Answer:
300.104
Step-by-step explanation:
Answer:
D) b = 
Step-by-step explanation:
First, move all the terms that do not include b to the left side with S:
S = 2ab + 2bc + 2ac
S - 2ac = 2ab + 2bc
Now, factor out 2b from the right side:
S - 2ac = 2b(a + c)
Divide both sides by 2(a + c):
b = 
Finally, multiply out the denominator:
b =
9514 1404 393
Answer:
Step-by-step explanation:
Let x and y represent the weights of the large and small boxes, respectively. The problem statement gives rise to the system of equations ...
x + y = 85 . . . . . combined weight of a large and small box
70x +50y = 5350 . . . . combined weight of 70 large and 50 small boxes
We can subtract 50 times the first equation from the second to find the weight of a large box.
(70x +50y) -50(x +y) = (5350) -50(85)
20x = 1100 . . . . simplify
x = 55 . . . . . . . divide by 20
Using this in the first equation, we can find the weight of a small box.
55 +y = 85
y = 30 . . . . . . . subtract 55
A large box weighs 55 pounds; a small box weighs 30 pounds.
So 16=heads
18=tails
after you add some tails
18+x
ratio of heads to tails is 1:1.5
heads=16, didn't change
tails=18+x
so
1:1.5=16:18+x
fractionify
1/1.5=16/(18+x)
1/1.5=2/3
2/3=16/(18+x)
mutiply both sides by (3)(18+x)
2(18+x)=3(16)
36+2x=48
subtract 36 from both sides
2x=12
divide 2
x=6
6 tails added
total=heads+tails
heads=16
tails=18+6=24
total=16+24
total=40
6 tails added
40 total coins
Considering that the addresses of memory locations are specified in hexadecimal.
a) The number of memory locations in a memory address range ( 0000₁₆ to FFFF₁₆ ) = 65536 memory locations
b) The range of hex addresses in a microcomputer with 4096 memory locations is ; 4095
<u>applying the given data </u>:
a) first step : convert FFFF₁₆ to decimal ( note F₁₆ = 15 decimal )
( F * 16^3 ) + ( F * 16^2 ) + ( F * 16^1 ) + ( F * 16^0 )
= ( 15 * 16^3 ) + ( 15 * 16^2 ) + ( 15 * 16^1 ) + ( 15 * 1 )
= 61440 + 3840 + 240 + 15 = 65535
∴ the memory locations from 0000₁₆ to FFFF₁₆ = from 0 to 65535 = 65536 locations
b) The range of hex addresses with a memory location of 4096
= 0000₁₆ to FFFF₁₆ = 0 to 4096
∴ the range = 4095
Hence we can conclude that the memory locations in ( a ) = 65536 while the range of hex addresses with a memory location of 4096 = 4095.
Learn more : brainly.com/question/18993173
