Answer:
B.x>4is a required interval.
Answer:
+4.245
Step-by-step explanation:
Before you can find the square root of x^2, you have to EVALUATE
64+49-2(8)(7)cos32, and THEN take the square root of the result:
64+49-2(8)(7)cos32 = 64 + 49 - 112(0.848) = 113 - 94.981 = 18.019
Then x^2 = 18.019.
Find the square root of 18.019: It is +4.245. A negative result here would make no sense.
Solve for the first variable in one of the equations, then substitute the result into the other equation.
Point Form:
(
3
5
,
4
5
)
,
(
−
7
25
,
−
24
25
)
(
3
5
,
4
5
)
,
(
-
7
25
,
-
24
25
)
Equation Form:
x
=
3
5
,
y
=
4
5
x
=
3
5
,
y
=
4
5
x
=
−
7
25
,
y
=
−
24
25
Answer:
V = ∫₀⁶⁰vdt/36 ft³
Step-by-step explanation:
The rate of change of the volume of the tank V is given by dV/dt. Since the volume of the tank is V = Ah where A = cross-sectional area of the square hole and h = height of tank.
Now dV/dt = dV/dh × dh/dt where dh/dt = v = velocity of water flowing through the tank = rate at which the height is changing.
dV/dt = dV/dh × dh/dt
dV/dt = dAh/dh × dh/dt
dV/dt = A × v
dV/dt = Av
Now the area of the dimension of the hole is 2 inches in feet is 2/12 feet. The area of the hole is thus A = (2/12)² = 1/6² = 1/36 ft².
So, dV/dt = v/36
dV = vdt/36
So, for the total volume lost in the first minute, we integrate t from t = 0 to t = 60 s.
V = ∫₀⁶⁰dV
V = ∫₀⁶⁰vdt/36 ft³
So, the definite integral that represents the total amount of water (in cubic feet) lost in the first minute is V = ∫₀⁶⁰vdt/36 ft³
