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asambeis [7]
3 years ago
6

3. Write out the prime factor of the following numbers. a. 15 b. 21 C. 38 d. 42

Mathematics
1 answer:
Minchanka [31]3 years ago
6 0

Answer:

The prime factors of 15 are: 1,3,5,15

The prime factors of 21 are: 1,3,7,21

The prime factors of 38 are: 1,3,13,39

The prime factor of 42 is : 1,2,3,6,7,14,21,42

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I need help! anyone know this?
xz_007 [3.2K]
Yeah, x= 7/2 and y= -5/2
Please mark as brainliest and I hope this helps
5 0
3 years ago
Please help math <br>question 1,2,3,4,5
kolezko [41]
The answers:

#1  C
#2 H maybe
#3 C
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6 0
3 years ago
What is the image point of (-6,-9)(−6,−9) after a translation left 1 unit and down 5 units?
Anna [14]

Answer:

(-7,-4)

Step-by-step explanation:

-6-1=-7

-9-(-5)=-4

7 0
3 years ago
The sum of two consecutive odd intergers is 32. Find the intergers.
elena-14-01-66 [18.8K]
So let the smaller integer be 2x+1 and the larger one be 2x+3. So 2x+1+2x+3 is 32. Simplifying, we get 4x+4=32. So subtracting 2 from both sides we get 4x+2=30. So dividing by 2 we get 2x+1=15. So 2x+3 is 17. So the numbers are 15 and 17
4 0
3 years ago
Factorise 24e^2-28e-12
Helga [31]

Answer:

4(2e - 3)(3e + 1)

Step-by-step explanation:

Given

24e² - 28e - 12 ← factor out 4 from each term

= 4(6e² - 7e - 3) ← factor the quadratic

Consider the factors of the product of the e² term and the constant term which sum to give the coefficient of the e- term.

product = 6 × - 3 = - 18 and sum = - 7

The factors are - 9 and + 2

Use these factors to split the e- term

6e² - 9e + 2e - 3 ( factor the first/second and third/fourth terms )

= 3e(2e - 3) + 1 (2e - 3) ← factor out (2e - 3) from each term

= (2e - 3)(3e + 1)

Then

24e² - 28e - 12 = 4(2e - 3)(3e + 1) ← in factored form

8 0
3 years ago
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