0
Add your answer:
Earn +20 pts
What is logically quivalent to q -- p?
It is: q+p because a double minus becomes a plus
How do you factor 2q squared minus 10?
2(q^2 - 5) or 2(q - the square root of 5)(q + the square root of 5)
Is the square root of pq irrational where p and q are primes?
Only if p and q are DIFFERENT primes.
Why is square root of 22 irrational?
Because it cannot be expressed in the form p/q where p and q are integers (q >0).
Why is the square root 24 irrational?
Because it cannot be expressed in the form p/q where p and q are integers and q > 0.
How is the sum of a rational and irrational number irrational?
This can easily be proved by contradiction. Without loss of generality, I will take specific numbers as an example. The proof can easily be extended to any rational + irrational number. Assumption: 1 plus the square root of 2 is rational. (It is a well-known fact that the square root of 2 is irrational. No need to prove it here; you can use any other irrational number will do.) This rational sum can be written as p / q, where "p" and "q" are whole numbers (this is basically the definition of a "rational number"). Then, the square root of 2, which is equal to the sum minus 1, is: p / q - 1 = p / q - q / q = (p - q) / q Since the difference of two whole numbers is a whole number, this makes the square root of 2 rational, which doesn't make sense.
If you take the square root of an irrational number.Will it be irrational too?
Certainly. Otherwise, there would be a rational number whose square was an irrational number; that is not possible. To show this, let p/q be any rational number, where p and q are integers. Then, the square of p/q is (p^2)/(q^2). Since p^2 and q^2 must both be integers, their quotient is, by definition, a rational number. Thus, the square of every rational number is itself rational.
What are P and Q if Square of P plus square of Q equals 123456789 in some order?
If p and q are both squared, then they would both be even numbers, and the sum of them couldn't end in 9, so not possible with whole numbers
Is the square root an exponent?
Yes, the square root is equivalent to an exponent of 1/2.Suppose the exponent is a rational number of the form p/q where p and q are integers and q > 0. Then x^(p/q) = (the qth root of x) raised to the power p or, equivalently, (the qth root of (x raised to the power p).
Is the square root of an irrational number rational?
No: Let r be some irrational number; as such it cannot be represented as s/t where s and t are both non-zero integers. Assume the square root of this irrational number r was rational. Then it can be represented in the form of p/q where p and q are both non-zero integers, ie √r = p/q As p is an integer, p² = p×p is also an integer, let y = p² And as q is an integer, q² = q×q is also an integer, let x = q² The number is the square of its square root, thus: (√r)² = (p/q)² = p²/q² = y/x but (√r)² = r, thus r = y/x and is a rational number. But r was chosen to be an irrational number, which is a contradiction (r cannot be both rational and irrational at the same time, so it cannot exist). Thus the square root of an irrational number cannot be rational. However, the square root of a rational number can be irrational, eg for the rational number ½ its square root (√½) is not rational.
Is the square root of 31 an irrational numbers?
Yes, the square root of 31 is an irrational number. Rational numbers are those which can be expressed in the form of p/q(where p, q are integers and q ≠0). Square root of 31 has non-terminating and non-repeating decimal so it can't be expressed in the form of p/q.
How would you go about solving this direct variation question and what is the answer- If P varies directly as the square of Q and the constant of variation is 6 what is the value of P when Q equals 12?
If P varies directly with the square of Q then the equation would be in the form of P = kQ2, where k is the constant of variation so the new equation would be: P = 6Q2, so when Q = 12 we have P=6*122, or P = 864
