2x + 3y = -6 (Subtract 2x from both sides)
3y = -2x - 6 (Divide both sides by 3 to isolate x) which will give you y = -2/3x - 2 and this graphed would cross the x-axis at (-3,0) and the y-axis at (0, -2). So a + b = -5
Answer:
Dh/dt = 0.082 ft/min
Step-by-step explanation:
As a perpendicular cross section of the trough is in the shape of an isosceles triangle the trough has a circular cone shape wit base of 1 feet and height h = 2 feet.
The volume of a circular cone is:
V(c) = 1/3 * π*r²*h
Then differentiating on both sides of the equation we get:
DV(c)/dt = 1/3* π*r² * Dh/dt (1)
We know that DV(c) / dt is 1 ft³ / 5 min or 1/5 ft³/min
and we are were asked how fast is the water rising when the water is 1/2 foot deep. We need to know what is the value of r at that moment
By proportion we know
r/h ( at the top of the cone 0,5/ 2) is equal to r/0.5 when water is 1/2 foot deep
Then r/h = 0,5/2 = r/0.5
r = (0,5)*( 0.5) / 2 ⇒ r = 0,125 ft
Then in equation (1) we got
(1/5) / 1/3* π*r² = Dh/dt
Dh/dt = 1/ 5*0.01635
Dh/dt = 0.082 ft/min
Answer:
diameter=18 in
radius=9 in
area=254.47 in
circumference=56.55 in
Step-by-step explanation:
<u>Key</u>
C=Circumference
D=Diameter
R=Radius
A=Area
π=pi=3.14159265359
<u>Formulas</u>
2r=d
d/2=r
πr^2=a
(2πr)=c
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