Answer:
-89
Step-by-step explanation:
The first term of the sequence is a1 = -12.
The common difference between terms is ...
d = -13 -(-12) = -1
The generic term is given by ...
an = a1 +d(n -1)
Then the 78th term is ...
a78 = -12 +(-1)(78 -1) = -12 -77
a78 = -89
The first thing you want to do is plug in x and y into both equations:
a(3) + b(4) = 4
b(3) + a(4) = 8
rearrange to line up a’s and b’s
3a + 4b = 4
4a + 3b = 8
now you want to choose a or b and multiply each equation by a number to make them have the same amount of a’s or b’s.
4(3a + 4b = 4) = 12a + 16b = 16
3(4a + 3b = 8) = 12a + 9b = 24
Now we subtract the bottom equation from the top and solve for b:
12a + 16b - (12a + 9b) = 16 - 24
7b = -8
b = -8/7
Now we plug back in for b to one of the original equations:
3a + 4(-8/7) = 4
3a + (-32/7) = 4
3a - (32/7) = 4
3a = 4 + (32/7)
3a = (28/7) + (32/7)
3a = 60/7
a = (60/7)/3 = 20/7.
Finally, plug a and b in together to double check using the second equation.
4a + 3b = 8
4(20/7) + 3(-8/7) = ?
(80/7) - (24/7) = ?
56/7 = 8.
5187 apples to be exact and if you need me to explain why let me know
20/5=4
25/7=<span>3.57142857143 Rounded to the nearest tenth: 3.6
4>3.6
If you want the nearest whole, you would round down to 3 because you would be short of 4.
3<4 </span>
Given :
On the first day of ticket sales the school sold 10 senior tickets and 1 child ticket for a total of $85 .
The school took in $75 on the second day by selling 5 senior citizens tickets and 7 child tickets.
To Find :
The price of a senior ticket and the price of a child ticket.
Solution :
Let, price of senior ticket and child ticket is x and y respectively.
Mathematical equation of condition 1 :
10x + y = 85 ...1)
Mathematical equation of condition 2 :
5x + 7y = 75 ...2)
Solving equation 1 and 2, we get :
2(2) - (1) :
2( 5x + 7y - 75 ) - ( 10x +y - 85 ) = 0
10x + 14y - 150 - 10x - y + 85 = 0
13y = 65
y = 5
10x - 5 = 85
x = 8
Therefore, price of a senior ticket and the price of a child ticket $8 and $5.
Hence, this is the required solution.
