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3 years ago

James says that 12 is a common factor of 3 and 4.june says that 12 is common multiple of 3 and 4.who is correct.

Mathematics

2 answers:

Sliva [168]3 years ago

7 0

June is correct because 12 is a multiple of both 3 and 4.

almond37 [142]3 years ago

4 0

June is correct.

When you are dealing with factors, you have to think of what numbers multiply to get the number I see.
Let's say the number is 10. What numbers can be multiplied to get 10.
Well, there is 1, 10, 2, 5 so all of those numbers are factors of 10.

When you are dealing with multiples, you have to think of the opposite. If the question asks is 12 a multiple of 3 and 4, than you have to think...If I multiply a number by 3 could I get 12 and... If I multiply a number by 4 could I get 12. The answer is yes you could multiply 3 and 4 and 4 and 3 to get the number 12. That makes 12 a multiple.

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What is the real part of the following complex number? -3+ 2iV2

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Answer:

Option (3)

Step-by-step explanation:

If a complex number is in the form of (a + ib), real part of the complex number is 'a' and imaginary part is 'b'.

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Comparing this number with (a + ib), real part of the number will be (-3) and imaginary part of the number will be 2.

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The triangles are similar, what is the value of x?

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X=5
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3 years ago

Evaluate the following expressions. a. 9!/8! b. 7!/5! c. 21!/19! d. 8!/4!

Nata [24]

Answer:

Remember, $n!=1*2*3*\cdots*n$ , then

$\frac{9!}{8!}=\frac{1*2*3*4*5*6*7*8*9}{1*2*3*4*5*6*7*8}=9$

$\frac{7!}{5!}=\frac{1*2*3*4*5*6*7}{1*2*3*4*5}=6*7=42$

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$\frac{8!}{4!}=\frac{1*2*3*4*5*6*7*8}{1*2*3*4}=5*6*7*8=1680$

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3 years ago

Solving systems of equations using substitution

wariber [46]

Y = 3 - x. this is equation 1
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as x = -1, substitute it back into equation 1,
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4 years ago