The line y = x and y = -x + 4 intersect when at the point (2, 2).
Expresing y = -x + 4 in terms of x, we have x = 4 - y.
Thus, the area of the region bounded by the <span>graphs of y = x, y = −x + 4, and y = 0 is given by
![\int\limits^2_0 {(y-(4-y))} \, dy = \int\limits^2_0 {(y-4+y)} \, dy \\ \\ = \int\limits^2_0 {(2y-4)} \, dy= \left[y^2-4y\right]_0^2 =|(2)^2-4(2)| \\ \\ =|4-8|=|-4|=4](https://tex.z-dn.net/?f=%20%5Cint%5Climits%5E2_0%20%7B%28y-%284-y%29%29%7D%20%5C%2C%20dy%20%3D%20%5Cint%5Climits%5E2_0%20%7B%28y-4%2By%29%7D%20%5C%2C%20dy%20%5C%5C%20%20%5C%5C%20%3D%20%5Cint%5Climits%5E2_0%20%7B%282y-4%29%7D%20%5C%2C%20dy%3D%20%5Cleft%5By%5E2-4y%5Cright%5D_0%5E2%20%3D%7C%282%29%5E2-4%282%29%7C%20%5C%5C%20%20%5C%5C%20%3D%7C4-8%7C%3D%7C-4%7C%3D4)
Therefore, the area bounded by the lines is 4 square units.
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<h3>The value of c is c = -12</h3>
This is because 3x^2+6x = 12 is the same as 3x^2+6x-12 = 0
Compare this to the general quadratic form ax^2 + bx + c = 0.
The other values are a = 3 and b = 6
(0.1x + 0.4y)^2 =
(0.1x + 0.4y)(0.1x + 0.4y)
FOIL method
F (first)...multiply the first terms in each set....0.1x * 0.1x = 0.01x^2
O (outside)...multiply outside terms in each set...0.1x * 0.4y = 0.04xy
I (inside)...multiply inside terms in each set....0.4y * 0.1x = 0.04xy
L (last)...multiply last terms in each set...0.4y * 0.4y = 0.16y^2
now we have : 0.01x^2 + 0.04xy + 0.04xy + 0.16y^2....combine like terms
0.01x^2 + 0.08xy + 0.16y^2 <===
2/4 of an hour is half an hour so 2/4 of an hour is 30 minutes. Hope it helps
Answer:
5000 grams
Step-by-step explanation: