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Featured 1 month ago

Use PEMDAS to help with the order of operations:

Pfor Parentheses

Efor Exponents

MDfor Multiplication and Division (left to right)

ASfor Addition and Subtraction (left to right)

So we find:

#(4xx4(6+4)-3(8+1)+2)-:3+(5+2)(5+2)-2xx2+(42-:6)(42-:6)+1#

#=(4xx4(10)-3(9)+2)-:3+(7)(7)-2xx2+(7)(7)+1#

#=(4xx40-27+2)-:3+49-2xx2+49+1#

#=(160-27+2)-:3+49-2xx2+49+1#

#=(133+2)-:3+49-2xx2+49+1#

#=135-:3+49-2xx2+49+1#

#=45+49-4+49+1#

#=94-4+49+1#

#=90+49+1#

#=139+1#

#=140#

Featured 1 month ago

As soon as there are indices, you cannot cancel any of the bases.

Working out each factor to its actual value is not practical - the values are far too big.

Write each base as the product of its prime factors.

You might have noticed that all the bases are powers of 2.

Multiply the indices:

Add the indices of like bases:

Featured 1 month ago

Isolate s by adding

#2/3# to both sides of the equation.Adding the same value to both sides retains the 'balance' of the equation.

#-2/5color(red)(+2/3)=cancel(-2/3)color(red)cancel(+2/3)+s#

#rArrs=-2/5+2/3larrcolor(magenta)"reversing the equation"# Before we can add these fractions we require them to have a

#color(blue)"common denominator"# Multiplying the numerator/denominator of the first fraction by 3 and multiplying the numerator/denominator of the second fraction by 5, gives a common denominator.

#rArrs=(-2/5xx3/3)+(2/3xx5/5)#

#color(white)(xxxx)=-6/15+10/15larrcolor(red)" common denominator"# Now the denominators are the same we can add the numerators.

#rArrs=(-6+10)/15=4/15#

Featured 1 month ago

In scientific notation, we write a number so that it has single digit to the left of decimal sign and is multiplied by an integer power of

Note that moving decimal

Hence, we should either divide the number by

In other words, it is written as

To write

Hence in scientific notation

Featured 4 weeks ago

To add 2 fractions we require the

#color(blue)"denominators"# to be the same value.In this case they are, both 8

We can therefore

#color(blue)"add the numerators"# while leaving the denominator as it is.

#rArr5/8+7/8#

#=(5+7)/8#

#=12/8# We can

#color(blue)"simplify"# the fraction by dividing the numerator/denominator by the#color(blue)"highest common factor"# of 12 and 8, which is 4

#rArr12/8=(12Ã·4)/(8Ã·4)=3/2larrcolor(red)" in simplest form"# This process is normally done using

#color(blue)"cancelling"#

#rArr12/8=cancel(12)^3/cancel(8)^2=3/2larrcolor(red)" in simplest form"# A fraction is in

#color(blue)"simplest form"# when no other factor but 1 divides into the numerator/denominator.

Featured 2 weeks ago

There are 2 possible approaches to this calculation, both made fairly 'awkward' due to the values on the denominators of the fractions.

#color(red)"Approach 1"# Change the

#color(blue)"mixed numbers " "to "color(blue)"improper fractions"#

#rArr5 1/16=81/16" and "1 13/18=31/18# The calculation is now.

#81/16-31/18# Before we can subtract the fractions we require them to have a

#color(blue)"common denominator"# We have to find the

#color(blue)"lowest common multiple"# ( LCM) of 16 and 18The LCM of 16 and 18 is 144

#rArr81/16xx9/9=729/144" and "31/18xx8/8=248/144#

#rArr729/144-248/144larrcolor(red)" is now the calculation"# Since the denominators are now common we can subtract the numerators, leaving the denominator as it is.

#rArr729/144-248/144=(729-248)/144#

#=481/144=3 49/144larrcolor(red)" returning a mixed number"#

#color(red)"Approach 2"#

#"Using the fact that "5 1/16=5+1/16;1 13/18=1+13/18#

#"Then "5 1/16-1 13/18#

#=5+1/16-(1+13/18)=5+1/16-1-13/18# We can now subtract the numbers and subtract the fractions separately.

#rArr5+1/16-1-13/18=(5-1)+1/16-13/18#

#=4+(1/16xx9/9-13/18xx8/8)#

#=4+(9/144-104/144)#

#=4+(-95/144)#

#=4-95/144#

#=576/144-95/144#

#=481/144#

#=3 49/144larrcolor(red)" as a mixed number"#

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