Answer:
First, you would have to convert fl oz to gallons: 1 fl oz = 0.0078125 gal. Therefore, 4 fl oz = 0.03125 gal.
Then, you would have to convert minutes to hours: 1 minute = 0.01666 hours, or 60 minutes = 1 hour.
From here, you can divide 0.03125 gal by 0.01666 hours OR multiply 0.03125 gal by 60 minutes, giving a rate of 1.87 gallons per hour.
Length of line segment <em>YX </em>is distance of both the ends of it. The length of line segment <em>YX </em>is 
units.
<h3>What is length of a line segment?</h3>
The length of a line segment is the measurement of the distance of both the ends of it.
Given information-
In the rectangle 
the length of line line segment YV is 24 units.
The rectangle is shown in the image below.
The angle given in the figure is,
Now in the figure of rectangle shown below, the triangle <em>XWV </em>is the right angle triangle.
In this right angle triangle the tan of angle is the ratio perpendicular to the base. thus,
As the opposite side of a rectangle are equal. Thus, the length of line segment is <em>VW </em>is equal to the length of the line segment <em>YX</em>. Therefore,
Hence, the length of line segment <em>YX </em>is units. The option B is the correct option.
Learn more about the line segment here;
brainly.com/question/2437195
Consider two lines in space `1 and `2 such that `1 passes through point P1 and is parallel to vector ~v1 and `2 passes through P2 and is parallel to ~v2. We want to compute the smallest distance D between the two lines.
If the two lines intersect, then it is clear that D = 0. If they do not intersect and are parallel, then D corresponds to the distance between point P2 and line `1 and is given by D = k−−−→ P1P2 ×~v1k k~v1k . Assume the lines are not parallel and do not intersect (skew lines) and let ~n = ~v1 ×~v2 be a vector perpendicular to both lines. The norm of the projection of vector −−−→ P1P2 over ~n will give us D, i.e., D = |−−−→ P1P2 ·~n| k~nk . Example Consider the two lines `1 : x = 0, y =−t, z = t and `2 : x = 1+2s, y = s, z =−3s. It is easy to see that the two lines are skew. Let P1 = (0,0,0), ~v1 = (0,−1,1), P2 = (1,0,0), and ~v2 = (2,1,−3). Then, −−−→ P1P2 = (1,0,0) and ~n = ~v1 ×~v2 = (2,2,2). We then get D = |−−−→ P1P2 ·~n| k~nk = 1 √3. Observe that the problem can also by solved with Calculus. Consider the problem of minimizing the Euclidean distance between two points on `1 and `2. Let Q1 = (x1,y1,z1) and Q2 = (x2,y2,z2) be arbitrary points on `1 and `2, and let F(s,t) = (x2 −x1)2 +(y2 −y1)2 +(z2 −z1)2 = (1+2s)2 +(s + t)2 +(−3s−t)2 = 14s2 +2t2 +8st +4s+1. Note that F(s,t) corresponds to the square of the Euclidean distance between Q1 and Q2. Let’s nd the critical points of F. Fs(s,t) = 28s+8t +4 = 0 Ft(s,t) = 4t +8s = 0 By solving the linear system, we nd that the unique critical point is (s0,t0) = (−1/3,2/3). Since the Hessian matrix of F, H =Fss Fst Fts Ftt=28 8 8 4, is positive denite, the critical point corresponds to the absolute minimum of F over all (s,t)∈R2. The minimal distance between the two lines is then D =pF(s0,t0) = 1 √3.
The width would be x.
the length would be x+6
there are two of each (since it’s a rectangle)
so x+x+x+6+x+6=140
simplify
x= 32
l= 38
w= 32
