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{
"cells": [
{
"cell_type": "markdown",
"metadata": {},
"source": [
"# Chapter - I : Introduction to powers of 10"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example No. I_9 Page No. 9"
]
},
{
"cell_type": "code",
"execution_count": 24,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"The addition of 170*10**3 and 23*10**4 =4.00E+05\n"
]
}
],
"source": [
"# Add 170*10**3 and 23*10**4. Express the final answer in scientific notation.\n",
"\n",
"# Given data\n",
"\n",
"A = 170*10**3# # Variable 1\n",
"B = 23*10**4# # Variable 2\n",
"\n",
"C = A+B#\n",
"print 'The addition of 170*10**3 and 23*10**4 =%0.2E'%C"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example No. I_10 Page No. 9"
]
},
{
"cell_type": "code",
"execution_count": 25,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"The substraction of 250*10**3 and 1.5*10**6 =1.25E+06\n"
]
}
],
"source": [
"# Substract 250*10**3 and 1.5*10**6. Express the final answer in scientific notation.\n",
"\n",
"# Given data\n",
"\n",
"A = 1.5*10**6# # Variable 1\n",
"B = 250*10**3# # Variable 2\n",
"\n",
"C = A-B#\n",
"print 'The substraction of 250*10**3 and 1.5*10**6 =%0.2E'%C"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example No. I_11 Page No. 10"
]
},
{
"cell_type": "code",
"execution_count": 26,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"The multiplication of 3*10**6 by 150*10**2 =4.50E+10\n"
]
}
],
"source": [
"# Multiply 3*10**6 by 150*10**2. Express the final answer in scientific notation.\n",
"\n",
"# Given data\n",
"\n",
"A = 3*10**6# # Variable 1\n",
"B = 150*10**2# # Variable 2\n",
"\n",
"C = A*B#\n",
"print 'The multiplication of 3*10**6 by 150*10**2 =%0.2E'%C"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example No. I_12 Page No. 10"
]
},
{
"cell_type": "code",
"execution_count": 27,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"The division of 5.0*10**7 by 2.0*10**4 =2.50E+03\n"
]
}
],
"source": [
"# Divide 5.0*10**7 by 2.0*10**4. Express the final answer in scientific notation.\n",
"\n",
"# Given data\n",
"\n",
"A = 5.0*10**7# # Variable 1\n",
"B = 2.0*10**4# # Variable 2\n",
"\n",
"C = A/B#\n",
"print 'The division of 5.0*10**7 by 2.0*10**4 =%0.2E'%C"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example No. I_13 Page No. 10"
]
},
{
"cell_type": "code",
"execution_count": 28,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"The reciprocal of 10**5 =1.00e-05\n",
"i.e 10**-5\n",
"The reciprocal of 10-3 = 1.00e+03\n",
"i.e 10**3\n"
]
}
],
"source": [
"# Find the reciprocals for the following powers of 10: (a) 10**5 (b) 10**-3.\n",
"\n",
"# Given data\n",
"\n",
"A = 10**5# # Variable 1\n",
"B = 10**-3# # Variable 2\n",
"\n",
"C = 1./A#\n",
"print 'The reciprocal of 10**5 =%0.2e'%C\n",
"print 'i.e 10**-5'\n",
"\n",
"D = 1./B#\n",
"print 'The reciprocal of 10-3 = %0.2e'%D\n",
"print 'i.e 10**3'"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example No. I_14 Page No. 11"
]
},
{
"cell_type": "code",
"execution_count": 29,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"The square of 3.0*10**4 =9.00E+08\n"
]
}
],
"source": [
"# Square 3.0*10**4. Express the answer in scientific notation.\n",
"\n",
"# Given data\n",
"\n",
"A = 3.0*10**4# # Variable 1\n",
"\n",
"B = A*A#\n",
"print 'The square of 3.0*10**4 =%0.2E'%B"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example No. I_15 Page No. 11"
]
},
{
"cell_type": "code",
"execution_count": 30,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"The squareroot of 4*10**6 = 2.00E+03\n"
]
}
],
"source": [
"from math import sqrt\n",
"# Find the squareroot of 4*10**6. Express the answer in scientific notation.\n",
"\n",
"# Given data\n",
"\n",
"A = 4*10**6# # Variable 1\n",
"\n",
"B = sqrt(A)#\n",
"print 'The squareroot of 4*10**6 = %0.2E'%B"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example No. I_16 Page No. 12"
]
},
{
"cell_type": "code",
"execution_count": 31,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"The squareroot of 90*10**5 =3.00E+03\n"
]
}
],
"source": [
"from math import sqrt\n",
"# Find the squareroot of 90*10**5. Express the answer in scientific notation.\n",
"\n",
"# Given data\n",
"\n",
"A = 90*10**5# # Variable 1\n",
"\n",
"B = sqrt(A)#\n",
"print 'The squareroot of 90*10**5 =%0.2E'% B"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example No. I_17 Page No. 12"
]
},
{
"cell_type": "code",
"execution_count": 32,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"The multiplication of 40*10**-3 by 5*10**6 =200000.00\n",
"i.e 200.000*10**03 OR 200E03\n"
]
}
],
"source": [
"# Show the keystrokes for multiplying 40*10**-3 by 5*10**6.\n",
"\n",
"# Given data\n",
"\n",
"A = 40*10**-3# # Variable 1\n",
"B = 5*10**6# # Variable 2\n",
"\n",
"C = A*B#\n",
"print 'The multiplication of 40*10**-3 by 5*10**6 =%0.2f'%C\n",
"print 'i.e 200.000*10**03 OR 200E03'"
]
}
],
"metadata": {
"kernelspec": {
"display_name": "Python 2",
"language": "python",
"name": "python2"
},
"language_info": {
"codemirror_mode": {
"name": "ipython",
"version": 2
},
"file_extension": ".py",
"mimetype": "text/x-python",
"name": "python",
"nbconvert_exporter": "python",
"pygments_lexer": "ipython2",
"version": "2.7.9"
}
},
"nbformat": 4,
"nbformat_minor": 0
}
|
