I am thinking of a 3-digit number When it is divided by 9, the
remainder is 3
When it is divided by 2, the
remainder is
When it is divided by 5, the
remainder is 4

THERE ARE YOUR ANSWERS ALL ARE CORRECT

129 = 9×14+3 = 2×64+1 = 5×25+4

219 = 9×24+3 = 2×109+1 = 5×43+4

309 = 9×34+3 = 2×154+1 = 5×61+4

399 = 9×44+3 = 2×199+1 = 5×79+4

489 = 9×54+3 = 2×244+1 = 5×97+4

579 = 9×64+3 = 2×289+1 = 5×115+4

669 = 9×74+3 = 2×334+1 = 5×133+4

759 = 9×84+3 = 2×379+1 = 5×151+4

849 = 9×94+3 = 2×424+1 = 5×169+4

939 = 9×104+3 = 2×469+1 = 5×187+4

Step-by-step explanation:

If we add a multiple of 9, 2 or 5, respectively to a multiple of 9, 2 or 5,

respectively, we'll get another multiple of 9, 2 or 5, respectively.  Any

multiple of all three, 9, 2, and 5 is a multiple of 90.

Therefore if we can find any number that satisfies 1,2, and 3 above, we can add

any multiple of 90 to it and get another solution.  This problem calls for a 3

digit solution, but if we can find a 1 or 2 digit solution, we can add 90 to it

one or more times and get a 3 digit number that will also satisfy 1, 2, and 3.

We look at property 1.  To get an odd number when adding 3, we must begin with

an even number

So we'll list the even multiples of 9, which are 18, 36, 54, 72, 90...

We add 3 to each and get 21, 39, 57, 75, 93 …

We look at property 3, that it is 4 more than a multiple of 5.

Every multiple of 5 ends in 0 or 5.  4 more than a number ending in 0 ends

with 4, which would make it even.  4 more than a number ending in 5 ends

with 9, which would make it odd.  We want an odd number, so it must end

in 9, and the only candidate listed above that ends with 9 is 39.  It's true

that 39 satisfies all three properties 1,2, and 3. That's becayuse

39 = 4×9+3 = 2×19+1 = 5×7+4

Trouble is, 39 only has two digits, so it doesn't qualify as a solution, but we

can add 90 to it any number of times we wish, as long as we get a 3-digit

number.  So adding 90 just once to 39 gives the three-digit number 129, and

129 = 4×18+3 = 2×64+1 = 5×25+4

We can keep getting more solutions by adding 90 over and over. There are 10

3-digit solutions to your problem:

129 = 9×14+3 = 2×64+1 = 5×25+4

219 = 9×24+3 = 2×109+1 = 5×43+4

309 = 9×34+3 = 2×154+1 = 5×61+4

399 = 9×44+3 = 2×199+1 = 5×79+4

489 = 9×54+3 = 2×244+1 = 5×97+4

579 = 9×64+3 = 2×289+1 = 5×115+4

669 = 9×74+3 = 2×334+1 = 5×133+4

759 = 9×84+3 = 2×379+1 = 5×151+4

849 = 9×94+3 = 2×424+1 = 5×169+4

939 = 9×104+3 = 2×469+1 = 5×187+4

just needed some points but try socratic

step-by-step explanation:

use pothagarean therom

a^2+b^2=c^2 so first ac

acm is a right triangle because c= 90°

and am (the hypotenuse or c) =16+9 =25

but there is not enough info we need another angle or side length