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2

Answer:

Yes

Explanation:

If you divide seven by nine, you get a repeating decimal of 0.77777777.
However, when you divide two by three, you also get a repeating decimal but smaller. 0.66666666

2

Answer:

Yes. If you change the fraction of 2/3 to something with the denominator of 9, you can compare the numerators easily and directly.

Explanation:

You would do this by multiplying 2/3 by 3/3 to make 6/9, which is smaller than 7/9.

3

Answer:

It is #"multiplication"#, not addition

It should be #1 xx 10^3#

Explanation:

#"10  cm"^2 = "1000  mm"^2#

In scientific notation, #1000 = 10^3#

So #"1000  mm"^2 = 1xx10^3" mm"^2"#

3

Answer:

Please see below.

Explanation:

.

There are infinite number of fractions that are equal to #1/3#.

You can multiply the numerator and denominator of #1/3# by any number and the resulting fraction would be equal to #1/3# so long as you are multiplying both by the same number.

For example:

#1/3=(1*2)/(3*2)=2/6#

#1/3=(1*4000)/(3*4000)=4000/12000#

and so on and so forth.

3

Answer:

#2xx2xx2xx5xx5xx7#

Explanation:

To find the prime factorization of #1400#, we need to break it down into prime factors.

Lets use these steps I found in here: https://www.wikihow.com/Find-Prime-Factorization Follow along!

Step 1: Understand factorization. Hopefully you do, but just in case I'll explain.

  • Factorization: the process of breaking a larger number into smaller numbers (algebraic definition)

Step 2: Know prime numbers. They are basically numbers that can only be factored by 1 and itself. e.g. 5 (#5xx1#), 47 (#47xx1#)

Step 3: Start with the number, which is #1400#. It is always helpful to rewrite the problem, for it is easy to make mistakes if you don't.

Step 4: Start by factoring the number into any two factors.

  • #1400#: #200xx7#

Step 5: If the factorization continues, start a factorization tree, so it is less vulnerable to mistakes.
- #1400#
-tttt^
- #200# #7#

Step 6: Continue factorization.

  • #1400#
  • tttt^
  • #200# #7#
  • ttt^
  • #100# #2#
  • ttt^
  • #50# #2#
  • ttt^
  • #25# #2#
  • t^
  • #5# #5#

Step 7: Note any Prime numbers.

  • #1400#
  • tttt^
  • #200# #color(red)7#
  • ttt^
  • #100# #color(red)2#
  • ttt^
  • #50# #color(red)2#
  • ttt^
  • #25# #color(red)2#
  • t^
  • #color(red)5# #color(red)5#

Step 8: Finish factorization. I already did this in the #6th# step, so...

Step 9: Finish by writing the line of prime factors neatly in increasing order.

  • #color(blue)(1400: 2xx2xx2xx5xx5xx7)#
1

Answer:

#540#

Explanation:

we have

#hcf(84,b)=12#

#lcm(84,b)=3780#

to find #b#

we have the well known relationship

#ab=hcf(a,b)xxlcm(a,b)#

#84b=12xx3780#

#b=(cancel(12)xx3780)/cancel(84)^7#

#b=3780/7=540#

1

Answer:

The other number is #540#

Explanation:

For any questions involving HCF and LCM,find each number as the product of the prime factors. That will tell you what you are working with.

#" "84 = color(blue)(2xx2xx3)xxcolor(red)(7)#
#" "? = ul(color(blue)(?xx?xx?)xx?xx?xx?xx?)#

#HCF = color(blue)(2xx2xx3) color(white)(wwwwwwwwwwwwww)= 12#
#LCM = color(blue)(2xx2xx3)xxcolor(red)(7)xxcolor(purple)(3xx3xx5)" " = 3780#

#3780 = color(blue)(HCF) xx color(red)(7) xx" another number"#
#3780 =" " color(blue)(12) xx color(red)(7)xxcolor(purple)((3xx3xx5))#
#3780 = " "color(blue)(12) xxcolor(red)(7) xxcolor(purple)(45)#

The other number cannot have #7# as a factor otherwise the #HCF# would be #84#

The other number is therefore #12 xx 45 =540#

1

Answer:

The prime factors of #1400 " are " 2,5,7#

#1400 = 2xx2xx2xx5xx5xx7#

Explanation:

The intention of the question is not absolutely clear....

Is it asking which of the factors of #1400# are prime numbers?

Or

Is it asking for #1400# to be written as the product of its prime factors.

It will help to write #1400# as the product of its prime factors anyway..

Divide #1400# by prime numbers which are factors until you get #1#

#2 |ul(color(white)(.)1400)#
#2 |ul(" "700)#
#2 |ul(" "350)#
#5 |ul(" "175)#
#5 |ul(" "35)#
#7 |ul(" "7)#
#color(white)(..ww...)1#

The prime factors of #1400 " are " 2,5,7#

As the product of its prime factors:

#1400 = 2xx2xx2xx5xx5xx7#

1

Answer:

#1/3 = 2/6 =3/9 = 4/12 = 5/15 = 9/27 = 14/42 = 17/51 = 50/150 .....#

Explanation:

There are many fractions which are equivalent to #1/3# but this is the simplest form.

To find an equivalent fraction, multiply #1/3 # by #1#, but with #1# written as #2/2, 3/3, 4/4# etc

Multiplying the numerator and denominator by the same number does not change the value of a fraction.

#1/3 xx 2/2 = 2/6#

#1/3 xx 7/7 = 7/21#

#1/3 = 2/6 =3/9 = 4/12 = 5/15 = 9/27 = 14/42 = 17/51 = 50/150 # etc

2

Answer:

#3/35#

Explanation:

When you divide #3//5# by #7#, you can think of it as making that #3//5# 7 times smaller.

If #1//5# is what we get when we take a whole and chop it up into 5 equal pieces:

https://www.mathsisfun.com/fractions-interactive.html

Then #1//(5*7)# or #1//35# is the value we get when we take #1//5# and divide it into #7# pieces:

https://www.mathsisfun.com/fractions-interactive.html

And, of course, we have 3 of these divided pieces, giving us a total of #3*1//35=3//35#.