<span>To solve these GCF and LCM problems, factor the numbers you're working with into primes:
3780 = 2*2*3*3*3*5*7
180 = 2*2*3*3*5
</span><span>We know that the LCM of the two numbers, call them A and B, = 3780 and that A = 180. The greatest common factor of 180 and B = 18 so B has factors 2*3*3 in common with 180 but no other factors in common with 180. So, B has no more 2's and no 5's
</span><span>Now, LCM(180,B) = 3780. So, A or B must have each of the factors of 3780. B needs another factor of 3 and a factor of 7 since LCM(A,B) needs for either A or B to have a factor of 2*2, which A has, and a factor of 3*3*3, which B will have with another factor of 3, and a factor of 7, which B will have.
So, B = 2*3*3*3*7 = 378.</span>
Answer:
YES
Step-by-step explanation:
The equation, , would be an identity if the equation remains true regardless of the value of x we choose to plug in into the equation.
Let's find out if we would always get a true statement using different value of x.
✍️Substituting x = 1 into the equation:
(TRUE)
✍️Substituting x = 2 into the equation:
(TRUE)
✍️Substituting x = 3 into the equation:
(TRUE)
Therefore, we can conclude that the equation, , is an identity.
It is awnser a or number one
Answer:
(2h-3k)(m-n)
Step-by-step explanation:
factor by grouping, you can group the first two terms and the last two terms:
2h(m-n) + (-3k)(m-n)
there will be one factor in common, use that factor in combination with the factors that are not duplicated:
(2h-3k)(m-n)
We know that
Two fractions are equivalent only if the product of the numerator (a) of the first fraction and the denominator (d) of the other fraction is equal to the product of the denominator (b) of the first fraction and the numerator (c<span>) of the other
that means
a/b=c/d-------> a*d=c*b
so
0.34=0.34/1=34/100---------> 0.34/1=34/100
then
0.34*100=1*34-------> 34=34------> fractions (0.34/1) and (34/100) are equivalent
34/100-------> 17//50
34*50=100*17---------> 1700=1700-----> fractions 34/100 and 17/50 are equivalent
the answer is
17/50</span>