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First, write out the prime factorization of each number

• #10 = 2 xx 5#
• #15 = 3 xx 5#
• #20 = 2^2 xx 5#
• #30 = 2 xx 3 xx 5#

We can rewrite the above with more clarity as

• #10 = 2^1 xx 3^0 xx 5^color(blue)(1)#
• #15 = 2^0 xx 3^color(blue)(1) xx 5^color(blue)(1)#
• #20 = 2^color(blue)(2) xx 3^0 xx 5^color(blue)(1)#
• #30 = 2^1 xx 3^color(blue)(1) xx 5^color(blue)(1)#

For each prime factor, take the one with the highest exponent. 2 is raised to the power of 2 in 20. 3 and 5 have both a maximum exponent of 1. Refer to the #color(blue)("blue")# colored exponents above.

Therefore,

#"LCM" = 2^2 xx 3^1 xx 5^1#

#= 60#

This algorithm is guaranteed to generate the least common multiple.

#1#. According to B.E.D.M.A.S. (brackets, exponents, division, multiplication, addition, subtraction), work on the exponents first, since the given problem does not contain any brackets. Recall that any number to the power of #0# is always #1#.

#6.73-2*color(teal)(5^0)-:2#

#=6.73-2*color(teal)1-:2#

#2#. Work on multiplication next since multiplication and division are ranked equally. Solve #color(darkorange)(2*1)# since multiplication appears first.

#=6.73-color(darkorange)(2*1)-:2#

#=6.73-color(darkorange)2-:2#

#3#. Work on division next. Solve #color(violet)(2-:2)#.

#=6.73-color(violet)(2-:2)#

#=6.73-color(violet)1#

#4#. Solve.

#=color(green)(|bar(ul(color(white)(a/a)color(black)(5.73)color(white)(a/a)|)))#

Given:
- Craig's older brother's shirts take up #1/6# of the basket
- Craig's younger brother's shirts take up #3/6# of the basket.

You are trying to find:
- how much of the basket belongs to Craig's older or younger brother

Thus, add #1/6# and #3/6# together. Since #1/6# and #3/6# have the same denominator, you can directly add the numerators of the fractions together.

#1/6+3/6#

#=(1+3)/6#

#=4/6#

Since the numerator and denominator have a common factor of #color(red)2#, the fraction is reduced.

#(4color(red)(-:2))/(6color(red)(-:2))#

#=color(green)(|bar(ul(color(white)(a/a)2/3color(white)(a/a)|)))#

The numbers to remember are 5, 9 and 32

Fahrenheit are the lager number when comparing like temperature for like. So to convert Centigrade to Fahrenheit, you need to make the numbers bigger. Thus you multiply by #9/5#. To go from Fahrenheit to centigrade you are making the numbers smaller so you multiply by #5/9#

All you need to remember now is that freezing point for each is different.

Centigrade freezing point is #0^o#
Fahrenheit freezing point is #32^o#

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Using this information to answer your question")#

We are going from Centigrade to Fahrenheit so it is small number to large. So we have #color(blue)(xx9/5)#

We are changing freezing point from 0 on 1 scale to 32 on a different scale so we now have:

#color(blue)(xx9/5+32)#

Thus the full conversion is:

#color(blue)("Fahrenheit temperature "=(29xx9/5)+32 = 84.2^o F )#

Bob's training can be broken down into three steps:

#1#. ran #14.6# miles away from home

#2#. ran #9.8# miles towards his home

#3#. ran #5.3# miles away from his home

Since Bob starts running from his home, he is #0# miles from his home.

If he starts running from his home for #14.6# miles, you would add #14.6# miles to #0# miles to indicate how far Bob ran for his first run. For this problem, when Bob travels away from home, you add the distance he travels. In contrast, when he travels towards his home, you subtract the distance he travels.

#0color(white)(i)"miles"+14.6color(white)(i)"miles"#

When Bob runs towards his home at a distance of #14.6# miles away from his home, you subtract #9.8# miles from #14.6# miles since he is now going in the opposite direction in which he was originally heading.

#0color(white)(i)"miles"+14.6color(white)(i)"miles"-9.8color(white)(i)"miles"#

Finally, when Bob runs #5.3# miles away from his home, opposite to the direction of his second run, you must add #5.3# miles.

#0color(white)(i)"miles"+14.6color(white)(i)"miles"-9.8color(white)(i)"miles"+5.3color(white)(i)"miles"#

If you solve the addition problem, you get #10.1# miles.

#:.#, Bob is #10.1# miles from his home.

Divisibility Rule for #11#

Divide the alternate digits in two different groups. Take the sum of alternate digits separately and find the difference of the two numbers. If the difference is #0# or is divisible #11#, the number is divisible by #11#.

Example: #86456293# is divided into two groups #{8,4,6,9}# and #{6,5,2,3}#. Sum of the groups is #27# and #16#, whose difference is #11# and the it is divisible by #11#, #86456293# is divisible by #11#.

Divisibility Rule for #12#

If the number is divisible by both #3# and #4#, the number is divisible by #12#. Divisibility rule of #3# is tat sum of digits is divisible by #3# and divisibility rule of #4# is that last two digits are divisible by #4#.

Example: In #185176368# sum of all the digits is #45# and is divisible by #3# and also last two digits #68# are divisible by #4#. As such the number #185176368# is divisible by #12#.

Divisibility Rule for #13#

Recall the divisibility rule of #7#, this works for #13# too.

Starting from right mark off the digits in groups of threes (just as we do when we put commas in large numbers).

Now add up alternate group of numbers and find the difference between the two. If the difference is divisible by #13#, entire number is divisible by #13#.

For example #123448789113#, these are grouped as #123#, #448#, #789# and #113#

and #123+789=912# and #448+113=561#.

As difference between #912-561=351#

As #351# is divisible by #13#, #123448789113# is divisible by #13#

If it is a terminating decimal, then multiply it by a power of #10# to make it into an integer, use that power of #10# as the denominator, then simplify it by dividing the numerator and denominator by any common factors.

For example:

#0.16 = 16/100 = (color(red)(cancel(color(black)(4)))*4)/(color(red)(cancel(color(black)(4)))*25) = 4/25#

If it is a repeating decimal, then multiply by a power of #10# to shift the repeating part to just after the decimal point and by a power of #10# corresponding to the length of the repeating section, minus #1#, to get an integer, then divide and simplify.

For example:

#0.2345345345... = 0.2bar(345)#

Multiply by #10(1000-1) = 10000-10# to get an integer:

#(10000-10)0.2bar(345) = 2345.bar(345) - 2.bar(345) = 2343#

So, dividing by #10000-10# we find:

#0.2bar(345) = 2343/(10000-10) = 2343/9990 = (color(red)(cancel(color(black)(3)))*781)/(color(red)(cancel(color(black)(3)))*3330) = 781/3330#

A percent is any number over 100, so 25% is 25 over 100, this means that 25% can be written as:

#25/100#

Now, any number in fractional form (a number over a number) is the same as division so 25% can also be written as a decimal, we just move the decimal point 2 places to the right of 25, like this:

#25/100= 0.25#

Now we have two ways that we can solve this problem! I'll start with the fractional method of solving.

---------------------------

The Fractional Method

So we have a fraction: #25/100#, lets think, how many 25's do we need to get to 100? Well:

#25+25= 50#

and

#50+50= 100#

so that means,

#25+25+25+25= 100#

or

#25*4= 100#

This means that we need four 25's to get 100.

or,

#25/100= 1/4#

(we could also divide both numerator and denominator by 5 to get the same result:

#25/100 div (5/5)= 5/20 div (5/5)= 1/4#

Now that we know 25% is equal to #1/4# we can either multiply 90 by #1/4# or we can divide 90 by 4, its easiest (for me anyway) to just divide 90 by 4.

#90 div 4= 22.5#

---------------------

The Decimal Method

As we found out above 25% equals 0.25 so we can solve this problem using decimals too! All we need to do is multiply 90 by 0.25, which gives us the same answer as above.

#0.25*(90)= 22.5#

---------------------

Despite the fact that that mathematics seemed to be a lot more involved in the fractional method, the calculations were a lot easier and personally I found it to be faster than the decimal method. The decimal method is great when you're working with fractions that aren't equal to 100 (For example: #3/7# of 21 is what number?) or when you have a calculator.

Sorry this is so long, I hope it helped though!

So lets first operate with positive numbers. We should go about this by thinking about a pie, like this:

Courtesy of: http://etc.usf.edu/clipart/40600/40610/pie_01-10a_40610.htm (ClipArt ETC Free Classroom License)

Lets say that the circle above is an apple pie. The apple pie has 10 slices, or parts. If no one takes a piece of the pie then we have all 10 slices of the pie. Since we have all ten slices of pie we can say that we have "10 of the 10 slices" or #10/10#.

#10/10# is a whole, in other words, it is equal to 1, and the equation looks like this:

#10/10= 1#

...or a full pie (this is sweet potato pie):

Courtesy of: http://culinaryphysics.blogspot.com/2015/11/patti-labelle-sweet-potato-pie-recipe-soul-food.html (Public Domain)

Or this (key lime):

My own photo (and baking), feel free to reuse, if you want the recipe shoot me a message

Okay, now lets say that 9 people enter the room and they all take a slice of the pie. That means there is only on piece of pie left #(10-9= 1)#. This means that we have #1/10# of a pie, since 9 slices were taken. Now #1/10# happens to be the same as #0.1#, as an equation it looks like this:

#1/10= 0.1#

We know this because #10/10= 1# and if we add 0.1 ten times we get 1, like this:

#0.1+0.1+0.1+0.1+0.1+0.1+0.1+0.1+0.1+0.1= 1#

There's an underlying tend here, which looks like this:

#1/10= 0.1#

_______

#2/10= 0.2#

_______

#3/10= 0.3#

_______

#4/10= 0.4#

_______

#5/10= 0.5#

_______

#6/10= 0.6#

_______

#7/10= 0.7#

_______

#8/10= 0.8#

_______

#9/10= 0.9#

_______

#10/10= 1#

_______

This exact same trend exists for negative fractions we just have to put a negative sign in front of the decimal. This means that:

#-3/10= -0.3#

I hoped this helped!

When talking about order of operation the PEDMAS rule applies. According to this rule first you have to calculate everything which is in the most inner brackets.

In this expression you have 2 kinds of brackets, but it makes no difference how they look like because you only have to look at those which contain no other brackets, and they are #(5+10*7)#

In those brackets you have 2 operations: addition and multiplication. According to PEDMAS rule you have to do multiplication first, then the addition, so you get:

#44-:2+9-[120-(5+color(red)(10*7))]=#
#44-:2+9-[120-(color(red)(5+70))]=#
#44-:2+9-[color(red)(120-75)]=#
#color(red)(44-:2)+9-45=#
#color(red)(22+9)-45=#
#31-45=-14#

In every line the next operation is marked with the red color exept last line with only one operation (substraction #31-45#).