25/100 in simplest form What do I divide it by

When would the product of the denominators and the least common denominator of the denominators be the same?

Bas_tet [7]

Answer:

Example 1:

Find the common denominator of the fractions.

16 and 38

We need to find the least common multiple of 6 and 8 . One way to do this is to list the multiples:

6,12,18,24−−,30,36,42,48,...8,16,24−−,32,40,48,...

The first number that occurs in both lists is 24 , so 24 is the LCM. So we use this as our common denominator.

Listing multiples is impractical for large numbers. Another way to find the LCM of two numbers is to divide their product by their greatest common factor ( GCF ).

Example 2:

Find the common denominator of the fractions.

512 and 215

The greatest common factor of 12 and 15 is 3 .

So, to find the least common multiple, divide the product by 3 .

12⋅153=3 ⋅ 4 ⋅ 153=60

If you can find a least common denominator, then you can rewrite the problem using equivalent fractions that have like denominators, so they are easy to add or subtract.

Example 3:

Add.

512+215

In the previous example, we found that the least common denominator was 60 .

Write each fraction as an equivalent fraction with the denominator 60 . To do this, we multiply both the numerator and denominator of the first fraction by 5 , and the numerator and denominator of the second fraction by 4 . (This is the same as multiplying by 1=55=44 , so it doesn't change the value.)

512=512⋅55=2560215=215⋅44=860

512+215=2560+860 =3360

Note that this method may not always give the result in lowest terms. In this case, we have to simplify.

=1130

The same idea can be used when there are variables in the fractions—that is, to add or subtract rational expressions .

Example 4:

Subtract.

12a−13b

The two expressions 2a and 3b have no common factors, so their least common multiple is simply their product: 2a⋅3b=6ab .

Rewrite the two fractions with 6ab in the denominator.

12a⋅3b3b=3b6ab13b⋅2a2a=2a6ab

Subtract.

12a−13b=3b6ab−2a6ab =3b − 2a6ab

Example 5:

Subtract.

x16−38x

16 and 8x have a common factor of 8 . So, to find the least common multiple, divide the product by 8 .

16⋅8x8=16x

The LCM is 16x . So, multiply the first expression by 1 in the form xx , and multiply the second expression by 1 in the form 22 .

x16⋅xx=x216x38x⋅22=616x

Subtract.

x16−38x=x216x−616x =x2 − 616x\

4 0

3 years ago

Help ASAP Given the function f(x) = 3x + 1, what is the value of f(-3)

vlada-n [284]

Answer:

-8

Step-by-step explanation:

f(x) = 3x + 1

f(-3) = 3(-3) + 1

F(-3)=-9+1

f(-3)= -8

3 0

3 years ago

How many integers between 10 and 60, inclusive, can be evenly divided by neither 2 nor 3?

erastovalidia [21]

Find total number of integers.

$a_1=10,a_n=60,d=1 \\a_n=a_1+(n-1)d \\60=10+(n-1)1 \\n-1=50 \\n=51$

Find how many integers is divisible by 2.

$a_1=10,a_n=60,d=2
\\a_n=a_1+(n-1)d
\\60=10+(n-1)2
\\2(n-1)=50
\\n-1=25
\\n=26$

Eliminate even numbers.

11, 13, 15,..., 57, 59

This array contains 51 - 26 = 25 numbers.

Eliminate numbers before the first number divisible by 3 and after the last number divisible by 3.

15, 17, 19,..., 55, 57

This array contains 25 - 3 = 22 numbers.

Now we should eliminate numbers divisible by 3: 15, 21, 27...

$a_1=15,a_n=57,d=6,n=?
\\a_n=a_1+(n-1)d
\\57=15+(n-1)6
\\6(n-1)=42
\\n-1=7
\\n=8$

There are 8 such numbers.

Therefore, there are 25 - 8 = 17 numbers that can be evenly divided by neither 2 nor 3

6 0

3 years ago

The sum of the digits of a two-digit number is 8. If the digits are interchanged ,new number is 10 more than double the original

vova2212 [387]

Answer:

Original number 26.

Step-by-step explanation:

xy - two-digit number

1) x + y = 8

2) Original two-digit number can be written as

10*x + y

3) If the digits interchanged yx,

then the new number can be written as

10*y + x

4) Double the original number is

2*(10*x + y)

5) New number is 10 more than double the original number

(10*y + x) - (2*(10*x + y)) = 10

6) Now we have the system of 2 equations:

x + y = 8

(10*y + x) - (2*(10*x + y)) = 10 -----> 10y + x - (20x + 2y) = 10 ---> 8y - 19x = 10

x = 8 - y

8y - 19(8 - y) = 10

8y - 152 +19y = 10

27y = 162

y = 6

x = 8 - y = 8 - 6 = 2

x = 2

So, x =2, y = 6.

Original number 26.

Check:

Original number 26.

New number 62.

Double of the original number = 2*26= 52.

New number is 10 more than double the original number :

62 - 52 = 10 True

6 0

3 years ago

Dan has bags of candy with two pieces in each. Bill has bags of candy with five pieces in each. Bill has six more bags than Dan.

Anna35 [415]

Answer:

Bill has bags of candy with five pieces in each. The Bill has 16 bags

Solution:

Given, Dan has bags of candy with two pieces in each.

Bill has bags of candy with five pieces in each.

Bill has six more bags than Dan.

Then, number of bags with bill = 6 + number of bags with dan.

The total number of pieces candy the two boys have 100.

Now, candies with bill + candies with dan = 100

⇒ 5 pieces per bag x number of bags with bill + 2 pieces per bag x number of bags with dan = 100

⇒ 5 x (6 + number of bags with dan) + 2 x number of bags with dan = 100

⇒ 30 + 5 x number of bags with dan + 2 x number of bags with dan = 100

⇒ (5 + 2) x number of bags with dan = 100 – 30

⇒ 7 x number of bags with dan = 70

⇒ Number of bags with dan = 10

So, number of bags with bill = 6 + 10 = 16

Hence, bill has 16 bags.

7 0

3 years ago