Two ways are...
1. 208,561
2. write it out as two hundred eight thousand five hundred sixty-one
#. you can also write it as 200,000 + 8,000 + 500 + 60 + 1
Count the number of 1 bits after the last decimal point. Call this number n.
There will be 2^n subnets and 2^(8-n) - 2 hosts per subnet. (You can mutilply the number of subnets by the number of hosts per subnet to find the total number of hosts.)
/25: 255.255.255.128, 2 subnets, 126 hosts per subnet
/26: 255.255.255.192, 4 subnets, 62 hosts per subnet
/27: 255.255.255.224, 8 subnets, 30 hosts per subnet
/28: 255.255.255.240, 16 subnets, 14 hosts per subnet
/29: 255.255.255.248, 32 subnets, 6 hosts per subnet
Answer:
- D(5, 4), E(14, 7), M(9.5, 5.5)
Step-by-step explanation:
As AD = 1/4AB and DE ║ AC, the ratio CE/CB = 1/4, or CE = 1/4CB
<u>Find the coordinates of D:</u>
- x = 1 + 1/4(17 - 1) = 1 + 4 = 5
- y = 5 + 1/4(1 - 5) = 5 - 1 = 4
<u>Find the coordinates of E:</u>
- x = 13 + 1/4(17 - 13) = 13 + 1 = 14
- y = 9 + 1/4(1 - 9) = 9 - 2 = 7
<u>Find the coordinates of the midpoint M of DE:</u>
- x = (5 + 14)/2 = 19/2 = 9.5
- y = (4 + 7)/2 = 11/2 = 5.5
Answer:
Step-by-step explanation:
Greatest multiplier of 4 that is less than 21 is 5 (4 x 5 = 20)
Put the remaining number as the numerator (21 - 20 = 1)
Answer: 15 meters
Step-by-step explanation:
To get the maximum distance, you first use the given numbers as basis.
10 meters/3 meters =x/4.5 meters
Where x = the maximum distance the projector can be placed that can produce an image 4.5 meters high
3x = 10 (4.5)
3x = 45
X= 45/3
X= 15 meters
So, the maximum distance that the projector can be placed from the screen which produces an image 4.5 meters high is 15 meters.