Answer:
79
Step-by-step explanation:
m<1 = 90 -38 = 52
m<3 = 90 - 52 = 38\
m<2 = 180-38-63 = 79
Answer:
The man has 6 GHc20.00 notes and 4 GHc10.00 left
Step-by-step explanation:
Given
Amount = 3 GH¢50.00, 7 GH¢20.00 and 5 GH¢10.00
Spendings = GH¢80,00
Required
Determine how many GH¢20.00 and GH¢10.00 notes left
The amount he has can be represented as:


This can be represented as:

Represent:
GHc50.00 with x, GHc20.00 with y and GHc10.00 with z
So, we have:





Subtract the spendings from the total to get amount left


Collect Like Terms


Recall that, we have represented:
GHc50.00 with x, GHc20.00 with y and GHc10.00 with z
<em>This means that, the man has 6 GHc20.00 notes and 4 GHc10.00 left</em>
Your simplified expression is 16s + 2
Since the cotangent function is defined as

we have that the cotangent equals zero at
, because we have

So, the limit simply becomes

Answer:
The duration of each of the three spacewalk is between 364 minutes and 480 minutes with a total duration of 1324 minutes
Step-by-step explanation:
A spacewalk also termed an EVA or extravehicular activity, is defined as the general activity of an astronaut outside of a vehicle in space
Spacewalks can last up to 5 to 8 hours
Here we have a word problem, and are required to find the number of minutes equivalent to 22 hour and 4 minutes
Number of minutes per hour = 60 minutes
Number of minutes in 22 hours = 60 × 22 = 1320 minutes
Total number of minutes in 22 hours 4 minutes = 1320 + 4 = 1324 minutes.
Whereby the duration of a space walk is up to 5 to 8 hours gives;
22 hours 4 minutes - 8 hour - 8 hour -x = 0
Hence x = 6 hour 4 minutes
Therefore, the duration of each spacewalk is between 6 hour 4 minutes and 8 hours or in minutes we have
The duration of each spacewalk is between (6 × 60 + 4) 364 minutes and 480 minutes.