An executive drove from home at an average rate of speed of 30 mph to an airport where a helicopter was waiting. The executive boarded the helicopter and flew to the corporate offices at an average speed of 60 mph. The entire distance was 150 miles. The entire trip took 3 hours. What is the distance from the airport to the corporate offices?

Distance = rate x time

let the time by helicopter be x hrs, then the time he drove would be 3-x hrs

Distance in helicopter = 60x
distance driven = 30(3-x)

so 60x + 30(3-x) = 150

solve for x, then sub into 60x

tell me what you got, I got 120

An instructor asked his/her class to evaluate the expression 1/2x for x = 5. Some students gave 0.1 for an answer while others gave 2.5. Which answer is correct?

Note: I cam eup with 0.15?

The way you wrote it as 1/2x, one has to follow the order of operation and it would be 1÷2*x which is then 1÷2*5 = 2.5

to obtain .1 you would have written it as
1/(2x)

I don't see how you could obtain .15 as an answer.

Therefore, the correct answer is 2.5.

Well, well, well, looks like we have a math wizard in the house! But, hey, don't worry if you didn't get it right. Math can be sneaky sometimes. Keep practicing, and soon you'll be solving equations like a pro!

I apologize for the confusion. It seems there was a mistake in my previous response.

The expression 1/2x can be interpreted in two ways:

1) If you mean (1/2)x, then the expression simplifies to 1/2 * 5 = 5/2 = 2.5. In this case, the answer is 2.5.

2) If you mean 1/(2x), then the expression simplifies to 1/(2*5) = 1/10 = 0.1. In this case, the answer is 0.1.

So, depending on the intended meaning of the expression, both 2.5 and 0.1 can be considered as correct answers.

To solve the equation 1/(2x) for x = 5:

1. Substitute x = 5 into the equation: 1/(2*5).
2. Simplify by multiplying 2 and 5: 1/10.
3. Therefore, the correct answer is 0.1.

If you obtained 0.15, it's possible that there was a mistake in the calculation or in the original equation. Double-check the expression and the calculation to ensure accuracy.