Did I do this equation correctly?

1. Mr. McClellan has a board game that requires the use of two 4-sided dice. According to his rules, you can only start the game

stealth61 [152]

(A)

1,4
1,3
1,2
1,1

2,1
2,2
2,3
2,4

3,1
3,2
3,3
3,4

4,1
4,2
4,3
4,4

5,1
5,2
5,3
5,4

6,4
6,3
6,2
6,1

(B)
3%

(C)
I'm not sure about the answer for C, Sorry. Hope this helps!

6 0

2 years ago

Height

dem82 [27]

Answer:

Step-by-step explanation:

Given the height of the discus can be modeled by the equation y=−16x ^2 +32x+4, where y represents the height in feet of the discus in seconds, the velocity of the discus at its maximum height is zero.

Velocity v = dy/dx = 0

dy/dx = -32x + 32

0 = -32x + 32

32x = 32

x = 1 secs

Hence it will take the discus 1 secs to reach its maximum height

8 0

3 years ago

What is the reciprocal of 3/4

Alex73 [517]

The reciprocal of 3/4 is 4/3

5 0

3 years ago

Read 2 more answers

Find the distance between the points given. (-3, -6) and (3, 2)

vlada-n [284]

Answer:

10 units

Step-by-step explanation:

Moving from (-3, -6) to (3, 2), x increases by 6. Draw a triangle and use this 6 as the length of the base. y increases by 8. Label the height of the triangle with this 8. Then find the hypotenuse (which is also the desired distance) by using the Pythagorean Theorem:

6^2 + 8^2 = d^2, where d is that distance:

36 + 64 = 100 = d^2, and so d = 10. The distance in question is 10 units.

5 0

3 years ago

A tank initially has 300 gallons of a solution that contains 50 lb. of dissolved salt. A brine solution with a concentration of

belka [17]

Let s(t) be the amount of salt in the tank at time t. Then s(0) = 50 lb.

Salt flows into the tank at a rate of

(2 gal/min) (6 lb/gal) = 12 lb/min

and flows out at a rate of

(2 gal/min) (s(t)/300 lb/gal) = s(t)/150 lb/min

so that the net rate at which salt is exchanged through the tank is

ds(t)/dt = 12 - s(t)/150 … (lb/min)

Solve for s(t). The DE is separable, so we have

ds/dt = 12 - s/150

150 ds/dt = 1800 - s

150/(1800 - s) ds = dt

Integrate both sides to get

-150 ln|1800 - s| = t + C

Solve for s :

ln|1800 - s| = -t/150 + C

1800 - s = exp(-t/150 + C )

1800 - s = C exp(-t/150)

s = 1800 - C exp(-t/150)

Now given that s(0) = 50, we solve for C :

50 = 1800 - C exp(-0/150)

50 = 1800 - C

C = 1750

Then the amount of salt in the tank at any time t ≥ 0 is

s(t) = 1800 - 1750 exp(-t/150)

To find the time it takes for the tank to hold 100 lb of salt, solve for t in

100 = 1800 - 1750 exp(-t/150)

1700 = 1750 exp(-t/150)

34/35 = exp(-t/150)

ln(34/35) = -t/150

t = -150 ln(34/35) ≈ 4.348

So it would take approximately 4.348 minutes.

By the way, we didn't have to solve for s(t), we could have instead stopped with

-150 ln|1800 - s| = t + C

Solve for C - this C is not the same as the one we found using the other method. s(0) = 50, so

-150 ln|1800 - 50| = 0 + C

C = -150 ln|1750|

==> t = 150 ln(1750) - 150 ln|1800 - s|

Then s(t) = 100 lb when

t = 150 ln(1750) - 150 ln(1700)

t = 150 ln(1750/1700)

t = 150 ln(35/34)

t ≈ 4.348

5 0

3 years ago

Did I do this equation correctly?​

Did I do this equation correctly?