Two positive integers have gcd (a, b) = 15 and lcm (a, b) = 90. Those two numbers are 15 and 90 or 30 and 45.
Suppose we have 2 positive integers, a and b, then:
gcd (a, b) = the greatest common divisor = common prime factors of a and b
lcm (a, b) = the least common multiple = multiplication of the greatest common prime factors of a and b
In the given problem:
gcd (a, b) = 15
prime factorization of 15:
15 = 3 x 5
Hence,
a = 3 x 5 x ....
b = 3 x 5 x ....
lcm (a, b) = 90
prime factorization of 90:
90 = 3 x 5 x 2 x 3
Therefore the possible pairs of a and b are:
Combination 1:
a = 3 x 5 = 15
b = 3 x 5 x 2 x 3 = 90
Combination 2:
a = 3 x 5 x 2 = 30
b = 3 x 5 x 3 = 35
We can conclude the two integers are 15 and 90 or 30 and 45.
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Simple....
you have:
3x-9=7x+12
To even solve this....isolate the variable your are trying to solve for....
3x-9=7x+12
-3x -3x
-9=4x+12
-9=4x+12
-12 -12
-21=4x
Simplify.....


Thus, your answer.
Answer:
4b ≥ 25
Step-by-step explanation:
Answer:
10in by 5in
Step-by-step explanation:
PERIMETER: p=2(a+b) p=30
AREA: A=a×b A=50
2(a+b)=30
a×b=50
Answer:
12 + 5 if not 12 * 5
Step-by-step explanation:
12 added to 5 =
13 + 4 =
14 + 3 =
15 + 2 =
16 + 1 =
17 + 0 =
17
and figure out the rest