Please solve this question and show your work! Thank you!
1 answer:
Answer:
Step-by-step explanation:
So we have the equation:
First, subtract -5 from both sides:
Divide both sides by -3. Since we are dividing by a negative, we will flip the sign:
Definition of absolute value:
Note that we flip the sign on the right because we multiplied 6 by a negative.
On both the left and right, subtract 2:
Divide both equations by -4. Since we're dividing by a negative, flip the signs:
So, our final solution is:
And we're done!
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<span>A.) the sum of a and b is never rational.
This is a true statement. Since an irrational umber has a decimal part that is infinite and non-periodical, when you add a rational number to an irrational number, the result will have the same infinite non periodical decimal part, so the new number will be irrational as well.
</span><span>B.) The product of a and b is rational
This one is false. Zero is a rational number, and when you multiply an irrational number by zero, the result is always zero.
</span><span>C.) b^2 is sometimes rational
This one is true. When you square an irrational number that comes from a square root like </span>
, you will end with a rational number: 
, but, if you square rationals from different roots than square root like ![\sqrt[3]{2}](https://tex.z-dn.net/?f=%20%5Csqrt%5B3%5D%7B2%7D%20)
, you will end with an irrational number: ![\sqrt[3]{2^{2} } = \sqrt[3]{2}](https://tex.z-dn.net/?f=%20%5Csqrt%5B3%5D%7B2%5E%7B2%7D%20%7D%20%3D%20%5Csqrt%5B3%5D%7B2%7D%20)
.
<span>D.) a^2 is always rational
This one is false. If you square a rational number, you will always end with another rational number.
</span><span>E.) square root of a is never rational
</span>This one is false. The square root of perfect squares are always rational numbers:
, 
,...
F.) square root of b is never rational
This one is true. Since the square root of any non-perfect square number is irrational, and all the irrational numbers are non-perfect squares, the square root of an irrational number is always irrational.
We can conclude that given that<span> a is a rational number and b be an irrational number, A, C, D, and F are true statements.</span>
This is a true statement. Since an irrational umber has a decimal part that is infinite and non-periodical, when you add a rational number to an irrational number, the result will have the same infinite non periodical decimal part, so the new number will be irrational as well.
</span><span>B.) The product of a and b is rational
This one is false. Zero is a rational number, and when you multiply an irrational number by zero, the result is always zero.
</span><span>C.) b^2 is sometimes rational
This one is true. When you square an irrational number that comes from a square root like </span>
<span>D.) a^2 is always rational
This one is false. If you square a rational number, you will always end with another rational number.
</span><span>E.) square root of a is never rational
</span>This one is false. The square root of perfect squares are always rational numbers:
F.) square root of b is never rational
This one is true. Since the square root of any non-perfect square number is irrational, and all the irrational numbers are non-perfect squares, the square root of an irrational number is always irrational.
We can conclude that given that<span> a is a rational number and b be an irrational number, A, C, D, and F are true statements.</span>