To justify the yearly membership, you want to pay at least the same amount as a no-membership purchase, otherwise you would be losing money by purchasing a yearly membership. So set the no-membership cost equal to the yearly membership cost and solve.
no-membership costs $2 per day for swimming and $5 per day for aerobic, in other words, $7 per day. So if we let d = number of days, our cost can be calculated by "7d"
a yearly membership costs $200 plus $3 per day, or in other words, "200 + 3d"
Set them equal to each other and solve:
7d = 200 + 3d
4d = 200
d = 50
So you would need to attend the classes for at least 50 days to justify a yearly membership. I hope that helps!
Steps?
A graph shows zeros to be ±3. Factoring those out leaves the quadratic
(x-2)² +1
which has complex roots 2±i.
The function has roots -3, 3, 2-i, 2+i.
-3y = x...so we sub in -3y for x in the other equation
-x + 7y = 70
-(-3y) + 7y = 70
3y + 7y = 70
10y = 70
y = 70/10
y = 7
so we sub in 7 for y in either of the original equations to find x
-3y = x
-3(7) = x
-21 = x
so ur solution is (-21,7)
<h3>
Answer: 31 games</h3>
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Explanation:
- 32/2 = 16 games will happen in round 1. Afterward, 16 teams are left.
- 16/2 = 8 games will happen in round 2. Afterward, 8 teams are left.
- 8/2 = 4 games happen in round 3.
- 4/2 = 2 games in round 4.
- 2/2 = 1 game as the final championship.
Count the number of times you divide by two and we have five occurrences of this. So there five rounds overall.
To get the total number of games played, we add up the quotients
16+8+4+2+1 = 31
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Or as a shortcut we can simply subtract off 1 since 1+2+4+...+2^n = 2^(n+1)-1
We can write that rule as

which is equivalent to

Hello,
Using vectors and scalar product:
![[(a+c)*\vec{i}+(b-0)*\vec{j} ].[(a-c)*\vec{i}+(b-0)*\vec{j}]=0](https://tex.z-dn.net/?f=%5B%28a%2Bc%29%2A%5Cvec%7Bi%7D%2B%28b-0%29%2A%5Cvec%7Bj%7D%20%5D.%5B%28a-c%29%2A%5Cvec%7Bi%7D%2B%28b-0%29%2A%5Cvec%7Bj%7D%5D%3D0%20)
Thus

By the way how can we make text larger in latex \larger{.....} don't work.
Answer A