How do you simplify $(-{{y}^{5}})({{y}^{2}})$ ?

Hint: In this question, we have to simplify the given algebraic term. Therefore, we will use the basic mathematical rule and the exponent rule to get the solution. We start solving this problem by applying the formula $(-1)(y)=-y$ in the given algebraic term. Then, we will apply the exponent rule which states when the bases are the same in the multiplication; the powers are added to each other. After that, we will make the necessary calculations to get the required solution.

Complete step by step solution:
According to the question, we have to find the simplified value of the given algebraic term.
Thus, we will apply the basic mathematical rules and the exponent rule to get the solution.
The algebraic term given to us is $\left( -{{y}^{5}} \right)\left( {{y}^{2}} \right)$ --------- (1)
Now, we will first use the formula $(-1)(y)=-y$ in the equation (1) to get the same bases in the given problem, we get
$\Rightarrow (-1)\left( {{y}^{5}} \right)\left( {{y}^{2}} \right)$
As we see in the above term, both the variables are the same, that is both the bases are equal to each other. Thus, we will apply the exponent rule which states when the bases are the same in the multiplication; the powers are added to each other, that is $\left( {{a}^{x}} \right).\left( {{a}^{y}} \right)={{a}^{x+y}}$ . thus, we will apply this formula in the above equation, we get
$\Rightarrow (-1)\left( {{y}^{5+2}} \right)$
On further simplification, we get
$\Rightarrow (-1)\left( {{y}^{7}} \right)$
Now, we will again apply the formula $(-1)(y)=-y$in the above equation, we get
$\Rightarrow \left( -{{y}^{7}} \right)$ which is the required solution for the problem.

Therefore, for the algebraic term $(-{{y}^{5}})({{y}^{2}})$, its simplified value is $\left( -{{y}^{7}} \right)$ .

Note: While solving this problem, do step-by-step calculations to avoid confusion and mathematical errors. At the time of applying the exponent rule, do add the exponent when the bases are the same instead of multiplying the exponents.