Answer:
Answer in explanation.
Step-by-step explanation:
Lets take
as L1.
Take L2 the line perpendicular to L1.
Standard form of equation of line:
y=mx+c, where m = slope and c = y-intercept.
Since L1 and L2 are perpendicular,
mL1 x mL2 = -1
Substitute mL1 into the equation,
3/4 x mL2 = -1
mL2 = -1 ÷ 3/4
mL2 = -4/3
L2 : y = mx+ c
Substitute y = 6, x = 8 and m = -4/3 into the equation,
therefore L2:
Lets take L3 as the line parallel to L1.
Since L3 and L1 are parallel,
mL3 = mL1 = 3/4
equation of line: y = mx+c
substitute y = 6, x = 8 and m = 3/4 into equation.
therefore L3:
For this case we must indicate the inequality represented in the graph.
It is noted that we have a closed point at -1, so the inequality at that point contains equality and is represented by "[". On the other hand, we observe an open point at +3, so inequality does not include equality (does not include +3) and is represented as ")".
So, the interval of the graph is:
This is equivalent to:
Answer:
Option B
This is a combination problem.
Given:
12 students
3 groups consisting of 4 students.
Mark can't be in the first group.
The combination formula that I used is: n! / r!(n-r)!
where: n = number of choices ; r = number of people to be chosen.
This is the formula I used because the order is not important and repetition is not allowed.
Since Mark can't be considered in the first group, the value of n would be 11 instead of 12. value of r is 4.
numerator: n! = 11! = 39,916,800
denominator: r!(n-r)! = 4!(11-4)! = 4!*7! = 120,960
Combination = 39,916,800 / 120,960 = 330
There are 330 ways that the instructor can choose 4 students for the first group
The error is that the student measured the wrong side. Acute angles are less than 90 degrees. So it isn’t possible for the measure to be over 90. The correct measure is 49
Answer:
I think it's graph 2 hope that helps
