Select the two statements that are true about the equation −10=60(−40)

y

-

10

=

60

(

x

-

40

)

The slope of the line is 60  

One point on the line is (40, 10)  

Step-by-step explanation:

The missing choices are :

1) The slope of the line is 10  

2)One point on the line is (40, 10)  

3)The slope of the line is 60  

4)One point on the line is (10, 60)

Given the equation y−10=60(x−40),  applying distributive property we get:

y−10=60x - 60*40

t = 60x - 2400 + 10

t = 60x - 2390

The slope is the number that multiplies x, that is, 60.

To check if the point (40, 10) belongs to the equation, replace the values x = 40 and y = 10 in the equation, as follows:

(10)−10=60((40)−40)

0 = 0

The same number appear at both sides of the equal sign,  then the point (40, 10) is on the line. Analogously, for point (10, 60):

(60)−10=60((10)−40)

50 = 60*(-30)

50 = -1800

The two numbers are different, then, the point (10, 60) is not on the line.

Step-by-step explanation:

a) The rate of change in volume is equal to the volume flow rate going in minus the volume flow rate going out.

W'(t) = F(t) − L(t)

At time t = 3, the slope of the line tangent to W(t) is W'(3).

W'(3) = F(3) − L(3)

W'(3) = arctan(π/2−3/10) − 0.03(20(3)−3²−75)

W'(3) ≈ 1.624

Using point-slope form, the linear approximation is:

y − 2.5 = 1.624 (x − 3)

y − 2.5 = 1.624 x − 4.872

y = 1.624 x − 2.372

The approximate volume at t = 3.5 is:

y = 1.624 (3.5) − 2.372

y = 3.312

b) W'(t) = arctan(π/2−t/10) − 0.03(20t−t²−75)

W"(t) = (-1/10) / (1 + (π/2−t/10)²) − 0.03(20−2t)

W"(8) = (-1/10) / (1 + (π/2−8/10)²) − 0.03(20−2(8))

W"(8) = -0.183

At t = 8 minutes, the flowrate is slowing down at a rate of 0.183 ft³/min².

c) If the rate of change of the volume (W'(t)) changes from positive to negative, then there must be a point where W'(t) = 0.

0 = arctan(π/2−t/10) − 0.03(20t−t²−75)

Solving with a calculator, t = 8.149 or t = 14.627.  Since we're only considering 5 < t < 10, a possible point is t = 8.149.

Evaluate W'(t) before and after t = 8.149:

W'(8) = 0.027

W'(9) = -0.129

So yes, there is a time where the rate of change of the volume changes from positive to negative.

d) The tub is rectangular, so the volume of water is equal to the area of the base times the depth of the water:

W(t) = (0.5) (4) h

W(t) = 2h

Taking derivative:

W'(t) = 2 dh/dt

Evaluate at t = 6:

arctan(π/2−6/10) − 0.03(20(6)−6²−75) = 2 dh/dt

0.501 = 2 dh/dt

dh/dt = 0.250

The depth of the water is increasing at 0.250 ft/min.