Suppose that a large mixing tank initially holds 700 gallons of water in which 50 pounds of salt have been dissolved. another brine solution is pumped into the tank at a rate of 3 gal/min, and when the solution is well stirred, it is then pumped out at a slower rate of 2 gal/min. if the concentration of the solution entering is 4 lb/gal, determine a differential equation for the amount of salt a(t) in the tank at time t > 0. (use a for a(t).)

A. In this case, they can be arranged at any order. Since there are 7 people all in all, therefore the number of arrangements is:

number of arrangements = 7P7 = 5040 ways

B. In this case, only the 5 people can be arranged in any order therefore 5P5. However the groom and bride can be interchanged on 2 places, therefore:

number of arrangements = 5P5 * 2 = 240 ways

C. In this case, only the 5 people can be arranged in any order while the groom and bride can no longer be interchanged, therefore:

number of arrangements = 5P5 = 120 ways

Final answer:

The arrangements for the wedding reception receiving line can be calculated using permutations: a) Any order will do, giving 7!, b) The bride and groom must be the last two in line, giving 5!, c) The groom must be last in line with the bride next to him, giving 6! x 2.

Explanation:

The question asks about the number of ways a bride and groom along with five attendants can be arranged in a receiving line at a wedding reception under different conditions. This is a permutations problem where we calculate the arrangements using factorial notation.

Case a: Any order will do.

In this case, we have a total of 7 people (bride, groom, and five attendants) who can be arranged in any order. The number of arrangements is the factorial of 7, which is calculated as 7! (7 factorial). So, the number of ways they can be arranged is 7! (which is 7 x 6 x 5 x 4 x 3 x 2 x 1 = 5,040 ways).

Case b: The bride and groom must be the last two in line.

Since the bride and groom's positions are fixed at the end of the line, we only need to arrange the five attendants. This is equivalent to 5! arrangements for the attendants. Thus, the number of ways they can be arranged is 5! (which is 5 x 4 x 3 x 2 x 1 = 120 ways).

Case c: The groom must be last in line with the bride next to him.

Here, the bride and groom are fixed in the last two positions, but the bride must be next to the groom. We consider the bride and groom as a single unit in addition to the five attendants. This gives us 6! arrangements for the attendants and the bride-groom unit, multiplied by 2 because the bride and groom can switch places within their unit. So, the number of ways they can be arranged is 6! x 2 (which is 6 x 5 x 4 x 3 x 2 x 1 x 2 = 1,440 ways).

Answer: the answer to this question is

choice c. 17

Final answer:

To find the markup on a pair of hiking boots with a cost of $80 and a 15% markup rate, multiply the cost by 0.15 to get a $12 markup. The selling price would then be $80 + $12, totaling $92.

Explanation:

To calculate the markup for a pair of hiking boots that the store buys for $80 with a markup rate of 15%, first convert the percentage to a decimal by dividing by 100.

Markup rate as a decimal: 0.15

Next, multiply the cost price by the decimal.

Markup amount = Cost price × Markup rate

Markup amount = $80 × 0.15

Markup amount = $12

So, the markup on the hiking boots is $12. To find the selling price, you add the markup to the cost price:

Selling price = Cost price + Markup amount

Selling price = $80 + $12

Selling price = $92

The store would sell the hiking boots for $92.

Answer:

The bird is at a height of 1.95 km from the ground.

Step-by-step explanation:

It is given that the distance between susan and mark is 7 km and bird is between  susan and mark. The angles of elevation they make are 20º and 50º, respectively.

Draw a perpendicular line from the bird on base.

Let the distance of susan from and the altitude be x.

In triangle ABS,

[tex]tan(20^{\circ})=\frac{AB}{x}[/tex]

[tex]tan(20^{\circ})x=AB[/tex]

[tex]0.364x=AB[/tex]       ..... (1)

In triangle ABM,

[tex]tan(50^{\circ})=\frac{AB}{7-x}[/tex]

[tex]1.192(7-x)=AB[/tex]     ..... (2)

From (1) and (2), we get

[tex]0.364x=1.192(7-x)[/tex]

[tex]1.556x=8342[/tex]

[tex]x=5.36[/tex]

The length of AB is,

[tex]AB=0.364(5.36)=1.95147\approx 1.95[/tex]

Therefore, the he bird is at a height of 1.95 km from the ground.

This is the answer. 200 + 90 + 0

The place value chart can help us to write a number in expanded notation. When we put 290 into the place value chart, we can recognize that our number is equal to

2 hundreds + 9 tens + 0 units

Image is provided.

Little Lake Rock is 1811.5 feet long, and Lake Big Rock is 3461.5 feet long.

Let's use algebra to solve this problem.

Let L be the length of Little Lake Rock in feet, and let B be the length of Lake Big Rock in feet.

We are given two pieces of information:

Lake Big Rock is 1650 ft longer than Little Lake Rock: B = L + 1650.

Together, they are 5273 ft long: B + L = 5273.

Now, we have a system of two equations with two variables:

B = L + 1650

B + L = 5273

We can use the first equation to express B in terms of L and then substitute this expression into the second equation:

B = L + 1650

Now, substitute B in the second equation:

(L + 1650) + L = 5273

Combine like terms:

2L + 1650 = 5273

Subtract 1650 from both sides:

2L = 5273 - 1650

2L = 3623

Now, divide by 2 to find the length of Little Lake Rock (L):

L = 3623 / 2

L = 1811.5 feet

Now that we know the length of Little Lake Rock, we can find the length of Lake Big Rock using the first equation:

B = L + 1650

B = 1811.5 + 1650

B = 3461.5 feet

So, Little Lake Rock is 1811.5 feet long, and Lake Big Rock is 3461.5 feet long.

for such more question on length

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Answer: 3

Step-by-step explanation:

Given : Juanita is making necklaces to give as presents.

She plans to put 15 beads on each necklace. Beads are sold in packages of 20.

To find the least common number of packages she can buy to make necklaces and have no beads left over,  First we need to find the least common multiple of 15 and 20.

Since , [tex]15=3\times5\\\\20=2\times2\times5[/tex]

Then least common multiple of 15 and 20 =[tex]2\times2\times3\times5=60[/tex]

Now, the least common number of packages she can buy to make necklaces and have no beads left over =[tex]\dfrac{60}{20}=3[/tex].

Hence, the least common number of packages she can buy to make necklaces and have no beads left over=3

The answer is B. 40/1 dollars per person; the cost is $40 for each person.

Answer:

The answer is 10,560 yards and 6 miles. The 5,280 yards is just from her home to the school. So you multiply it by 2 and divide by 1760.

Step-by-step explanation: