Advertisements
Advertisements
प्रश्न
At age of 2 years, a cat or a dog is considered 24 “human” years old. Each year, after age 2 is equivalent to 4 “human” years. Fill in the expression [24 +
(a – 2)] so that it represents the age of a cat or dog in human years. Also, you need to determine for what ‘a’ stands for. Copy the chart and use your expression to complete it.
| Age | [24 + (a – 2)] |
Age (Human Years) |
| 2 | ||
| 3 | ||
| 4 | ||
| 5 | ||
| 6 |
Advertisements
उत्तर
The expression is [24 + 4(a – 2)]
Here, ‘a’ represents the present age of dog or cat.
| Age | [24 + 4 (a – 2)] | Age (Human Years) |
| 2 | [24 + 4(2 – 2)] | 24 |
| 3 | [24 + 4(3 – 2)] | 28 |
| 4 | [24 + 4(4 – 2)] | 32 |
| 5 | [24 + 4(5 – 2)] | 36 |
| 6 | [24 + 4(6 – 2)] | 40 |
APPEARS IN
संबंधित प्रश्न
Add the following:
2p2q2 − 3pq + 4, 5 + 7pq − 3p2q2
Simplify combining like terms: (3y2 + 5y - 4) - (8y - y2 - 4)
Subtract: 5a2 - 7ab + 5b2 from 3ab - 2a2 -2b2
Add the following algebraic expression:
3a2b, − 4a2b, 9a2b
Add the following algebraic expression: \[\frac{7}{2} x^3 - \frac{1}{2} x^2 + \frac{5}{3}, \frac{3}{2} x^3 + \frac{7}{4} x^2 - x + \frac{1}{3}, \frac{3}{2} x^2 - \frac{5}{2}x - 2\]
Subtract:
\[x^2 y - \frac{4}{5}x y^2 + \frac{4}{3}xy \text { from } \frac{2}{3} x^2 y + \frac{3}{2}x y^2 - \frac{1}{3}xy\]
Add:
−3y2 + 10y − 16; 7y2 + 8
Add: 8x, 3x
Multiply the following:
(ab + c), (ab + c)
How much is y4 – 12y2 + y + 14 greater than 17y3 + 34y2 – 51y + 68?
(a – 2)] 