Answer:

Step-by-step explanation:
The parabola's are


So



So, the points at which the parabola's intersect each other is
and 
The three points of the triangle are
,
and 
Area of a triangle is given by
![A=\dfrac{1}{2}[x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2)]\\\Rightarrow A=\dfrac{1}{2}[2(5-0)+(-2)(0-5)+0(5-5)]\\\Rightarrow A=10\ \text{sq. units}](https://tex.z-dn.net/?f=A%3D%5Cdfrac%7B1%7D%7B2%7D%5Bx_1%28y_2-y_3%29%20%2B%20x_2%28y_3-y_1%29%20%2B%20x_3%28y_1-y_2%29%5D%5C%5C%5CRightarrow%20A%3D%5Cdfrac%7B1%7D%7B2%7D%5B2%285-0%29%2B%28-2%29%280-5%29%2B0%285-5%29%5D%5C%5C%5CRightarrow%20A%3D10%5C%20%5Ctext%7Bsq.%20units%7D)
Area of the triangle formed is
.
Answer:
Step-by-step explanation:
1.
Dependent variable: a
Independent variable: c
2.
(in this order)
x = 9
x = 15
x = 4
x = 36 (assuming that says x = 12 * 3)
combine like terms, so 6m + 5m =11m
then 7n - 3n = 4n
ANSWER: 11m+4n
The function increases in the interval (-∞, -3) and the function also increases in the interval (-1,∞) .
The given function is of the form
f(x) = x³ + 6x² + 8x
Now we take the first differentiation of the function
f'(x) = 2x² + 12x + 8
f'(x) = 2 (x² + 6x + 9) -10
f'(x) = 2(x+3)² - 10
Therefore at x = -3 , f'(x) = -10.
Hence the function is increasing in the interval of (-∞, -3)
Again f'(x) = 2x² + 12x + 8 , so after first differentiation we get :
That the function is also increasing in the interval (-1,∞)
Now for the interval (-4,-2), we can say that the graph of the function is positive as the y value increases and then decreases but all y values are positive as illustrated in the graph.
In the interval (0,∞) the function is strictly increasing and has positive values only.
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