To solve this problem you must apply the proccedure shown below:
1. You must make a system of equations, as below:
2. Let's call:
x: the number of first class tickets bought.
y: the number of coach tickets bought.
3. Then, you have:
x+y=7 (First equation)
970x+370y=3790 (Second equation)
x=7-y
4. By substitution, you have:
970x+370y=3790
970(7-y)+370=3790
y=5
x=7-y
x=7-5
x=2
Therefore, the answer is:
- Number of first class tickets bought=2
- Number of coach tickets bought=5
Answer:
three consecutive odd integers: 2n-1 2n+1 2n+3
that the sum of the smaller two is three
times larger increased by seven: 2n-1 + 2n+1 = 3(2n+3) +7
4n = 6n+ 9 +7
4n = 6n+ 16
4n -6n = 16
-2n = 16
n = 16/(-2)
n=-8
a) 2n -1 = 2(-8) -1 = -17
b) 2n+1 = 2(-8)+1 = -15
c) 2n+1 = 2(-8)+3 = -13
Ans. -17 ; -15 ; -13
Answer:
125/6(In(x-25)) - 5/6(In(x+5))+C
Step-by-step explanation:
∫x2/x1−20x2−125dx
Should be
∫x²/(x²−20x−125)dx
First of all let's factorize the denominator.
x²−20x−125= x²+5x-25x-125
x²−20x−125= x(x+5) -25(x+5)
x²−20x−125= (x-25)(x+5)
∫x²/(x²−20x−125)dx= ∫x²/((x-25)(x+5))dx
x²/(x²−20x−125) =x²/((x-25)(x+5))
x²/((x-25)(x+5))= a/(x-25) +b/(x+5)
x²/= a(x+5) + b(x-25)
Let x=25
625 = a30
a= 625/30
a= 125/6
Let x= -5
25 = -30b
b= 25/-30
b= -5/6
x²/((x-25)(x+5))= 125/6(x-25) -5/6(x+5)
∫x²/(x²−20x−125)dx
=∫125/6(x-25) -∫5/6(x+5) Dx
= 125/6(In(x-25)) - 5/6(In(x+5))+C
