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3

Answer:

The least common multiple (LCM) of #6# and #9# is #18# and the explanation is down below!

Explanation:

I always take the first 5 multiples of each number and write them down. And if we don't our LCM in there, then I write down the next five multiples! So let's start, shall we?

(First 5) Multiples of #6#:
#6, 12, 18, 24, 30#

(First 5) Multiples of #9#:
#9, 18, 27, 36, 45#

Now, we have to see if there is any common number in the numbers. If we do, we have to make sure that there isn't any other number lower than that that #6# and #9# go into!

(First 5) Multiples of #6#:
#6, 12,##color(blue)(18)##, 24, 30#

(First 5) Multiples of #9#:
#9, ##color(blue)(18)##, 27, 36, 45#

There isn't any number before #18# that #6# and #9# both go into, so #18# is the LCM of #6# and #9#! Now we don't need to write down anymore multiples. :)

Source: My knowledge!

I hope that helps you!

3

Answer:

See a solution process below:

Explanation:

We can rewrite this as:

#(1 xx 1,000,000,000) xx (6 xx 1,000,000,000) =>#

#(1 xx 6) xx (1,000,000,000 xx 1,000,000,000) =>#

#6 xx (1,000,000,000 xx 1,000,000,000)#

To multiply the numbers on the right we combine the number of #0#s #xx 1# giving:

#6 xx (1,color(red)(000,000,000) xx 1,color(blue)(000,000,000))#

#6 xx (1,color(red)(000,000,000),color(blue)(000,000,000)) =>#

#6,000,000,000,000,000,000#

3

Answer:

Alternatively, using scientific notation makes things easier (even though it's technically the same method).

The answer is #" "6 xx 10^18 " " or " " 6,000,000,000,000,000,000#

Explanation:

#1,underbrace(000,000,000)_(color(red)"9 ""zeros") = 1 xx 10^color(red)9#

#6,underbrace(000,000,000)_(color(red)"9 ""zeros") = 6 xx 10^color(red)9#

Therefore:

#1 xx 10^color(red)9 xx 6 xx 10^color(red)9#

#= (1 xx 6) xx (10^color(red)9 xx 10^color(red)9)#

#= 6 xx 10^(color(red)9 + color(red)9)#

#= 6 xx 10^color(blue)18#

#= 6,underbrace(000,000,000,000,000,000)_(color(blue)"18 ""zeros")#

Final Answer

0

Answer:

#5/7#

Explanation:

#20-0=20#
#33-5=28#

#20/28#

Divide both the numerator and the denominator by #4#.
#20-:4=5 #
#28-:4=7#

#:.20/28 = 5/7#

1

Answer:

#83 and 89#

Explanation:

To list the prime numbers between #80 and 90# we can use the rules of divisibility as a start;

#80, 81," "82," "83," "84," "85," "86," "87," "88," "89," "90#

All even numbers are divisible 2, they are not prime. apart from #2#
This leaves all the odd numbers.

#81," "83," "85," "87," "89,#

A number ending in #5# is divisible by #5#

#81," "83," "87," "89,#

If the sum of the digits is divisible by #3#, #3# is a factor

#81" "rarr 8+1=9." "# 81 is not prime

#83" "rarr 8+3 = 11" "# 83 might be prime

#87" "rarr 8+7=15" "# 87 is not prime

#89" "rarr 8+9 = 17" "#89 might be prime

As factors are always in pairs we only need to consider factors less than the square roots.

#sqrt83 = 9.....#

#2,3,4,5,6,7,8,9# are not factors: #83# is prime

#sqrt89 = 9......#

#2,3,4,5,6,7,8,9# are not factors: #89# is prime

1

Answer:

See a solution process below:

Explanation:

First, evaluate the expression within the right parenthesis:

#10 - (-8) - (color(red)(3 - 7))#

#10 - (-8) - (-4)#

Next, rewrite both terms within parenthesis. Remember, minus a minus is a plus:

#10 - color(red)((-8)) - color(blue)((-4))#

#10 + 8 + 4#

Now, evaluate the addition operations left to right:

#color(red)(10 + 8) + 4#

#color(red)(18 + 4)#

#22#

1

Answer:

#(2sqrt6)^6=13824#

Explanation:

#(2sqrt6)^6#

= #2^6xx(sqrt6)^6#

= #(2xx2xx2xx2xx2xx2)xx(sqrt(6xx6xx6xx6xx6xx6))#

= #64xxsqrt(ul(6xx6)xxul(6xx6)xxul(6xx6))#

= #64xx6xx6xx6#

= #64xx216#

= #13824#

1

Answer:

#13.824#

Explanation:

Use the laws of indices.

#(2 sqrt6)^6 = 2^6 xx (sqrt6)^6" "larr ((ab)^m = a^m b^m)#

#=2^6 xx (6)^(6/2)" "larr sqrtx = x^(1/2)#

#=2^6 xx 6^3#

#=(2xx2xx2xx2xx2xx2)xx (6xx6xx6)#

It helps to know the powers less than 1000 by heart.

#= 64 xx 216#

#=13,824#

1

Answer:

#"57.2 ft/s"#

Explanation:

In order to solve this, we will use dimensional analysis.

First, to eliminate the #"mi"# unit, multiply by #"5280 ft" / "1 mi"#. Since #"1 mi = 5280 ft"#, you are essentially multiplying by #1#.

#("156" cancel("mi")) / "4 hr" * color(blue)("5280 ft" / ("1" cancel("mi")))#

Now, we are left with #"ft"/"hr"#. We want the final unit to be #"ft"/"s"#, so we can use the conversion factors #"1 hr = 60 min"# and #"1 min = 60 s"#.

#("156" cancel("mi")) / ("4" cancel("hr")) * "5280 ft" / ("1" cancel("mi")) * color(blue)(("1" cancel("hr")) / "60 min")# #-># cancels out the #"hr"# unit

#("156" cancel("mi")) / ("4" cancel("hr")) * "5280 ft" / ("1" cancel("mi")) * ("1" cancel("hr")) / ("60" cancel("min")) * color(blue)(("1" cancel("min"))/ "60 s")# #-># cancels out the #"min"# unit

Now, we are left with our desired unit of #"ft"/"s"#. Finally, multiply and divide the remaining numbers.

#(156 * "5280 ft") / (4 * 60 * "60 s") = "57.2 ft/s"#

So, #"156 mi" / "4 hr"# #=# #"57.2 ft/s"#.

2

Answer:

#57.2# feet per second

Explanation:

By 'unit rate', we mean how many of the first quantity for ONE of the second unit.

So, by 'feet per second, we mean, how many feet are covered in ONE second?

We can convert the given quantities to the required units:

#156 " miles: " 156 xx 1760 xx 3 = 823,680 " feet"#

miles #rarr# yards #rarr# feet

#4 " hours: " 4 xx 60 xx 60 =14,400 " seconds"#

hours # rarr # minutes # rarr # seconds

#(823,680" feet ")/(14,400" seconds ") =57.2 # feet per second.