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2

Answer:

4

Explanation:

For this problem, we will need to have knowledge of the order of operations.

Parentheses
Exponents
Multiplication
Division
Addition
Subtraction.

In order from top to bottom.

..................................................................................................................

Using the order of operations we know that first we need to solve what is inside the parentheses.

#22- 2( 13- 7+ 3)#

#22- 2( 6+3)#

#22- 2( 9)#

..................................................................................................................

Now, we are going to do the multiplication, because that is before subtraction on our order of operations.

#22- 2( 9)#

#22- 18#

..................................................................................................................

Now, we just have to do basic subtraction.

#22-18=4#

This brings us to our final answer of 4.

2

Answer:

Find factor pairs.

#F_44= {1," "2," "4," "11," "22," "44}#

Explanation:

A factor is a number which divides into a bigger number without leaving a remainder. A number is always a factor of itself.

Factors are always in pairs. Dividing by one factor will give you the other.

To find all the factors of a number, start from 1 and consider all the possible factors up to the square root of the number.
(The square root lies in the middle of the factors arranged in order.)

#sqrt44 = 6. ....# The only numbers we need to consider are

#1," "2" "3" "4" "5" "6#

#1# is a factor: The factor pair is #color(blue)(1 xx44)#
#2# is a factor: The factor pair is #color(blue)(2xx22)#
#3# is not a factor of 44
#4# is a factor: The factor pair is #color(blue)(4xx11)#
#5# is not a factor of 44
#6# is not a factor of 44

The factors of 44 are therefore:

#color(blue)(1," "2," "4," "11," "22," "44)#
#color(white)(......................)color(red)(uarr)#
#color(white)(.....................)color(red)(sqrt44)#

Note that:

1 and the number itself are always factors of any number

Using a factor tree or prime factors in this case would not help, because they do not indicate all the possible combinations of the prime factors.

1

Answer:

12

Explanation:

I like to answer these questions by first doing prime factorizations:

#60=2xx30=2xx2xx15=2xx2xx3xx5#
#72=2xx36=2xx2xx18=2xx2xx2xx9=2xx2xx2xx3xx3#

The GCF will have all the numbers that are common to both 60 and 72.

#GCF=?#

Let's start with 2's first. 60 has two 2s, while 72 has three. Two 2s are common to both, so our GCF will have two 2s:

#GCF=2xx2xx?#

Now 3s. 60 has one 3 and 72 has two 3s. One 3 is common to both, so we'll put that in:

#GCF=2xx2xx3xx?#

And lastly to 5s - 60 has one but 72 doesn't, so there's nothing common there. And so we stop here:

#GCF=2xx2xx3=12#

#60=12xx5#
#72=12xx6#

Tony B

2

Answer:

#x=48#

Explanation:

#color(blue)("Shortcut method that jumps steps from first principle method")#

#color(brown)("Shortcut methods are much faster than first principles.")##color(brown)("Which is why people use them. However the shortcut outcome is")##color(brown)("based on the outcome of the first principle method.")#

#x/6=8#

The 6 has the process of divide applied to it in that
#x/6 = x xx1/6 = x-:6#

As the 6 has divide applied to it on the left of = it becomes the opposite of multiply when we move it to the other side of the =. So we have:

#x-:6=8" "->" "x=8xx6#

#x=48#

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("3 very important facts used in first principle method")#

  1. An equation by definition is stating that what is on one side of = has the same value as on the other side. It is just that they may look different. So if you change one side in some way you have to change the other side in exactly the same way

  2. For multiply and divide: to move something to the other side of the = change it to 1 as #color(brown)(1xx"something = something")#.

  3. For add and subtract: to move something to the other side of = change it to 0 as #color(brown)(0+" something = something")#.
    ...................................................................................................................
    #color(blue)("Answering your question using first principles")#

#x/6=8#

#x/6# is the same as #1/6xx x#

So to get #x# on its own we need to change #1/6" into "1#

Multiply both sides by #color(red)(6)#

#color(green)(x/6=8" "->" "color(red)(6xx)x/6" "=" "color(red)(6xx)8 )#

#" "color(green)((color(red)(6))/6 xx x" "=" "48#

#" "color(green)(1xx x" "=" "48)#

But #1xx x" gives just "x#

#" "color(green)(x" "=" "48#

1

Answer:

LCM is 120

Explanation:

#color(blue)("Method 1")#

Just looking at and considering the numbers:

#color(brown)("Observation 1")#

A multiple of 10 by any whole positive number will have 0 as the last digit. So the least common multiple has 0 for the last digit.
....................................................................................................

#color(brown)("Observation 2")#
All the numbers are even so the common multiple has to be even thus divisible by 2. This is what makes the derivation more straightforward.
.........................................................................................................

#color(brown)("Observation 3")#

Consider the 2 from 12: the multiple of 12 has to end in 0

note the least value times 12 to give 0 as a last digit is 5 and #5xx12=60#

But 60 does not divide exactly by 8 so using observation 2 multiply 60 by 2 giving 120
++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
#color(blue)("Method 2")#

You would probably find it very useful to remember some of the prime numbers. Gradually build your memory up to the number 101.
Do an image search in google for Prime Factors.

The first few are: #2;3;5;7;11;13#
The method with prime factors is to use the maximum count of factors for any one the numbers given and multiply them all together. So we have:
Tony B

So we have:#" " (2xx2xx2)xx(3)xx(5) = 120#

1

Answer:

A slightly different approach

#x=28#

Explanation:

#color(blue)("Introduction to the idea of the method")#

Treating it like a ratio

Given:#" "15/7=60/x#

Notice that 5 is divisible into both 60 and 15.

If we can change the 15 into 60 and maintain proportionality then we can determine the value of #x#

To convert 15 into 60 first turn it into 1 then multiply by 60 to turn the 1 into 60 #-> 15xx1/15xx60" " ->" "15xx60/15#

But #60/15" is the same as "4#

To maintain proportionality what you do to the top you do to the bottom (for multiply and divide).
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("The actual calculation")#

#(15xx4)/(7xx4) =60/28 = 60/x#

#x=28#

1

Answer:

I found:
Oranges: #$0.26#
Apples: #$0.32#

Explanation:

Let us call the cost of oranges #Or# and apples #Ap#. We can write:

#12Or +7Ap=5.36#
and
#8Or +5Ap=3.68#

we can extract from the first:

#Or=(5.36-7Ap)/12#

we substitute this into the second equation for #Or#:

#cancel(8)^2*(5.36-7Ap)/cancel(12)^3+5Ap=3.68#

solve for #Ap#:

#10.72-14Ap+15Ap=11.04#

#Ap=0.32#

use this value bach into the first equation:

#Or=(5.36-7Ap)/12=(5.36-7*0.32)/12=0.26#

1

Answer:

They both have the same value so one is not greater than the other

Explanation:

A fraction consists of: #("count")/("size indicator")->("numerator")/("denominator")#

You can not directly compare counts unless the size indicators are the same.

Multiply by 1 and you do not change the value. However 1 comes in many forms so you can change the way a number looks without changing its true value.

#color(green)([3/5color(red)(xx1)]" compare to "9/15#

#color(green)([3/5color(red)(xx3/3)]" compare to "9/15#

# " "color(green)(9/15" compare to "9/15#

They both have the same value so one is not greater than the other

1

Answer:

here is how...

By the way, it is #{1,2,5,7,10,14,35,70}#

Explanation:

division is needed here, but it needs to be done multiple times so I will put in a diagram

Here is a pic of how...
Created by me!

Sorry if the glitter effects get in the way...I had too much fun creating it...

anyway...now you just put them in fancy brackets that look like this...#{put, stuff, here}#

1

Answer:

#-16#

Explanation:

When evaluating an expression with #color(blue)"mixed operations"# there is a particular order that must be followed.

Following the order as set out in the acronym PEMDAS

[Parenthesis(brackets) , Exponents(powers), Multiplication, Division, Addition, Subtraction ]

#rArr8+6xx(4-20)/2^2#

#=8+6xx(-16)/4larrcolor(red)"brackets/powers"#

When multiplication/division are in the calculation, as here, we evaluate from left to right.

#rArr8+(6xx(-16))/4#

#=8+(-96)/4larrcolor(red)" multiplication"#

#=8+(-24)larrcolor(red)" division"#

#=8-24=-16#