The basis to respond this question are:
1) Perpedicular lines form a 90° angle between them.
2) The product of the slopes of two any perpendicular lines is - 1.
So, from that basic knowledge you can analyze each option:
a.Lines s and t have slopes that are opposite reciprocals.
TRUE. Tha comes the number 2 basic condition for the perpendicular lines.
slope_1 * slope_2 = - 1 => slope_1 = - 1 / slope_2, which is what opposite reciprocals means.
b.Lines s and t have the same slope.
FALSE. We have already stated the the slopes are opposite reciprocals.
c.The product of the slopes of s and t is equal to -1
TRUE: that is one of the basic statements that you need to know and handle.
d.The lines have the same steepness.
FALSE: the slope is a measure of steepness, so they have different steepness.
e.The lines have different y intercepts.
FALSE: the y intercepts may be equal or different. For example y = x + 2 and y = -x + 2 are perpendicular and both have the same y intercept, 2.
f.The lines never intersect.
FALSE: perpendicular lines always intersept (in a 90° angle).
g.The intersection of s and t forms right angle.
TRUE: right angle = 90°.
h.If the slope of s is 6, the slope of t is -6
FALSE. - 6 is not the opposite reciprocal of 6. The opposite reciprocal of 6 is - 1/6.
So, the right choices are a, c and g.
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.
Answer:
1/3 or 3.3<u>3</u>
Step-by-step explanation:
There are 6 sides to dice, and to land on one side is a 1/6th of a chance. to land on 2 possible sides, you do 1/6+1/6 which equals 2/6 which is equivalent to 1/3 or 3.3<u>3</u> as a decimal.
I hope this helps and good luck!
Answer:
348
Step-by-step explanation:
6 8 10 22 75 78 81 354
subtract the highest number to the lowest number
Answer
It would take them the same amount of time as it normally took.
Which would be 30 and 45 minutes.
please feel free to ask me some more questions, or clarify.
