The domain and range of the given function are equal to (0, 3.85) and (0, 18.75) respectively.
<h3>How to calculate the domain of the function?</h3>
In this exercise, you're given the following function h(t) = -4.87t² + 18.75t. Next, we would equate the function to zero (0) to determine its domain as follows:
0= -4.87t² + 18.75t.
4.87t(-t + 3.85) = 0
t = 0 or t = 3.85.
Therefore, the domain is 0 ≤ t ≤ 3.85 or (0, 3.85).
<h3>How to calculate the range of the function?</h3>
h(t) = -4.87t² + 18.75t
h(t) = -4.87(t² - 3.85t + 3.85 - 3.85)
h(t) = -4.87(t - 1.925)² + 18.05
Therefore, the range is 0 ≤ h ≤ 18.05 or (0, 18.75).
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Answer:
120 pages can be printed in 1 hour.
Step-by-step explanation:
First divide 36 by 18 to figure out how many pages can be printed in a minute.
Like this: 36 pages ÷ 18 minutes = 2 pages in one minute.
Now, multiply 2 by 60 to figure out how many pages can be printed out in a hour.
Like this: 2 pages in one minute × 60 minutes ( one hour) = 120 pages in one hour can be printed.
Hope this helps!
 
        
             
        
        
        
Answer:
y
=1
/3
x
+
14
/3
Step-by-step explanation:
 
        
             
        
        
        
10.0.0.0/8 is variably subnetted, 3 subnets, 2 masksC10.1.1.0/30 is directly connected, Serial0/0/0L10.1.1.1/32 is directly connected, Serial0/0/0R10.2.2.0/30 [120/1] via 10.1.1.2, 00:00:21, Serial0/0/0172.30.0.0/16 is variably subnetted, 2 subnets, 2 masksC172.30.10.0/24 is directly connected, GigabitEthernet0/1L172.30.10.1/32 is directly connected, GigabitEthernet0/1R3 only displays its own subnets for the 172.30.0.0 network. R3 does not have any routes for the172.30.0.0 subnets on R1.