Advertisements
Advertisements
प्रश्न
Multiply the following:
–5a2bc, 11ab, 13abc2
योग
Advertisements
उत्तर
We have,
–5a2bc, 11ab and 13abc2
∴ –5a2bc × 11ab × 13abc2 = (–5 × 11 × 13)a2bc × ab × abc2
= –715a4b3c3
shaalaa.com
क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 7: Algebraic Expression, Identities and Factorisation - Exercise [पृष्ठ २३०]
APPEARS IN
संबंधित प्रश्न
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 product of a, 2b, 3c, 6abc.
Multiply: x + 4 by x − 5
Multiply: 5a − 1 by 7a − 3
Multiply: 12a + 5b by 7a − b
Multiply: a2, ab and b2
Solve: (-12x) × 3y2
Multiply the following:
15xy2, 17yz2
Multiply the following:
–3x2y, (5y – xy)
Multiply the following:
7pqr, (p – q + r)
