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3 years ago

What are the first three common multiples of 5 and 15

1 answer:

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!

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Can you please help me out on this...

ss7ja [257]

Answer:

The answer is either a, b, c, or d.

Step-by-step explanation:

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3 years ago

Mei Su had 80 coins. She gave most of them to her friends in such a way that each of her friends got at least one coin and no tw

Mama L [17]

Answer: 12 friends.

Step-by-step explanation:

the data we have is:

Mei Su had 80 coins.

She gave the coins to her friends, in such a way that every friend got a different number of coins, then we have that:

The maximum number of friends that could have coins is when:

friend 1 got 1 coin

friend 2 got 2 coins

friend 3 got 3 coins

friend N got N coins

in such a way that:

(1 + 2 + 3 + ... + N) ≥ 79

I use 79 because "she gave most of the coins", not all.

We want to find the maximum possible N.

Then let's calculate:

1 + 2 + 3 + 4 + 5 = 15

15 + 6 + 7 + 8 + 9 + 10 = 55

now we are close, lets add by one number:

55 + 11 = 66

66 + 12 = 78

now, we can not add more because we will have a number larger than 80.

Then we have N = 12

This means that the maximum number of friends is 12.

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3 years ago

Explain why f(x) = x2+4/x2-x-2 is not continuous at x = 2.

xxTIMURxx [149]

Answer: Option a.

Step-by-step explanation:

Make the denominator equal to zero and solve for x:

Factor the quadratic equation. Find two number whose sum is -1 and whose product is -2. Then:

$x^{2}-x-2=0\\(x-2)(x+1)=0\\x=2\\x=-1$

Then, as you can see the value x=2 makes the denominator equal to zero and the division by zero does not exist. Therefore you can conclude that the function shown in the problem is not defined at x=2

The answer is the option a.

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