Ask question

Ask question

All categories

4 years ago

8

What are the first three common multiples of 5 and 15

1 answer:

Artist 52 [7]4 years ago

8 0

The first 3 common multiples of 5 and 15 are 15, 30, and 45. Remember that the number itself is also a multiple of any number, since you can multiply it by 1.

5 - 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85, 90, 95, 100
15 - 15, 30, 45, 60, 75, 90, 105

Hope this helps!

You might be interested in

2 and 3/4 x 1 and 1/3

aivan3 [116]

Answer is 3.667
or in fraction form 3 and 2/3

4 0

4 years ago

Which pattern follows the given rule? Rule: Start with 10 and then multiply each term by 3 to get the next term. A. 10, 13, 20,

Ghella [55]

B....is the correct one

5 0

3 years ago

Read 2 more answers

HELP ONLY DO THE FIRST ONE

igomit [66]

Answer:

For A : The quotient of the sum of a and b

For B : The sum of A and the quotient of B and 2

5 0

4 years ago

How do you solve this equation

Nezavi [6.7K]

Remember the order of operations: Parentheses, Exponents, Multiplication/Division, Addition/Subtraction

Solve the expressions in parentheses first.

There is an expression within parentheses that is also in an expression within brackets. Solve that expression first:

$6 - 10 = -4$

The expression in brackets will now read:

$[4 - 3(-4)]$

Multiply 3(-4):

$3 \cdot -4 = -12$

$[4 - (-12)]$

Solve the expression:

$4 - (-12) = 4 + 12 = 16$

The problem should now look like this:

$(8-2)^2 + \frac{1}{4}(16)$

Solve the expression in parentheses that has an exponent:

$8 - 2 = 6$

$(6)^2 = 6 \cdot 6 = 36$

Solve the term with 1/4 as a coefficient:

$\frac{1}{4}(16) = 4$

The expression should now look like this:

$36 + 4 = \boxed{40}$

The answer is 40.

8 0

3 years ago

Read 2 more answers

Jing jing works in a store and she spends some of her time organizing the stationary section. If there are nine different items

Alexus [3.1K]

Answer:

282240 ways

Step-by-step explanation:

number of items = 9 different items

Ranking based on worth

<u>determine number of ways to arrange the 9 items such that the 2 most expensive are apart</u>

first step: consider the 2expensive items as 1

∴ number of permutation = [ ( 9 - 2 ) + 1 ]! * 2!

= [ 8 ] ! * 2! = 80640

Total number of permutation/arrangement = 9! = 362880

hence to arrange the 9 items without having the two most expensive items together

= 362,880 - 80640

= 282240 ways

4 0

3 years ago