Saturday
May 18, 2013

# Posts by Melinda

Total # Posts: 149

Math
i got 0.25 answer, I wish I could show you the graph... T.T

Math
I need help with this question, it says to find the instantaneous rate at distance t = 6s with the given graph First of all, it is a linear line, do i need to find the equation of the graph? After finding the equation, i plug in 5.99 and 6.01? and solve ?

precalculus
I believe its between (10,6) and (12,6) because we know that the starting point is at 6.. I don't know after that

precalculus
What about question 4? 4. A student is walking in a straight line in front of a sensor. The sensor begins collecting data when the student is 6m away. The student walks toward the sensor for 4 s at a rate of 1 m/s. Then she walks away from the sensor for 8 s at a rate of 0.5 m...

precalculus
For problem 2, it is A? x= -1

precalculus
Then the answer must be c) h(x) = -x^2 + 9/4-1 = -(4)^2 + 9/4-1 = -7/4-1 =-x(1)^2 + 9 = 8 = -7 - 8/3 =-5 there c is negative

precalculus
This is the correct question, sorry I forgot to type in the one part of the question. 4. A student is walking in a straight line in front of a sensor. The sensor begins collecting data when the student is 6m away. The student walks toward the sensor for 4 s at a rate of 1 m/s....

precalculus
Why is problem 1 not A? I plugged in 4 and 1 into each of the equations given and A gave me negative avg rate T.T

precalculus
For Number 1, the answer is A wow, totally miscalculated. What about other questions am I right? For number 3 4. A student is walking in a straight line in front of a sensor. The sensor begins collecting data when the student is 6m away. The student walks toward the sensor for...

precalculus
2. For which value of x is the instantaneous rate of change of h(x) = 0.5x^2 + x - 2 closest to 0? a) x= -1 b) x= 0 c) x= 1 d x= 3 sorry mistake on #2 its B?

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