Question

If $y>0$, find the range of all possible values of $y$ such that $\lceil{y}\rceil\cdot\lfloor{y}\rfloor

Answer 1

The range of possible values of y such that ceil(y) * floor(y) is explained by understanding the behavior of the ceiling and floor functions.

Step-by-step explanation:

The range of all possible values of y such that ceil(y) * floor(y) is defined by the behavior of the ceiling and floor functions.

When we take the ceiling of y, we round up to the nearest integer, and when we take the floor of y, we round down to the nearest integer.

Therefore, the possible values of y that satisfy the given condition are all real numbers greater than 0, because the ceiling and floor functions will always yield the same value for those numbers.

The complete question is:If $y>0$, find the range of all possible values of $y$ such that $\lceil{y}\rceil\cdot\lfloor{y}\rfloor

Answer 2

This interval indicates that
`\( y \)` can be any value greater than 6.48, extending to infinity.

To solve the given equation for
`\( y \)`, where
`\( y > 0 \)` and the equation is:

`\[ |y| \cdot |y| = 42 \]`

We will work through the solution step-by-step. Given that
`\( y > 0 \)`, the absolute value of
`\( y \)`, which is
`\( |y| \)`, will simply be
`\( y \)`. This simplifies our equation to:

`\[ y \cdot y = 42 \]`

or

`\[ y^2 = 42 \]`

Now, let's solve for
`\( y \)`. Since we are only considering positive values for
`\( y \)`, we only need the positive square root of 42. Let's calculate that.

The positive value of
`\( y \)` that satisfies the equation
`\( y^2 = 42 \)` is approximately 6.48. Since we are asked for the range in interval notation and considering only positive values of
`\( y \)`, the range would be:

`\[ (0, \infty) \]`

However, if we are to provide the smallest interval that contains all the possible values of
`\( y \)` given that
`\( y > 0 \)`, then it would be:

`\[ (6.48, \infty) \]`

This interval indicates that
`\( y \)` can be any value greater than 6.48, extending to infinity.