A----------D----------C
AD = 2x + 6
DC = 4x - 7
the midpoint D splits AC into 2 equal parts....therefore, AD = DC
2x + 6 = 4x - 7
2x - 4x = -7 - 6
-2x = - 13
x = -13/-2
x = 6.5
DC = 4x - 7......4(6.5) - 7 = 19 <===
AD = 2x + 6.....2(6.5) + 6 = 19
Answer:
nsgs6
68÷9
02_3
73929
Step-by-step explanation:
8282×627=628373
Answer:
The number of 32 Gigabit keys that can be fitted on the hard drive is 375.
Step-by-step explanation:
The question is:
If my hard drive has a capacity of 1.5 Terabytes, how many 32 Gigabit keys can fit on that hard drive?
Solution:
1 Terabyte = 8000 Gigabits
Then 1.5 Terabytes in Gigabits is:
1.5 Terabytes = (8000 × 1.5) Gigabits
= 12000 Gigabits
One key is of 32 Gigabits.
Compute the number of 32 Gigabit keys that can be fitted on the hard drive as follows:

Thus, the number of 32 Gigabit keys that can be fitted on the hard drive is 375.