Answer:
l = 2.25 cm
Step-by-step explanation:
given l is inversely proportional to w² then the equation relating them is
l = 
← k is the constant of proportion
(i)
to find k use the condition w = 1.5 , l = 16 , then
16 = 
= 
( multiply both sides by 2.25 )
36 = k
l = 
← equation of proportion
(ii)
when w = 4 , then
l = 
= 
= 2.25 cm
Step-by-step explanation:
We want to find two things-- the speed of the boat in still water and the speed of the current. Each of these things will be represented by a different variable:
B = speed of the boat in still water
C = speed of the current
Since we have two variables, we will need to find a system of two equations to solve.
How do we find the two equations we need?
Rate problems are based on the relationship Distance = (Rate)(Time).
Fill in the chart with your data (chart attached)
The resulting speed of the boat (traveling upstream) is B-C miles per hour. On the other hand, if the boat is traveling downstream, the current will be pushing the boat faster, and the boat's speed will increase by C miles per hour. The resulting speed of the boat (traveling downstream) is B+C miles per hour. Put this info in the second column in the chart. Now plug it into a formula! <u>Distance=(Rate)(Time) </u>Now solve using the systems of equations!
(3x19.95+23.5+124.95)1.05=
59.85+23.5+124.95 =208.3
208.3×1.05=218.715
Answer: 277 ways
Step-by-step explanation:
Let’s start bycreating 10-unit pieces using the 4-unit piece.
The arrangements are:
1). 4-3-3 (3 permutations)
2). 4-3-2-1 = 4! = 24 permutations.
3). 4-3-1-1-1 (5*[4!/(3!1!)]
= 5*4
= 20permutations
4). 4-2-2-2 (4 permutations)
5). 4-2-2-1-1 (5 *[4!/(2!2!)]
= 5*6
= 30 permutations
6). 4-2-1-1-1-1(6*[5!/(4!1!)]
= 6*5
= 30 permutations
Let’s-consider the arrangements using one or more3-unit pieces and no 4-unit piece:
7). 3-3-2-2 (4!/(2!2!)
8). 3-3-2-1-1 (5*(4!/(2!2!)
= 5*6
= 30 permutations.
9). 3-3-1-1-1-1 (6!/(4!2!) = 0
10). 3-2-2-2-1 (5*4!/(3!1!)
= 5*4
= 20permutations
11). 3-2-2-1-1-1 (6*5!/(3!2!)
= 6*10
= 60 permutations
Finally we would look at the arrangements using only 1-and 2-unit pieces:
12). 2-2-2-1-1-1-1 (7!/(4!3!)
= 35 permutations
add them all up:
(3 + 24 + 20 + 4) + (30 + 30 + 6 + 30) + (15 + 20 + 60 + 35)
=51 + 96 + 130
= 277ways
Answer:
92 attendees had activity cards
Step-by-step explanation:
Let x be the number of students with activity cards. Then 130-x is the number without, and the total revenue is ...
7x +10(130 -x) = 1024
7x +1300 -10x = 1024 . . . . eliminate parentheses
-3x = -276 . . . . . . . . . . . . . collect terms; subtract 1300
x = 92 . . . . . . divide by 3
92 students with activity cards attended the dance.
_____
<em>Comment on the solution</em>
Often, you will see such a problem solved using two equations. For example, they might be ...
Let 'a' represent the number with an activity card; 'w' the number without. Then ...
- a+w = 130 . . . . the total number of students
- 7a +10w = 1024 . . . . the revenue from ticket sales
The problem statement asks for the value of 'a', so you want to eliminate w from these equations. You can do that using substitution. Using the first equation to write an expression for w, you have ...
w = 130-a
and making the substitution into the second equation gives ...
7a +10(130 -a) = 1024
This should look a lot like the equation we used above. There, we skipped the extra variable and went straight to the single equation we needed to solve.
