Answer:
(5,18) & (-5 , 38)
Step-by-step explanation:
<h3>Solutions of linear and quadratic system of equations:</h3>
A linear equation represented by a line in a graph intersect the quadratic equation represented by a parabola zero, one or two times.
f(x) = x² - 2x + 3 -------------------(I)
f(x) = -2x + 28 ---------------------(II)
x² - 2x + 3 = -2x + 28
x² - 2x + 2x + 3 - 28 = 0
x² - 25 = 0
x² = 25
x = √25 = √5*5
x = ± 5
Plugin x = 5 in equation (II),
f(5) = -2*5 + 28
= -10 + 28
= 18
(5 , 18)
Plugin x = -5 in equation (II),
f(-5) = -2*(-5) + 28
= 10 + 28
= 38
(-5, 38)
1.55 i did that math too it was kinda hard but i did it
Answer:
![\sqrt{\left(a+b\right)^{2}+c^{2}}](https://tex.z-dn.net/?f=%5Csqrt%7B%5Cleft%28a%2Bb%5Cright%29%5E%7B2%7D%2Bc%5E%7B2%7D%7D)
Step-by-step explanation:
The distance between the line joining points A (<em>-b, c</em>) and C (<em>a, o</em>) will be the length of the diagonal of given trapezoid.
Now, ![x_{1} = - b,\ y_{1} = c\ and\ x_{2} = a, y_{2} = 0](https://tex.z-dn.net/?f=x_%7B1%7D%20%3D%20-%20b%2C%5C%20y_%7B1%7D%20%3D%20c%5C%20and%5C%20x_%7B2%7D%20%3D%20a%2C%20y_%7B2%7D%20%3D%200)
Now, using distance formula,
AC = ![\sqrt{\left (x_{2}-x_{1}\right )^{2}+\left (y_{2}-y_{1}\right )^{2}}](https://tex.z-dn.net/?f=%5Csqrt%7B%5Cleft%20%28x_%7B2%7D-x_%7B1%7D%5Cright%20%29%5E%7B2%7D%2B%5Cleft%20%28y_%7B2%7D-y_%7B1%7D%5Cright%20%29%5E%7B2%7D%7D)
AC = ![\sqrt{\left(a-(-b) \right)^{2}+\left(0-c\right)^{2}}](https://tex.z-dn.net/?f=%5Csqrt%7B%5Cleft%28a-%28-b%29%20%5Cright%29%5E%7B2%7D%2B%5Cleft%280-c%5Cright%29%5E%7B2%7D%7D)
AC = ![\sqrt{\left(a+b\right)^{2}+\left(-c\right)^{2}}](https://tex.z-dn.net/?f=%5Csqrt%7B%5Cleft%28a%2Bb%5Cright%29%5E%7B2%7D%2B%5Cleft%28-c%5Cright%29%5E%7B2%7D%7D)
AC = ![\sqrt{\left(a+b\right)^{2}+c^{2}}](https://tex.z-dn.net/?f=%5Csqrt%7B%5Cleft%28a%2Bb%5Cright%29%5E%7B2%7D%2Bc%5E%7B2%7D%7D)
Therefore, the length of diagonal AC of trapezoid = ![\sqrt{\left(a+b\right)^{2}+c^{2}}](https://tex.z-dn.net/?f=%5Csqrt%7B%5Cleft%28a%2Bb%5Cright%29%5E%7B2%7D%2Bc%5E%7B2%7D%7D)
So, option (4) is the right option.
The answer to this question is 650m.
For a regular n-agon with side length "s" and apothem "h", the area will be
.. A = n*(1/2)*s*h
Your area is
.. A = 10*(1/2)*(3.25 m)*(5 m) ≈ 81.3 m^2