Wouldn't this be in literature?
This is a quadratic equation because it has degree 2.
As known as a parabola
Use compound interest formula F=P(1+i)^n twice, one for each deposit and sum the two results.
For the P=$40,000 deposit,
i=10%/2=5% (semi-annual)
number of periods (6 months), n = 6*2 = 12
Future value (at end of year 6),
F = P(1+i)^n = 40,000(1+0.05)^12 = $71834.253
For the P=20000, deposited at the START of the fourth year, which is the same as the end of the third year.
i=5% (semi-annual
n=2*(6-3), n = 6
Future value (at end of year 6)
F=P(1+i)^n = 20000(1+0.05)^6 = 26801.913
Total amount after 6 years
= 71834.253 + 26801.913
=98636.17 (to the nearest cent.)
12) LCM of 2 & 12 is 12
Answer: (F)
13) (25 x 8) x 4
Step 1: Order of operations do parenthesis first: (25 x 8) = 200
Step 2: Multiply: 200 x 4 = 800
Answer: 800
14) I would plug each value in to see which works
D) 10x + 5y = 10(6) + 5(9) = 60 + 45 = 105
Answer is (D)
15) 2 x 12 - 8 (divided by) 2^2
Exponent first: 2 x 12 - 8 divided by 4
Multiplication/ Division: 2 x 12 = 24 and -8 divided by 4 = -2
Add/Subtract: 24 - 2 = 22
Answer: (I)
Answer: Lattice parameter, a = (4R)/(√3)
Step-by-step explanation:
The typical arrangement of atoms in a unit cell of BCC is shown in the first attachment.
The second attachment shows how to obtain the value of the diagonal of the base of the unit cell.
If the diagonal of the base of the unit cell = x
(a^2) + (a^2) = (x^2)
x = a(√2)
Then, diagonal across the unit cell (a cube) makes a right angled triangle with one side of the unit cell & the diagonal on the base of the unit cell.
Let the diagonal across the cube be y
Pythagoras theorem,
(a^2) + ((a(√2))^2) = (y^2)
(a^2) + 2(a^2) = (y^2) = 3(a^2)
y = a√3
But the diagonal through the cube = 4R (evident from the image in the first attachment)
y = 4R = a√3
a = (4R)/(√3)
QED!!!
