Algebra Homework: Equations, Calculus, and Radioactive Decay Problems

This document presents a detailed solution to an algebra homework assignment. The solution begins by demonstrating the substitution method to solve a system of linear equations. It then proceeds to solve problems involving derivatives and integrals, including calculations of gradients and areas under curves. The assignment also includes problems related to radioactive decay, calculating the activity of a sample and determining the flux of gamma rays emitted from a mixture of radioisotopes with different half-lives. The solutions provide step-by-step explanations and calculations, making it a valuable resource for students studying algebra and related concepts. The document covers a range of topics, from basic algebraic manipulations to more advanced calculus and physics applications, providing a comprehensive approach to problem-solving.

Solving by substitution method
3x + 2y = 6
Making x the subject of the formula
3x = 6 – 2y
x = 2 - 2
3 y
Substitute to
6x – y = 9
6(2 - 2
3 y) – y = 9
12 –4y – y = 9
12 – 5y = 9
-5y = -3
Y = 3
5
Substituting to x = 2 - 2
3 y
x = 2 -
2
3∗3
5
x = 8
5
The value of the variable y
y = 3
5

d
dx [ cos ( x)
1−sin(x ) ]
=
d
dx [ cos ( x ) ]∗( 1−sin ( x ) ) −cos ( x ) ∗d
dx [1−sin ( x ) ]
( 1−sin ( x ) ) 2
= (−sin ( x ) ) ( 1−sin ( x ) )−( d
dx [ 1 ]− d
dx [ sin ( x ) ] )cos (x )
( 1−sin ( x ) )2
= cos2 ( x )− ( 1−sin ( x ) ) sin ( x )
( 1−sin ( x ) ) 2
= cos2 (x )
( 1−sin ( x ) )2 − sin ( x)
1−sin (x )
= sin2 ( x)−sin ( x ) +cos2 (x)
( sin ( x ) −1 ) 2
X = -0.97
Substituting
= sin2 (−0.97)−sin ( −0.97 ) +cos2 (−0.97)
( sin ( −0.97 ) −1 )
2
= 0.983352931
= 0.98

Question 3
Y = 8x + 4x2
d
dx [ 4 x2 +8 x ]
= 4∗d
dx [ x2 ] + 8∗d
dx [ x ]
= 4*2x + 8*1
=8x +8
At x = 1
=8*1 +8
= 17
Question 4
d
dx [ x3+ 7 x+97]
= d
dx [ x3 ] + 7∗d
dx [ x ] + d
dx [97]

= 3x2 +7*1 + 0
= 3x2 + 7
At x = 1
The gradient will be
= 3*12 + 7
= 10
Question 5
Area = ∫
0.04
0.09
e4 x+1 dx
Let u = 4x + 1, dx = 1
4 du
Solving
∫ eu du
Applying exponential rule
∫ eu du = au
ln ( a) with a = e
= eu
Plugging in
1
4 ∫
0.04
0.09
eu du
= [ eu
4 ]0.040.09

Substitute u = 4x + 1
= [ e4 x +1
4 ]0.040.09
= [ e4∗0.09+1
4 +c ]− [ e4∗0.04+1
4 +c ]
= 0.974048 – 0.797483
= 0.176565
= 0.18 unit square
Question 6
The decay equation is
N = N0e− λt
If A is the activity then the equation will be
A = A0e− λt
A1/A2 = e−λt 1
e−λt 2
1000/A2 = e
216∗0.693
275
A2 = 580.2 Bq
Question 7

E* = E
1+ ∝(1−cosθ)
Maximum angle of scatter θ=1800
∝=m0C2 = 511 kev
E* = 389
1+ 511(1−cos 1800 )
= 0.38025
= 0.4 kev
Question 8
Area = ∫
1.5
2.9
1
5 x−1 dx
Let u = 5x – 1, dx = 1
5 du
¿ 1
5∗∫
1.5
2.9
1
u du

Solving
∫
1.5
2.9
1
u du
= ln(u)
Plugging in
1
5∗∫
1.5
2.9
1
u du
= [ ln (u)
5 ]1.52.9
Substituting
u = 5x – 1
= [ ln (5 x−1)
5 ]1.52.9
= [ ln (5∗2.9−1)
5 + c ]− [ ln (5∗1.5−1)
5 +c ]
= 0.52054 – 0.37436
= 0.1462
= 0.15 unit square
Question 9
d
dx [ln ( 4 x−1 ) ]
=
1
4 x−1∗d
dx [4 x−1]

=
4∗d
dx [ x ] + d
dx [−1]
4 x−1
= 4∗1+0
4 x−1
= 4
4 x−1
Substituting
X = 4
= 4
4∗4−1
= 0.26667
= 0.3
Question 10
A sample contains a mixture of two radioisotopes, both of which are pure
gamma emitters, emitting one gamma ray per disintegration. Isotope A has
a half-life of 3 minutes and isotope B has a half-life of 6 minutes. At five
past two the total flux of gamma rays being emitted from the sample is 1000
per second. At ten past two this has fallen to 400 per second. Calculate the
flux at two o’clock. (12 marks)
T1/2 =
−ln 2∗t
ln ( Nt
N 0 )
From the data given period = 10 -5 = 5 seconds
Overall half-life
T1/2 =
−ln2∗5
ln ( 400
1000 )
= 3.7824
But
Nt = N0*
0.5t
t
1
2

Period is t = 10 seconds
Nt = 400
Nt = N0*
0.5t
t
1
2
400 = N0*0.5
10
3.7824
N0 = 2,500 per second
The flux at to o’clock
= 2,500 per second