First you want to find the length and width of the rectangle using the distance formula:
d=√(x2-x1)²+(y2-y1)²
AB=√(6-3)²+ (-2 - -2)²
AB=√3² + 0
AB=√9
AB=3
BC=<span>√(6-6)²+ (5 - -2)²
BC=</span>√0 + 7²
BC=√49
BC=7
We can find the area by multiplying these two distances together:
A=(3)(7)
A=21 units²
We know that
[area of triangle]=b*h/2
b=15 units
h=8 units
[area of triangle]=15*8/2---------> 60 units ²
see the attached figure
the answer is
60 units ²
Answer:
I think square
Cause it is having 4 sides and in ur question the were 4 vertuces so i think its a square
Answer:
x = -18, Y = -44
Step-by-step explanation:
Since Y = 2x - 8, we can substitute that into the first equation:
2x - 8 = 3x + 10
3x - 2x = -8 - 10
x = -18
Now we plug x = -18 back into the second equation:
Y = 2(-18) - 8
Y = -36 - 8
Y = -44
Answer:
Check the explanation
Step-by-step explanation:
(a) The appropriate test is the matched-pairs test because a student’s score on Try 1 is certainly correlated with his/her score on Try 2. Using the differences, we have xbar = 29 and s = 59.
(b) To test H0: mu=0 vs. H1 mu > 0, we compute

with df = 426. This is certainly significant, with P < 0.0005. Coached students do improve their scores on average
(a) H0: μ1 = μ2 vs. Ha: μ1 > μ2, where μ1 is the mean gain among all coached students and μ2 is the mean gain among uncoached students. H0 and Ha. Using the conservative approach, df = 426 is rounded down to df = 100 in (t table) and we obtain 0.0025 < P < 0.005. Using software, df = 534.45 and P = 0.004. There is evidence that coached students had a greater average increase.
(b) 8 ± t*(3.0235) where t* equals 2.626 (using df = 100 with (t table) ) or 2.585 (df = 534.45 with software). This gives either 0.06 to 15.94 points, or 0.184 to 15.816 points, respectively.
(c) Increasing one’s score by 0 to 16 points is not likely to make a difference in being granted admission or scholarships from any colleges.