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Answer:

If the denominators are the same just divide the numerators.

#3-:5->3/5#

Explanation:

A fraction consists of #("count")/("size indicators of what you are counting")#

Using the allocated names we have:

#("count")/("size indicator")->("numerator")/("denominator")#

Consider whole numbers. For example 6 and 3

These can, and may, be written as #6/1 and 3/1# They are rational numbers. It is not normally done but never the less it is correct.

Now consider # 6-:3->6/1-:3/1#

#color(blue)("You just divide the counts as the 'size indicators' are the same")#

#6-:3->6/1-:3/1->6/3=2#

#color(magenta)("You have been applying this principle for a long time without even realizing that you were.")#

Answer:

8

Explanation:

The lowest common multiple is the lowest number which is a multiple of both numbers.

Since 8 * 1 = 8 and 2 * 4 = 8, 8 is the lowest common multiple.

An easy way to find the LCM is to list all multiples of each number, up to the product of both numbers (in this case, 16).

For 2:
- 2, 4, 6, 8, 10, 12, 14, 16

For 8:
- 8, 16,

8 is the lowest number they both have in common.

The highest possible value of an LCM is both numbers multiplied by each other. This is why for both lists i stopped at 16, because
2*8 = 16

Answer:

His average speed is #"12 km/hr"#.

Explanation:

Remember the triangle of Speed, Distance and Time. If you remember it, you'll ace these kinda questions.

http://www.bbc.co.uk/bitesize

I had trouble with these formulas but the triangle helped me a LOT! Anyways, let's get back to the question. The formula for average speed is pretty much the same as the formula for just speed. The average speed formula is

#"Average speed" = "Total Distance"/"Total Time"#

So......

#"48 km"/"4 hr" = "12 km/hr"#

My source

I hope this explanation helps you!

Answer:

#550/3000 = 11/60#

Explanation:

When you write a fraction you are showing :

#("part")/("whole")#

In order to do this the units have to be the same, so you can write #3km # as #3000m#

#550m# is the 'part' and #3000m# is the 'whole'.

The fraction is : #(550m)/(3000m)#

Note:

  • the units are the same, so they will cancel.
    The fraction does not have a unit, it merely indicates how the two quantities compare with each other.

  • 'fraction' generally refers to a common fraction.

  • fractions should be given in the simplest form, Divide the numerator and denominator by the highest common factor

#(550 div 50)/(3000 div 50) = 11/60#

Answer:

#1/3# lies in the range #(1/4,2/5)#, #11/12# lies in the range #(13/15,1)#, but #4/7# does not lie in any of the given range.

Explanation:

If #a/b < c/d# then #(color(blue)ma+color(red)nc)/(color(blue)mb+color(red)nd)# lies between #(a/b,c/d)#, if #color(blue)m>0# and #color(red)n>0#

Observe that #1/3=3/9=(color(blue)1xx1+color(red)1xx2)/(color(blue)1xx4+color(red)1xx5)#

Hence #1/3# lies in the range #(1/4,2/5)#

Similarly as #13/15<1# i.e. #13/15<9/9#, #(color(blue)1xx13+color(red)1xx9)/(color(blue)1xx15+color(red)1xx9)=(13+9)/(15+9)=22/24=11/12#

Hence #11/12# lies in the range #(13/15,1)#.

Although above is an interesting way for such problems, the easiest is to convert every number to decimal form.

Hence numbers are #4/7=0.57#, #1/3=0.33# and #11/12=0.916#

and ranges are #(0,0.25)#, #(0.866,1)# and #(0.25,0.40)#

and as is seen #1/3# lies in the range #(1/4,2/5)#, #11/12# lies in the range #(13/15,1)#, but #4/7# does not lie in any of the given range.

Additional Information #-# If in #(color(blue)ma+color(red)nc)/(color(blue)mb+color(red)nd)#, one of the #m# or #n# is negative, #(color(blue)ma+color(red)nc)/(color(blue)mb+color(red)nd)# gives out a number which is outside the range.

Observe that #(color(blue)((-9))xx0+color(red)4xx1)/(color(blue)((-9))xx1+color(red)4xx4)=(color(blue)3xx13+color(red)((-3))xx1)/(color(blue)3xx15+color(red)((-3))xx1)=(color(blue)((-2))xx1+color(red)3xx2)/(color(blue)((-2))xx4+color(red)3xx5)=4/7#

Answer:

The three integers are #40#, #41#, and #42#.

Explanation:

There are two ways to solve this problem.

A] Since we are looking for three consecutive integers that add up to #123#, if we divide the total by #3#, we get the average, which is also the middle integer.

It follows that the first and third integers will be that integer minus and plus one respectively.

#123/3=41#

#41# is the middle integer.

#41-1=40#

#40# is the first integer.

#41+1=42#

#42# is the third integer.

Hence, the three integers are #40#, #41#, and #42#.


B] The other way to solve this problem involves very basic algebra (even though this question is in Prealgebra), in which we represent the first integer as #x#, in which case the second and third integers will be #x+1# and #x+2#.

Since we know the sum of these three integers is #123#, we can write:

#x+x+1+x+2=123#

Adding the #x# separately and the numbers separately, we get:

#3x+3=123#

Subtract #3# from each side.

#3x+3-3=123-3#

#3x+cancel3-cancel3=120#

#3x=120#

Divide each side by #3#.

#(3x)/3=120/3#

#(cancel3x)/(cancel3)=40#

#x=40#

#:.x+1=40+1=41# and #x+2=40+2=42#

Hence the three consecutive integers are #40#, #41#, and #42#.

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