Advertisements
Advertisements
Question
Multiply the following:
–5a2bc, 11ab, 13abc2
Advertisements
Solution
We have,
–5a2bc, 11ab and 13abc2
∴ –5a2bc × 11ab × 13abc2 = (–5 × 11 × 13)a2bc × ab × abc2
= –715a4b3c3
APPEARS IN
RELATED QUESTIONS
Find the product of the following pair of monomial.
4p3, − 3p
Complete the table of products.
|
First monomial→ |
2x |
–5y |
3x2 |
–4xy |
7x2y |
–9x2y2 |
|
Second monomial ↓ |
||||||
| 2x | 4x2 | ... | ... | ... | ... | ... |
| –5y | ... | ... | –15x2y | ... | ... | ... |
| 3x2 | ... | ... | ... | ... | ... | ... |
| – 4xy | ... | ... | ... | ... | ... | ... |
| 7x2y | ... | ... | ... | ... | ... | ... |
| –9x2y2 | ... | ... | ... | ... | ... | ... |
Obtain the volume of a rectangular box with the following length, breadth, and height, respectively.
2p, 4q, 8r
Obtain the volume of a rectangular box with the following length, breadth, and height, respectively.
xy, 2x2y, 2xy2
Obtain the product of a, − a2, a3
Multiply: −5cd2 by − 5cd2
Multiply: abx, −3a2x and 7b2x3
Three consecutive integers, when taken in increasing order and multiplied by 2, 3 and 4 respectively, total up to 74. Find the three numbers.
Multiply the following:
–7pq2r3, –13p3q2r
Multiply the following:
–3x2y, (5y – xy)
