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lozanna [386]
3 years ago
5

The Symphony

Mathematics
1 answer:
lord [1]3 years ago
5 0
Answer is 12.50. u can comment on my answer if you would like to see how that is the correct answer :)
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1/2 divided by n = 1/8
Leto [7]
1/2 divide by 4/1 = 1/8
when u divide u flip the second fraction around and multiply
1/2 x 1/4 = 1/8
3 0
3 years ago
What is the value of the function at x=−2?
adell [148]

x = -2

function value at x = -2 is intersection of the x value and graph

which is, 3

Hope it is clear:)

5 0
3 years ago
Find the area of the triangle.
Maslowich

Answer:

Area of triangle = 1/2*b*h

\frac{1}{2} * 2.8* 7.8\\\frac{1}{2} * 21.84\\\frac{21.84}{2}\\= 10.92

4 0
3 years ago
Read 2 more answers
a teacher and 10 students are to be seated along a bench in the bleachers at a basketball game. In how many ways can this be don
Veronika [31]

Wow !

OK.  The line-up on the bench has two "zones" ...

-- One zone, consisting of exactly two people, the teacher and the difficult student.
   Their identities don't change, and their arrangement doesn't change.

-- The other zone, consisting of the other 9 students.
   They can line up in any possible way.

How many ways can you line up 9 students ?

The first one can be any one of 9.   For each of these . . .
The second one can be any one of the remaining 8.  For each of these . . .
The third one can be any one of the remaining 7.  For each of these . . .
The fourth one can be any one of the remaining 6.  For each of these . . .
The fifth one can be any one of the remaining 5.  For each of these . . .
The sixth one can be any one of the remaining 4.  For each of these . . .
The seventh one can be any one of the remaining 3.  For each of these . . .
The eighth one can be either of the remaining 2.  For each of these . . .
The ninth one must be the only one remaining student.

     The total number of possible line-ups is 

               (9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1)  =  9!  =  362,880 .

But wait !  We're not done yet !

For each possible line-up, the teacher and the difficult student can sit

-- On the left end,
-- Between the 1st and 2nd students in the lineup,
-- Between the 2nd and 3rd students in the lineup,
-- Between the 3rd and 4th students in the lineup,
-- Between the 4th and 5th students in the lineup,
-- Between the 5th and 6th students in the lineup,
-- Between the 6th and 7th students in the lineup,
-- Between the 7th and 8th students in the lineup,
-- Between the 8th and 9th students in the lineup,
-- On the right end.

That's 10 different places to put the teacher and the difficult student,
in EACH possible line-up of the other 9 .

So the total total number of ways to do this is

           (362,880) x (10)  =  3,628,800  ways.

If they sit a different way at every game, the class can see a bunch of games
without duplicating their seating arrangement !

4 0
3 years ago
What is x? Please help fast
maks197457 [2]

Answer:

72

Step-by-step explanation:

180-108

5 0
3 years ago
Read 2 more answers
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