How do you solve this math problem?
h= -8t^2+40t
then it says:
when t=0 and when t= 5
there are 2 questions that i have to answer:
1.) find the value of h
2.) graph this equation
- i don't know if i did the problem right but when i answered #1 i got h=0,
- and when i answered #2 the 8 and y coordinates that i was supossed to point were too high, like (1,32), (2,48,), (3, 48), etc. but our graph only goes up to 15-20, so either i did it wrong, or I need to know how to change the interval values on the a ti-84 graphing calculator
~ so just show me how to solve each problem, and if you have to change the intervals tell me what buttons to press! thanks so much!
1. is correct
2. not sure, sorry
To solve the math problem, we need to find the value of h when t = 0 and t = 5 for the given equation h = -8t^2 + 40t.
1. Finding the value of h:
Substitute t = 0 into the equation h = -8t^2 + 40t:
h = -8(0)^2 + 40(0)
h = 0 + 0
h = 0
So when t = 0, the value of h is 0.
Substitute t = 5 into the equation h = -8t^2 + 40t:
h = -8(5)^2 + 40(5)
h = -8(25) + 200
h = -200 + 200
h = 0
So when t = 5, the value of h is also 0.
Therefore, the value of h is 0 for both t = 0 and t = 5.
2. Graphing the equation:
To graph the equation h = -8t^2 + 40t, you can use a graphing calculator, such as the TI-84.
1) Press the "Y=" button on the calculator.
2) Enter the equation as: -8x^2 + 40x.
3) Press the "Graph" button to plot the graph.
To adjust the interval values on the TI-84 graphing calculator, you can follow these steps:
1) After pressing "Y=" to enter the equation, press the "WINDOW" button.
2) Adjust the values for "Xmin", "Xmax", "Ymin", and "Ymax" to increase or decrease the interval.
For example, you can set "Xmin" to -10, "Xmax" to 10, "Ymin" to -20, and "Ymax" to 50 to fit the desired range.
3) Press "GRAPH" to display the graph within the adjusted interval.
By adjusting the interval values, you should be able to see the graph within the desired range, which in this case is around 15-20 in the y-coordinate.