Question

find the product of monomial
a) -4p,7pq​

Answer 1

Answer:

Let us multiply the given monomials −4p and 7pq as shown below:(−4p)×7pq

=(−4×7)×(p×p)×q

=(−28)×(p1+1)×q[∵ax×ay=ax+y]

=−28p2q

Hence, (−4p)×7pq=−28p2q.

Step-by-step explanation:

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Answer 2

Given: A monomial is$-4p,7pq$.

We have to find the product of a monomial$-4p,7pq$.

As we know that $\left[ a^{x} \times a^{y}=a^{x+y}\right]$

We are solving in the following way:

We have,

The monomial is$-4p,7pq$.

$\begin{array}{l}(-4 \mathrm{p}) \times 7 \mathrm{pq} \\=(-4 \times 7) \times(\mathrm{p} \times \mathrm{p}) \times \mathrm{q} \\=(-28) \times\left(\mathrm{p}^{1+1}\right) \times \mathrm{q} \quad\left[\because \mathrm{a}^{\mathrm{x}} \times \mathrm{a}^{\mathrm{y}}=\mathrm{a}^{\mathrm{x}+\mathrm{y}}\right] \\=-28 \mathrm{p}^{2} \mathrm{q}\end{array}$

Hence,  the product of a monomial$-4p,7pq$ will be$-28p^{2} q.$