Answer:
3 cars
Step-by-step explanation:
If 12 cars are needed to carry 36 students,
<em>12</em><em>c</em><em>a</em><em>r</em><em>s</em><em> = 36</em>
the number of cars to carry 9 students will be:
xcars = 9
cross the two equations
36x = 12 X 9
36x = 108
x = 108/36
x= 3
Therefore, 3 cars are needed to carry 9 students
OR
12 cars will carry 36 students
so 1 car will carry, (36/12)students
therefore, 1 car will carry 3 students
So, for 9 students,
9students = 9/3 = 3
Answer:
circumference of circle = 2π×r
here r = 7
so 2 ×22÷7 × 7
= 2×22
= 44
therefore the circumference is 44 inches
Sin(35)x14 =8.03007...
So answer is 8
Let us assume the first odd integer to be = x
Then
The consecutive middle odd integer = x + 2
The third consecutive odd integer = x + 4
So we can now write the equation as
x + x + 2 + x + 4 = 5(x + 2) - 18
3x + 6 = 5x + 10 - 18
3x + 6 = 5x - 8
3x - 5x = - 8 - 6
- 2x = - 14
2x = 14
x = 14/2
= 7
So the first odd integer is 7
The second consecutive odd integer = (x + 2)
= (7 + 2)
= 9
The third consecutive odd integer = (x + 4)
= 7 + 4)
= 11
So the three consecutive odd integers are 7, 9, 11.
No because he only surveyed the members of the chess club
