Answer:
The answer is Ari, Beth, Cindy
Explanation
Ari
- 2¾ miles is 2.75 miles
- speed of 2½ miles per hour is 2.5 mi/h
DISTANCE = SPEED × TIME
t = d/s
= 2.75/2.5
= 1.1hr
= 66mins
Beth
- 1¾ miles is 1.75 miles
- speed of 1¼ miles per hour is 1.25 mi/h
t = d/s
= 1.75/1.25
=1.4hr
=84mins
Cindy
(Correct me if i am wrong)
You cannot add row 2 to column 3 because they have different dimensions. You can do any of the other operations, but the only one that makes any sort of sense is ...
Multiply row 2 by -1 and add it to row 3
_____
It makes no sense to multiply a row by zero. That makes the entire row zero and makes the matrix useless for finding any sort of solution.
You can switch columns, but that doesn't get you any closer to a solution here.
If I were trying to find a solution, I might
switch rows 1 and 2
multiply the new row 1 by -3 and add it to the new row 2
multiply the new row 1 by 2 and add it to row 3
This sequence of operations will make the first column [1 0 0], reducing the problem to 2×2 from 3×3.
Answer:
The height of deepness of submarine is 424.1 feet
Step-by-step explanation:
Given as :
The angle at which submarine drives of the surface of water = Ф = 13°
The speed at which submarine travel = s = 760 feet per minute
Let The height of deepness of submarine = h feet
The time taken to reach that deepness = t = 5 minutes
<u>According to question</u>
The height of deepness of submarine = 
i.e h × cos(90 - Ф) = 
Or, h × Sin(Ф) = 
Or, h × Sin(13°) = 95 feet
Or, h × 0.224 = 95 feet
∴ h = 
i.e h = 424.10 feet
So, The height of deepness of submarine = h = 424.10 feet
Hence, The height of deepness of submarine is 424.1 feet Answer
The dilation factor Sophia is using appears to be
(image coordinate)/(original coordinate) = -9/-3 = 15/5 = 3
Then the problem tells us
(pre-image point)×3 = (3, 1)
so we conclude
(pre-image point) = (3, 1)/3 = (1, 1/3)
