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r-ruslan [8.4K]
3 years ago
6

At a wedding reception an equal number of guests were seated at each of three large tables, and 7 late arriving desks were seate

d at a smaller table. There were 37 guests in all. If n represents the number of guests seated at each of the large tables, what equation represents the situation?
Mathematics
1 answer:
nikitadnepr [17]3 years ago
8 0

Answer:

10 guests a table

Step-by-step explanation:

3 tables

7 late arrivals

37 total people

37 minus 7 is 30

30 divided by 3 is 10

answer is 10

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7 + 7x; 7(x+1)<br><br> Are these equivalent? <br> TRUE or FALSE?
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I NEED HELP WITH THIS AND U GOTTA EXPLAIN HOW U GOT THE ANSWER q2
Fantom [35]

Answer:

27 degrees

Step-by-step explanation:

5t - 13 = 3t + 3

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Can you help me find t value in this problem?
Eduardwww [97]

we have a maximum at t = 0, where the maximum is y = 30.

We have a minimum at t = -1 and t = 1, where the minimum is y = 20.

<h3>How to find the maximums and minimums?</h3>

These are given by the zeros of the first derivation.

In this case, the function is:

w(t) = 10t^4 - 20t^2 + 30.

The first derivation is:

w'(t) = 4*10t^3 - 2*20t

w'(t) = 40t^3 - 40t

The zeros are:

0 = 40t^3 - 40t

We can rewrite this as:

0 = t*(40t^2 - 40)

So one zero is at t = 0, the other two are given by:

0 = 40t^2 - 40

40/40 = t^2

±√1 = ±1 = t

So we have 3 roots:

t = -1, 0, 1

We can just evaluate the function in these 3 values to see which ones are maximums and minimums.

w(-1) = 10*(-1)^4 - 20*(-1)^2 + 30 = 10 - 20 + 30 = 20

w(0) = 10*0^4 - 20*0^2 + 30    = 30

w(1) =  10*(1)^4 - 20*(1)^2 + 30 =  20

So we have a maximum at x = 0, where the maximum is y = 30.

We have a minimum at x = -1 and x = 1, where the minimum is y = 20.

If you want to learn more about maximization, you can read:

brainly.com/question/19819849

6 0
2 years ago
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