1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
creativ13 [48]
3 years ago
14

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

Mathematics
1 answer:
Bas_tet [7]3 years ago
4 0

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\

You might be interested in
Billy earned $150 from a summer job to buy new clothes for school. He found several items he would like to buy, but is trying to
lisabon 2012 [21]

Answer:

Billy went from rich to broke.

Step-by-step explanation:

8 0
3 years ago
I need help with this please! ASAP<br> (PLEASE MULTIPLY AND SIMPLIFY COMPLETELY)
ivanzaharov [21]

Answer:

6/25

Step-by-step explanation:

7 0
3 years ago
Read 2 more answers
A tank contains 60 kg of salt and 1000 L of water. Pure water enters a tank at the rate 6 L/min. The solution is mixed and drain
MissTica

Answer:

(a) 60 kg; (b) 21.6 kg; (c) 0 kg/L

Step-by-step explanation:

(a) Initial amount of salt in tank

The tank initially contains 60 kg of salt.

(b) Amount of salt after 4.5 h

\text{Let A = mass of salt after t min}\\\text{and }r_{i} = \text{rate of salt coming into tank}\\\text{and }r_{0} =\text{rate of salt going out of tank}

(i) Set up an expression for the rate of change of salt concentration.

\dfrac{\text{d}A}{\text{d}t} = r_{i} - r_{o}\\\\\text{The fresh water is entering with no salt, so}\\ r_{i} = 0\\r_{o} = \dfrac{\text{3 L}}{\text{1 min}} \times \dfrac {A\text{ kg}}{\text{1000 L}} =\dfrac{3A}{1000}\text{ kg/min}\\\\\dfrac{\text{d}A}{\text{d}t} = -0.003A \text{ kg/min}

(ii) Integrate the expression

\dfrac{\text{d}A}{\text{d}t} = -0.003A\\\\\dfrac{\text{d}A}{A} = -0.003\text{d}t\\\\\int \dfrac{\text{d}A}{A} = -\int 0.003\text{d}t\\\\\ln A = -0.003t + C

(iii) Find the constant of integration

\ln A = -0.003t + C\\\text{At t = 0, A = 60 kg/1000 L = 0.060 kg/L} \\\ln (0.060) = -0.003\times0 + C\\C = \ln(0.060)

(iv) Solve for A as a function of time.

\text{The integrated rate expression is}\\\ln A = -0.003t +  \ln(0.060)\\\text{Solve for } A\\A = 0.060e^{-0.003t}

(v) Calculate the amount of salt after 4.5 h

a. Convert hours to minutes

\text{Time} = \text{4.5 h} \times \dfrac{\text{60 min}}{\text{1h}} = \text{270 min}

b.Calculate the concentration

A = 0.060e^{-0.003t} = 0.060e^{-0.003\times270} = 0.060e^{-0.81} = 0.060 \times 0.445 = \text{0.0267 kg/L}

c. Calculate the volume

The tank has been filling at 6 L/min and draining at 3 L/min, so it is filling at a net rate of 3 L/min.

The volume added in 4.5 h is  

\text{Volume added} = \text{270 min} \times \dfrac{\text{3 L}}{\text{1 min}} = \text{810 L}

Total volume in tank = 1000 L + 810 L = 1810 L

d. Calculate the mass of salt in the tank

\text{Mass of salt in tank } = \text{1810 L} \times \dfrac{\text{0.0267 kg}}{\text{1 L}} = \textbf{21.6 kg}

(c) Concentration at infinite time

\text{As t $\longrightarrow \, -\infty,\, e^{-\infty} \longrightarrow \, 0$, so A $\longrightarrow \, 0$.}

This makes sense, because the salt is continuously being flushed out by the fresh water coming in.

The graph below shows how the concentration of salt varies with time.

3 0
3 years ago
Polygon Q is a scaled copy of Polygon P using a scale factor of 1/2. Polygon 1 point
DiKsa [7]

Answer:

  1/4

Step-by-step explanation:

The scale factor for area is the square of the scale factor for linear dimensions.

  area scale factor = (1/2)² = 1/4

Polygon Q's area is 1/4 of the area of Polygon P.

7 0
4 years ago
Hurt just hurry and answer please
Luden [163]

Answer:

84

Step-by-step explanation:

Solve:

6*7= 42

42*2.00=84

Hope this helps!

3 0
3 years ago
Other questions:
  • The temperature of a subatance decreased by 24 per minute for 3 minutes
    15·1 answer
  • The ratio of the number of cups of red paint to the number of cups of yellow paint that Garcia mixed to make a shade of orange p
    8·1 answer
  • Please help 1,2,3<br> 25 points*
    12·2 answers
  • Chris drove 265 miles using 10 gallons of gas. At this rate, how many gallons of gas would he need to drive 424 miles?
    5·1 answer
  • The actual length of the rectangle below is 41 cm. Using the scale drawing below, find the actual width.
    5·2 answers
  • (Help with this Please?)
    12·1 answer
  • 3 pages in 7 minutes or 8 pages in 14 minutes, which is better
    13·2 answers
  • If a=1 b=2 c=3 find the value of 4a3b2
    5·2 answers
  • SOMEONE PLEASE HELP ME I’m really confused
    12·1 answer
  • What is the approximate length of arc s on the circle? 5. 23 in. 41. 87 in. 52. 36 in. 104. 67 in.
    7·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!