The equation of a line is y = mx + b where m is the slope and b is the y-intercept (the point where the line crosses the y-axis). we can look at the graph and see that the line crosses the y axis at y = 4. Therefore, the y-intercept, b, is 4.
Now we must calculate the slope. To do so, we divide the "rise" of the line,
y₂ - y₁ by the "run" of the line x₂ - x₁. We must choose two points (x₁, y₁) and (x₂, y₂) to find the slope. Two points are already given to us, (1, 2) and (4, -4):
m = (-4 - 2)/(4 - 1)
m = -6/3
m = -2
Now that we know the slope and the y-intercept, we can simply plug them back into the equation to find the equation of the line:
y = -2x + 4
Z=20, y=4,x=0.8
have a good day
Answer:
2346
Step-by-step explanation:
4263x462x346x634x62=3
∆BOC is equilateral, since both OC and OB are radii of the circle with length 4 cm. Then the angle subtended by the minor arc BC has measure 60°. (Note that OA is also a radius.) AB is a diameter of the circle, so the arc AB subtends an angle measuring 180°. This means the minor arc AC measures 120°.
Since ∆BOC is equilateral, its area is √3/4 (4 cm)² = 4√3 cm². The area of the sector containing ∆BOC is 60/360 = 1/6 the total area of the circle, or π/6 (4 cm)² = 8π/3 cm². Then the area of the shaded segment adjacent to ∆BOC is (8π/3 - 4√3) cm².
∆AOC is isosceles, with vertex angle measuring 120°, so the other two angles measure (180° - 120°)/2 = 30°. Using trigonometry, we find
![\sin(30^\circ) = \dfrac{h}{4\,\rm cm} \implies h= 2\,\rm cm](https://tex.z-dn.net/?f=%5Csin%2830%5E%5Ccirc%29%20%3D%20%5Cdfrac%7Bh%7D%7B4%5C%2C%5Crm%20cm%7D%20%5Cimplies%20h%3D%202%5C%2C%5Crm%20cm)
where
is the length of the altitude originating from vertex O, and so
![\left(\dfrac b2\right)^2 + h^2 = (4\,\mathrm{cm})^2 \implies b = 4\sqrt3 \,\rm cm](https://tex.z-dn.net/?f=%5Cleft%28%5Cdfrac%20b2%5Cright%29%5E2%20%2B%20h%5E2%20%3D%20%284%5C%2C%5Cmathrm%7Bcm%7D%29%5E2%20%5Cimplies%20b%20%3D%204%5Csqrt3%20%5C%2C%5Crm%20cm)
where
is the length of the base AC. Hence the area of ∆AOC is 1/2 (2 cm) (4√3 cm) = 4√3 cm². The area of the sector containing ∆AOC is 120/360 = 1/3 of the total area of the circle, or π/3 (4 cm)² = 16π/3 cm². Then the area of the other shaded segment is (16π/3 - 4√3) cm².
So, the total area of the shaded region is
(8π/3 - 4√3) + (16π/3 - 4√3) = (8π - 8√3) cm²
Answer:
78.4pi or 246.30
Step-by-step explanation:
144/360=2/5
EFG=2/5 area of circle
Area of circle=14^2*pi
Area=196pi
78.4pi
I think that it is 246.30 to the nearest hundredth