ground
The way to go about this is to notice the kind of shape or trajectory the kicked football will take.
The shape it will take is non-linear. The shape it will take would be a parabolic curve. These parabolic curve models a quadratic equation and not a linear function. Hence we can conclude that based on the kind of path in which the ball traces, we can conclude that the function that results from its trajectory would not be a linear kind of equation or representation.