Answer:
(0,-3)
Step-by-step explanation:
hi, with this we can just do substitution (substitute x for y+3)
6(y+3)-5y=15
evaluate
6y+18-5y=15
y+18=15
y=-3
then we plug that in
x=-3+3
x=0
First let's define variables:
r: reading
s: skimming
We write the system of equations that adapts to the problem:
r + s = 275
(15/10) r + (10/15) s = 355
We solve the system of equations:
r = 206
s = 69
Note: see attached image for system solution.
Answer:
Jayden had to skim 69 pages through
Answer: D) reflection across y = -x
Explanation:
When we reflect over y = x, we basically swap x and y. So for instance, the point (3,1) becomes (1,3).
When reflecting over y = -x, we will do the same thing but we'll make each coordinate swap in sign from positive to negative (or vice versa). The rule for reflecting over y = -x is
So if we apply that rule to point A(3,1) then it becomes A ' (-1, -3).
Similarly, B(1,5) moves to B ' (-5, -1)
Finally, C(6,9) becomes C ' (-9, -6)
Answer:
Let's suppose that each person works at an hourly rate R.
Then if 4 people working 8 hours per day, a total of 15 days to complete the task, we can write this as:
4*R*(15*8 hours) = 1 task.
Whit this we can find the value of R.
R = 1 task/(4*15*8 h) = (1/480) task/hour.
a) Now suppose that we have 5 workers, and each one of them works 6 hours per day for a total of D days to complete the task, then we have the equation:
5*( (1/480) task/hour)*(D*6 hours) = 1 task.
We only need to isolate D, that is the number of days that will take the 5 workers to complete the task:
D = (1 task)/(5*6h*1/480 task/hour) = (1 task)/(30/480 taks) = 480/30 = 16
D = 16
Then the 5 workers working 6 hours per day, need 16 days to complete the job.
b) The assumption is that all workers work at the same rate R. If this was not the case (and each one worked at a different rate) we couldn't find the rate at which each worker completes the task (because we had not enough information), and then we would be incapable of completing the question.
Answer:
B. BC ≅ LM
Step-by-step explanation:
BC corresponds to LM. These are the only two sides without the hash marks that represent congruence. If the marks were there to indicate that the sides are congruent, then the triangles could be proved congruent by SSS.