Since the distance is 560 miles and the bus goes 268 miles per hour, the answer can be given as:
If 268=1 hour
560=?
560/268*1 hour =2(rounded to nearest ones)
Therefore it will take the train approximately 2 hours to travel between both cities.
Answer:
The new point is at (5, -5)
Step-by-step explanation:
When a point is being translated down, they are being subtracted on the y-value. So all you have to do is subtract 9 from 4 and we end up with -5, our new y-value.
(5, -5)
Answer:
its a glitch
Step-by-step explanation:
Let one odd integer = x
other odd integer = x +2
Sum = x + x+2 = -44
=> 2x + 2 = -44
=> 2x = -44 -2 = -46
=> x = -46/2 = -23
x+2 = -23 + 2 = -21
Integers are -23 and -21
You're trying to find constants

such that

. Equivalently, you're looking for the least-square solution to the following matrix equation.

To solve

, multiply both sides by the transpose of

, which introduces an invertible square matrix on the LHS.

Computing this, you'd find that

which means the first choice is correct.