Answer:
C
Step-by-step explanation:
Given
![\left[\begin{array}{ccc}-2&3\\2&4\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-2%263%5C%5C2%264%5C%5C%5Cend%7Barray%7D%5Cright%5D)
+ ![\left[\begin{array}{ccc}3&1\\-1&2\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D3%261%5C%5C-1%262%5C%5C%5Cend%7Barray%7D%5Cright%5D)
Add corresponding elements to obtain the sum, that is
= ![\left[\begin{array}{ccc}-2+3&3+1\\2-1&4+2\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-2%2B3%263%2B1%5C%5C2-1%264%2B2%5C%5C%5Cend%7Barray%7D%5Cright%5D)
= ![\left[\begin{array}{ccc}1&4\\1&6\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%264%5C%5C1%266%5C%5C%5Cend%7Barray%7D%5Cright%5D)
→ C
Answer:
n = 2/3
Step-by-step explanation:
So set up an equation!
7* n/4 > 7
So trying to get the variable alone...
n/4 > (7/7) aka 1
n> 1*4
n> 4
Try it out!
7* 6/4 = 42/4 = 21/2= 10 and 1/2
It is greater than 7
Answer:
Step-by-step explanation:
(6/5)√x = x - 6
(36/25)x = x² - 12x + 36
0 = x² - 13.44x + 36
x = (13.44 ± √(13.44² - 4(1)(36))) / (2(1))
x = (13.44 ± 6.052569702...) / 2
x = 3.69371... zebras
or
x = 9.74628... zebras
In either case we have some odd fractions of zebras hanging around. This makes me highly suspect that the question is misreported
Answer:
The rate at which the distance between them is changing at 2:00 p.m. is approximately 1.92 km/h
Step-by-step explanation:
At noon the location of Lan = 300 km north of Makenna
Lan's direction = South
Lan's speed = 60 km/h
Makenna's direction and speed = West at 75 km/h
The distance Lan has traveled at 2:00 PM = 2 h × 60 km/h = 120 km
The distance north between Lan and Makenna at 2:00 p.m = 300 km - 120 km = 180 km
The distance West Makenna has traveled at 2:00 p.m. = 2 h × 75 km/h = 150 km
Let 's' represent the distance between them, let 'y' represent the Lan's position north of Makenna at 2:00 p.m., and let 'x' represent Makenna's position west from Lan at 2:00 p.m.
By Pythagoras' theorem, we have;
s² = x² + y²
The distance between them at 2:00 p.m. s = √(180² + 150²) = 30·√61
ds²/dt = dx²/dt + dy²/dt
2·s·ds/dt = 2·x·dx/dt + 2·y·dy/dt
2×30·√61 × ds/dt = 2×150×75 + 2×180×(-60) = 900
ds/dt = 900/(2×30·√61) ≈ 1.92
The rate at which the distance between them is changing at 2:00 p.m. ds/dt ≈ 1.92 km/h
