Mr. Hayes has Ill straw bundles to use for bedding in his horse stalls. What is the

1 answer:

4 0

We have been given that Mr. Hayes has 111 straw bundles to use for bedding in his horse stalls. We are asked to find the greatest number of bundles that he can fill if each stall holds exactly 4 bundles of straw.

To solve our given problem, we will divide 111 by 4.

$\text{Greatest number of stalls that can be filled}=\frac{111}{4}$

$\text{Greatest number of stalls that can be filled}=27.75$

Since 0.75 is not an integer, so Mr. Hayes cannot fill 0.75 of a stall. Therefore, the greatest possible number of stalls will be 27.

You might be interested in

5. Find the value(s) of x so that the line containing the points (2x + 3, x + 2) and (0, 2) is

Dvinal [7]

Answer:

x = -2 or -9

Step-by-step explanation:

You want the values of x such that the line defined by the two points (2x+3, x+2) and (0, 2) is perpendicular to the line defined by the two points (x+2, -3-3x) and (8, -1).

<h3>Slope</h3>

The slope of a line is given by the slope formula:

m = (y2 -y1)/(x2 -x1)

Using the formula, the slopes of the two lines are ...

m1 = (2 -(x+2))/(0 -(2x+3)) = (-x)/(-2x-3) = x/(2x +3)

and

m2 = (-1 -(-3-3x))/(8 -(x+2)) = (2+3x)/(6 -x)

<h3>Perpendicular lines</h3>

The slopes of perpendicular lines have product of -1:

$\dfrac{x}{2x+3}\cdot\dfrac{2+3x}{6-x}=-1\\\\x(3x+2)=(2x+3)(x-6)\qquad\text{multiply by $(2x+3)(6-x)$}\\\\3x^2+2x=2x^2-9x-18\qquad\text{eliminate parentheses}\\\\x^2+11x+18=0\qquad\text{put in standard form}\\\\(x+2)(x+9)=0\qquad\text{factor}$