Evaluate h(x)=2.8x^3+0.01x^2 - 1 for x=1 and x=2

h(1) = 1.81 and h(2) = 21.44

Step-by-step explanation:

* Lets read the problem and solve it

- Evaluate means find the value, so evaluate h(x) means find the value

 of it at the given values of x

∵ h(x) = 2.8x³ + 0.01x² - 1

∵ x = 1 and x = 2

- Then find h(1) by substitute x by 1 and find h(2) by substitute x by 2

# At x = 1

∴ h(1) = 2.8(1)³ + 0.01(1)² - 1

∴ h(1) = 2.8(1) + 0.01(1) - 1

∴ h(1) = 2.8 + 0.01 - 1

∴ h(1) = 1.81

# At x = 2

∴ h(2) = 2.8(2)³ + 0.01(2)² - 1

∴ h(2) = 2.8(8) + 0.01(4) - 1

∴ h(2) = 22.4 + 0.04 - 1 ⇒ simplify

∴ h(2) = 21.44

* h(1) = 1.81 and h(2) = 21.44

answer: 20

step-by-step explanation: subtract the minimum date value from the maximum data value to find the date range. in this casem the date range is 42 - 22 = 20

hope this ! : ) ~zane