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3 years ago

Factoring has nothing to do with finding the least common denominator.

Mathematics

1 answer:

Nadusha1986 [10]3 years ago

6 0

Answer:

probably FALSE

Step-by-step explanation:

The least common denominator (LCD) of two fractions is the least common multiple (LCM) of their denominators.

One of the ways to find the LCM of numbers is to factor them, then multiply the highest powers of each of the unique factors. Using this method makes your statement FALSE.

Another way to find the LCM is to multiply the numbers and divide that product by their greatest common denominator (GCD). Again, finding the GCD can involve factoring, as it is the product of all of the common factors of the numbers.

However, Euclid's algorithm is a method of finding the GCD that does not involve factoring. That is, it is entirely possible to find the LCM of two numbers without doing any factoring, per se. (Though, finding a divisor could be considered factoring.) Using this method makes your statement TRUE.

Our guess is that the desired answer is FALSE. Some methods of finding the LCD do involve factoring.

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What is the total area of this composite figure rounded to the nearest hundredth?

Alex777 [14]

Answer:

39.53 ft^2

Step-by-step explanation:

We can find the area of the rectangle

A = l*w

= 12*3 = 36

And the area of the semicircle

A = 1/2 pir^2

The diameter is 3, the radius is 3/2

=1/2 pi* (3/2)^2

= 1/2 *3.14 * 9/4

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Add the two areas together

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Rounding to the nearest hundredth

39.53 ft^2

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3 years ago

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Answer:

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Step-by-step explanation:

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3 years ago

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3 years ago

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stellarik [79]

Answer:

The numbers on the axis need to follow a repeating pattern

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3 years ago

A basketball coach will select the members of a five-player team from among 9 players, including John and Peter. If the five pla

vivado [14]

Answer:

Step-by-step explanation:

Given that a basketball coach will select the members of a five-player team from among 9 players, including John and Peter.

Out of nine players five are chosen at random.

The team consists of John and Peter.

Hence we can sort 9 players as I group, John and Peter and II group 7 players.

Now the selection is 2 from I group and remaining 3 from II group.

Hence no of ways of selecting a team that includes both John and Peter= $2C2*7C3$ =35

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= $\frac{2C2*7C3}{9C5}$ = $\frac{35}{126} =\frac{5}{18}$

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3 years ago