Check the picture below.
a)
so the perimeter will include "part" of the circumference of the green circle, and it will include "part" of the red encircled section, plus the endpoints where the pathway ends.
the endpoints, are just 2 meters long, as you can see 2+15+2 is 19, or the radius of the "outer radius".
let's find the circumference of the green circle, and then subtract the arc of that sector that's not part of the perimeter.
and then let's get the circumference of the red encircled section, and also subtract the arc of that sector, and then we add the endpoints and that's the perimeter.
b)
we do about the same here as well, we get the full area of the red encircled area, and then subtract the sector with 135°, and then subtract the sector of the green circle that is 360° - 135°, or 225°, the part that wasn't included in the previous subtraction.
C
A million apologies if I’m incorrect
Answer:
- time: t = -0.3
- minimum: v = 0.55
Step-by-step explanation:
For quadratic ax^2 + bx + c, the extreme value is found at x=-b/(2a). For your quadratic, the minimum is found at ...
t = -(3)/(2(5))
t = -0.3 . . . . . time of minimum velocity
__
The value of velocity at that time is ...
v = 5(-0.3)^2 +3(-0.3) +1 = 5(.09) -.9 +1
v = 0.55 . . . . . value of minimum velocity
Answer:
B
Step-by-step explanation:
B'(7, -4)
Let the pre-image coordinates be (x,y)
Consider x-coordinate,
x + 4 = 7
x = 3
Consider y-coordinates,
y + 5 = -4
y = -9
pre-image coordinates is (3, -9)
Answer:
um it is 3344434565657554545656757656
Step-by-step explanation:
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