What is the smallest integer $k>2000$ such that both $\dfrac{17k}{66}$ and $\dfrac{13k}{105}$ are terminating decimals? please help ty（＾-＾）

2079

Step-by-step explanation:

Let's look at the factorization of the denominators first. (I'm doing this to choose k so that everything cancels in the denominator that isn't a power of 2 or power of 5.)

66=2(3)(11) or 2(33)

105=5(3)(7) or 5(21)

Now, the reason I grouped the factors 3 and 11 together is because I don't recall anything divided by 3 and 11 giving me a decimal that terminates other than 3 and 11 itself.

I grouped 3 and 7 in the second factorization for similar reasoning.

So we want k to possibly be a multiple of

3(7)(11)=3(77)=231.

I didn't need two factors of 3 because each fraction contains k in numerator and each fraction as exactly one factor of 3 to cancel.

Let's try the first multiple.

17(231)/66=59.5

13(231)/105=28.6

The 1st multiple was a success.

We also want our number choice to be greater than 2000.

So we want to find m so that 231m>2000..

. m>2000/231

We want m to be an integer so choose smallest integer number in this set for m..

That's 9 since 2000/231 is approximately 8.7.

231(9)=2079

for the first question it would be the second answer for the second question it would be the third answer and for the third question it would be the first answer

step-by-step explanation:

answer: jasmine will bake 4/3 or 1 1/3 cookies in an hour

step-by-step explanation:

if u add 1/3 every 15 minutes after an hour u will have 4/3 or 1 1/3