Here is the solution of the given problem above.
First, let's analyze the question.
Given: 1 bag = 9/10 pound
2/3 bag = ? pound
What we are going to do is to divide 9/10 pound to 3.
so 9/10 divided by 3 and we get 9/30 and to simplify that, 3/10.
So per 1/3 of the bag, there is 3/10 pound.
To get the weight for 2/3 of the bag, we multiply 3/10 by 2 and we get 6/10 or to simplify it, it is 3/5. Therefore, the 2/3 bag weights 3/5 pound. Hope this answer helps.
y = mx + b
m = slope and b = y-intercept
We can arrange 6y = x - 12 in the form of y = mx + b
6y = x - 12
y = 1/6(x) - 2
Slope of y = 1/6(x) - 2 is 1/6. Taking the negative reciprocal of the slope we get the slope for the perpendicular line.
Negative reciprocal of 1/6 is -6.
The equation for the perpendicular line is
y = -6x + b
To find b we can plug in the x and y values of (4,-4) into it since it passes through those coordinates
-4 = -6(4) + b
b = -4 + 6(4)
b = -4 + 24
b = 20
So the equation for the perpendicular line is y = -6x + 20
600 + 9.5x (less than or equal to) 900
600 fixed amount
9.5 is the variable changing amount and differs among number of students taken
Answer:
There may be 1 or 3 tricycles in the parking lot.
Step-by-step explanation:
Since at any point in time, there could be bicycles, tricycles, and cars in the school parking lot, and today, there are 53 wheels in total, if there are 15 bicycles, tricycles, and cars in total, to determine how many tricycles could be in the parking lot, the following calculation must be performed:
13 x 4 + 1 x 3 + 1 x 2 = 57
11 x 4 + 1 x 3 + 3 x 2 = 53
10 x 4 + 3 x 3 + 2 x 2 = 53
8 x 4 + 5 x 3 + 2 x 2 = 51
10 x 2 + 1 x 3 + 4 x 4 = 39
9 x 3 + 1 x 2 + 5 x 4 = 49
Therefore, there may be 1 or 3 tricycles in the parking lot.
