The sum or product of a non-zero rational number and an irrational number is always
2 answers:
Answer:
Option 2 - Irrational
Step-by-step explanation:
To find : The sum or product of a non-zero rational number and an irrational number is always
Solution :
By the statement,
→ "The sum of a rational number and an irrational number is irrational."
Example : Let a rational number 3 and irrational number
Sum of rational and irrational number
5.236067977.... is a irrational number as if you add a non-repeating and non-terminating decimal to a repeating decimal, you will have a repeating decimal.
→ "The product of a non-zero rational number and an irrational number is irrational."
Example : Let a rational number 3 and irrational number
Sum of rational and irrational number
6.708203931...... is a irrational number as If you multiply any irrational number by the rational number it gives you irrational except zero which is rational.
Therefore, Option 2 is correct.
The sum or product of a non-zero rational number and an irrational number is always irrational.
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Answer:
(t, t -1, t)
Step-by-step explanation:
You have three unknowns but only 2 equations, so you can't really SOLVE this...you can get a solution with a variable still in it (I forget what this is called. I think it refers to infinite many solutions). Here's how it works:
Set up your matrix:
You want to change the number in position 21 (the 2 in the scond row) to a 0 so you have y and z left. Do this by multiplying the top row by -2 then adding it to the second row to get that 2 to become a 0. Multiplying in a -2 to the top row gives you:
Then add, keeping the first row the same and changing the second to reflect the addition:
The second equation is this now:
3y - 3z = -3. Solving for y gives you y = z - 1. Let's let z = t (some random real number that will make the system true. Any number will work. I'll show you at the end. Just bear with me...)
lf z = t, and if y = z - 1, then y = t - 1. So far we have that y = t - 1 and z = t. Now we solve for x:
From the first equation in the original system,
x - 2y + z = 2. Subbing in t - 1 for y and t for z:
x - 2(t - 1) + t = 2. Simplify to get
x - 2t + 2 + t = 2 and x - t = 0, and x = t. So the solution set is (t, t - 1, t). Picking a random value for t of, let's say 2, sub that in and make sure it works. If:
x - 2y + z = 2, then t - 2(t - 1) + t = 2 becomes t - 2t + 2 + t = 2, and with t = 2, 2 - 2(2) + 2 + 2 = 2. Check it: 2 - 4 + 4 = 2 and 2 = 2. You could pick any value for t and it will work.
Answer: =−3/5
Step-by-step explanation: Hope this help :D
