Can someone please help me
1 answer:
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Answer:
Step-by-step explanation:
A V-shaped graph is the graph of an absolute value function.
The parent absolute value function is
Among the given options, a V-shaped graph is
This graph is obtained by shifting the parent absolute value function 2 units right.
The correct option is A
Option D
The sum of two irrational numbers is always rational is false statement
<em><u>Solution:</u></em>
<h3><u>The sum of two rational numbers is always rational</u></h3>"The sum of two rational numbers is rational."
So, adding two rationals is the same as adding two such fractions, which will result in another fraction of this same form since integers are closed under addition and multiplication. Thus, adding two rational numbers produces another rational number.
For example:
rational number + rational number = rational number
Hence this statement is true
<h3><u>The product of a non zero rational number and an irrational number is always irrational</u></h3>If you multiply any irrational number by the rational number zero, the result will be zero, which is rational.
Any other situation, however, of a rational times an irrational will be irrational.
A better statement would be: "The product of a non-zero rational number and an irrational number is irrational."
So this statement is correct
<h3><u>The product of two rational numbers is always rational</u></h3>The product of two rational numbers is always rational.
A number is said to be a rational number if it is of the form p/q,where p and q are integers and q ≠ 0
Any integer is a rational number because it can be written in p/q form.
Hence it is clear that product of two rational numbers is always rational.
So this statement is correct
<h3><u>The sum of two irrational numbers is always rational</u></h3>"The sum of two irrational numbers is SOMETIMES irrational."
The sum of two irrational numbers, in some cases, will be irrational. However, if the irrational parts of the numbers have a zero sum (cancel each other out), the sum will be rational.
Thus this statement is false