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sesenic [268]
3 years ago
13

what types of problems can be solved using the greatest common factors? What types of problems can be solved using the least com

mon multiple
Mathematics
2 answers:
gayaneshka [121]3 years ago
7 0
What types of problems can be solved using the greatest common factor? What types of problems can be solved using the least common multiple? Complete the explanation. 
<span>*** Use the words 'same' and 'different' to complete the following sentences.*** </span>

<span>Problems in which two different amounts must be split into (the same) number of groups can be solved using the GCF. Problems with events that occur on (different) schedules can be solved using the LCM.</span>
Lorico [155]3 years ago
4 0

Answer:

The greatest common factor or GCF is the greatest factor any two numbers share or we can say that GCF is the greatest factor that divides two different numbers.

Like the greatest common factor of 12 and 8 is 4.

There can be problems like - Kesha has 8 cans of normal soda and 16 cans of diet soda. She wants to create few refreshment tables that will operate during the basketball game soon to be held. She does not want any sodas left over. What is the greatest number of refreshment tables that she can stock?

-------------------------------------------------------------------------------------

The least common multiple or the LCM is the smallest common number or multiple of any two numbers.

Like the least common multiple of 3 and 5 is 15.

There can be problem like - Brian has painting lessons every fifth day and swimming lessons every third day. If he had both the lessons on May 7, when will be the next date on which he has both lessons?

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7 t-shirts cost £67.55. How much do 11 t-shirts cost?
Oksanka [162]

Answer:

11 tshirts would cost £106.15

Step-by-step explanation:

67.55 divided by 7 = 9.65

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Therefore the total would be £106.15.

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3 years ago
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Altitudes AA1 and BB1 are drawn in acute △ABC. Prove that A1C·BC=B1C·AC
Sophie [7]

Answer:

See the attached figure which represents the problem.

As shown, AA₁ and BB₁ are the altitudes in acute △ABC.

△AA₁C is a right triangle at A₁

So, Cos x = adjacent/hypotenuse = A₁C/AC ⇒(1)

△BB₁C is a right triangle at B₁

So, Cos x = adjacent/hypotenuse = B₁C/BC ⇒(2)

From (1) and  (2)

∴  A₁C/AC = B₁C/BC

using scissors method

∴ A₁C · BC = B₁C · AC

7 0
3 years ago
Image is below, please help me fast i have 2 minutes left, thank you very much
Yuki888 [10]

Answer:

45

Step-by-step explanation:

1/3 of 90

= 1/3 * 90

= 90/3

= 30

Let the missing number be x.

2/3 of x

= 2/3 * x

= 2x/3

30 = 2x/3

Cross multiply,

2x = 30 * 3

2x = 90

x = 90/2

x = 45

Hence,

the missing number is 45.

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2 years ago
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