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3 years ago

6

Which two numbers have a least common multiple of 48?

1 answer:

tekilochka [14]3 years ago

6 0

While neither 9 nor 8 are prime numbers, the factors of 9 are 3*3<span> and of 8 are 2*2*2. Since they share no factors in common, their product will be the least common multiple. (Multiples of 9 are 18, 27, </span>36<span>, 45, 54, 63, and 72; Multiples of 8 are </span>16<span>, </span>24,32<span>, </span>40<span>, 48, </span>56<span>, </span>64<span>, and 72).

This is sited on google! </span>

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Anyone know the answer to this?

jeyben [28]

3x-5+19-x+2x
5x-x=14
4x=14

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3 years ago

Read 2 more answers

I need help writing the first 5 terms of the sequence!

Angelina_Jolie [31]

The first five terms of the sequence are 1, 4, 7, 10, 13.

Solution:

Given data:

$a_{1}=1$

$a_{n}=a_{n-1}+3$

General term of the arithmetic sequence.

$a_{n}=a_{n-1}+d$ , where d is the common difference.

d = 3

$a_{n}=a_{n-1}+3$

Put n = 2 in $a_{n}=a_{n-1}+3$ , we get

$a_{2}=a_1+3$

$a_{2}=1+3$

$a_2=4$

Put n = 3 in $a_{n}=a_{n-1}+3$ , we get

$a_{3}=a_2+3$

$a_{3}=4+3$

$a_3=7$

Put n = 4 in $a_{n}=a_{n-1}+3$ , we get

$a_{4}=a_3+3$

$a_{4}=7+3$

$a_4=10$

Put n = 5 in $a_{n}=a_{n-1}+3$ , we get

$a_{5}=a_4+3$

$a_{5}=10+3$

$a_5=13$

The first five terms of the sequence are 1, 4, 7, 10, 13.

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3 years ago

Nataly_w [17]

Answer:

I don't know

Step-by-step explanation:

I don't know

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3 years ago

Question 7: Timani bought a video game console and some games for $350. The video game

lutik1710 [3]

None of these are right

y=$350 because that is her total.

m=$25 because each game cost that.

B=$200 because that's how much the console cost.

$350=$25x+$200

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3 years ago

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Aleksandr [31]

Answer:

.

Step-by-step explanation:

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