The product of two rational numbers is always rational because (ac/bd) is the ratio of two integers, making it a rational number.
We need to prove that the product of two rational numbers is always rational. A rational number is a number that can be stated as the quotient or fraction of two integers : a numerator and a non-zero denominator.
Let us consider two rational numbers, a/b and c/d. The variables "a", "b", "c", and "d" all represent integers. The denominators "b" and "d" are non-zero. Let the product of these two rational numbers be represented by "P".
P = (a/b)×(c/d)
P = (a×c)/(b×d)
The numerator is again an integer. The denominator is also a non-zero integer. Hence, the product is a rational number.
learn more about of rational numbers here
brainly.com/question/29407966
#SPJ4
1/8 of 32 is 4 so if Ryan is 4 then Brandon is 2
so Ryan = 4 and Brandon = 2
9u means you're multiplying 9 into that vector, both components. Same with the 2v. 9*3 = 27 and 9*-1 = -9, so your new vector u is <27, -9>. Now let's do v. 2* -6 (twice) = -12, so your new v vector is <-12, -12>. Add those together now, first components of each and second components of each. 27 + (-12) = 15; -9+(-12)=-21. So the addition of those gives us a final vector with a displacement of <15, -21>
Answer:
C
Step-by-step explanation:
If I had to make a choice it would be C. Its strange because all numbers can be divided by 4 so they all can make a square. But the closest was C.
Answer:
The last one
Step-by-step explanation:
f(2) = 4x-5 = 3
f(-4) = 4x - 5 = -21
F(2) has a greater value than f(-4)
