The answer is B, and here's why. Set up a table for "there" and "back" and use the distance = rate * time formula, like this:
d r t
there d 450 t
back d 400 1-t
Let me explain this table to you. The distance is d, we don't know what it is, that's what we are actually looking for. We only know that if we go somewhere from point A to point B, then back again to point A, the distance there is the same as the distance back. Hence, the d in both spaces. There he flew 450 mph, back he flew 400 mph. If the total distance was 1 hour, he flew an unknown time there and one hour minus that unknown time back. For example, if he flew for 20 minutes there, one hour minus 20 minutes means that he flew 60 minutes - 20 minutes = 40 minutes back. See? Now, because the distance there = the distance back, we can set the rt in both equal to each other. If d = rt there and d = rt back and the d's are the same, then we can set the rt's equal to each other. 450t = 400(1-t) and
450t = 400 - 400t and 850t = 400. Solve for t to get t = .47058. Now, t is time, not the distance and we are looking for distance. So multiply that t value by the rate (cuz d = r*t) to get that the distance one way is
d = 450(.470580 and d = 211. 76 or, rounded like you need, 212.
First you can solve for b: Subtract a on both sides and get 2b=-a+10, then divide by 2 to get b by itself and get: b=-1/2a+5
Then you can plug in this equation for b in the other: This would give you: 2a+-1/2a+5=6. Then you can use that equation to solve for a and get: 1.5a+5=6, subtract 5, 1.5a=1, divide by 1.5, a=1/1.5
Then you can plug in the value of a to solve for b.
Answer:
21-2x
Step-by-step explanation:
11-2(x-5)
11-2x+10
21-2x
not completely sure if im right... if i am can you mark me brainliest
The cross product of two vectors gives a third vector

that is orthogonal to the first two.

Normalize this vector by dividing it by its norm:

To get another vector orthogonal to the first two, you can just change the sign and use

.
Answer:
Step-by-step explanation:
Let price of milgrade gas be x and price of lawnmower gas be y.
18x + 5y=59.51
14x+y=39.75
Isolate one variable and substitute to solve for other variable.