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3 years ago

What is the least common denominator (LCD) of 12 and 23?

Mathematics

1 answer:

leva [86]3 years ago

7 0

Answer: none of those, 276...?

Step-by-step explanation:

You need to know the least common denominator (LCD) of 12 and 23 if you want to add or subtract two fractions with 12 and 23 as denominators.

The least common denominator, also called lowest common denominator (LCD), of 12 and 23 is 276.

Here is a math problem example where you need to know the LCD of 12 and 23 to solve:

3/12 + 2/23 = ?

Step 1) Take the LCD and divide each denominator by it as follows:

276/12 = 23

276/23 = 12

Step 2) Multiply each nominator with the respective answers from Step 1:

3 x 23 = 69

2 x 12 = 24

Step 3) Put it all together to solve the problem:

69/276 + 24/276 = 93/276

= 3/12 + 2/23 = 93/276

It's that easy! Once again, the lowest common denominator (LCD) of 12 and 23 is as follows:

276

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Find the number of elements in A 1 ∪ A 2 ∪ A 3 if there are 200 elements in A 1 , 1000 in A 2 , and 5, 000 in A 3 if (a) A 1 ⊆ A

lina2011 [118]

Answer:

a. 4600

b. 6200

c. 6193

Step-by-step explanation:

Let $n(A)$ the number of elements in A.

Remember, the number of elements in $A_1 \cup A_2 \cup A_3$ satisfies

$n(A_1 \cup A_2 \cup A_3)=n(A_1)+n(A_2)+n(A_3)-n(A_1\cap A_2)-n(A_1\cap A_3)-n(A_2\cap A_3)-n(A_1\cap A_2 \cap A_3)$

Then,

a) If $A_1\subseteq A_2, n(A_1 \cap A_2)=n(A_1)=200$ , and if $A_2\subseteq A_3, n(A_2\cap A_3)=n(A_2)=1000$

Since $A_1\subseteq A_2\; and \; A_2\subseteq A_3, \; then \; A_1\cap A_2 \cap A_3= A_1$

$n(A_1 \cup A_2 \cup A_3)=\\=n(A_1)+n(A_2)+n(A_3)-n(A_1\cap A_2)-n(A_1\cap A_3)-n(A_2\cap A_3)-n(A_1\cap A_2 \cap A_3)=\\=200+1000+5000-200-200-1000-200=4600$

b) Since the sets are pairwise disjoint

$n(A_1 \cup A_2 \cup A_3)=\\n(A_1)+n(A_2)+n(A_3)-n(A_1\cap A_2)-n(A_1\cap A_3)-n(A_2\cap A_3)-n(A_1\cap A_2 \cap A_3)=\\200+1000+5000-0-0-0-0=6200$

c) Since there are two elements in common to each pair of sets and one element in all three sets, then

$n(A_1 \cup A_2 \cup A_3)=\\=n(A_1)+n(A_2)+n(A_3)-n(A_1\cap A_2)-n(A_1\cap A_3)-n(A_2\cap A_3)-n(A_1\cap A_2 \cap A_3)=\\=200+1000+5000-2-2-2-1=6193$

8 0

3 years ago

Please someone help me...need answer but also need understanding if you can PLEASEE

alexdok [17]

20 cups of 10 oz each it's 200 oz
1 gallon it's 4(32)=128 oz
2 gallons is 256 oz. she will need to buy two gallons

7 0

3 years ago

2. Multiply and simplify: (x - 2)(2x + 3) Answer

inysia [295]

Step-by-step explanation:

(x - 2)(2x + 3)

=2x²+3x-4x-6

=2x²-x-6

8 0

3 years ago

Read 2 more answers

31,680 ft = mi A. 6 B. 18 C. 60 D. 2,640

Vlad1618 [11]

5280 feet = 1 mile
31,680 feet = x miles

31,680/5,280 = 6

31,680 feet = 6 miles.

Your answer is A. 6.

6 0

3 years ago

Leslie determined that the system of equations below has infinitely many solutions. Is she correct?

Nookie1986 [14]

Answer:

Leslie determined that the system of equations below has infinitely many solutions. Is she correct?

x=4y-4

2x-8y=-24

A.Yes, Leslie is correct.

B. No, the solution is (-8,-24)

C. No, the solution is (0,-16)

D. No, the system of equations has no solution. 

Step-by-step explanation:

you have to solve they as a pair of simultaneous equations and their is no solutions

if you need a more detailed explanation post the question again and i will write a detailed explanation

plz mark as brainliest

6 0

3 years ago

Read 2 more answers