If you would like to solve <span>(8r^6s^3 – 9r^5s^4 + 3r^4s^5) – (2r^4s^5 – 5r^3s^6 – 4r^5s^4), you can do this using the following steps:
</span>(8r^6s^3 – 9r^5s^4 + 3r^4s^5) – (2r^4s^5 – 5r^3s^6 – 4r^5s^4) = 8r^6s^3 – 9r^5s^4 + 3r^4s^5 – 2r^4s^5 + 5r^3s^6 + 4r^5s^4 = 8r^6s^3 – 5r^5s^4 + r^4s^5<span> + 5r^3s^6
</span>
The correct result would be 8r^6s^3 – 5r^5s^4 + r^4s^5<span> + 5r^3s^6.</span>
A. -3+4i Square both numbers, add them, then find the square root. Essentially, use the Pythagorean theorem. -3^2 + 4^2 9 + 16=25 Square root of 25 is 5.
Answer:
B and C are true
Step-by-step explanation:
hope this helps :)
Answer:
Table C
Step-by-step explanation:
Given
Required
Which table is correct
We have:
For 7 packages, it will be:
--- i.e. 6 * 7
For 8:
--- i.e. 8 * 7
For p packages
<em>Using the above formula, we can conclude that table (C) is correct</em>
(9)^4 × (27)^3 × (81)^2 / (3)^24 = 3.
Explanation:
- All the numbers in the numerator i.e. 9, 27, 81 are multiples of 3.
- 9 = 3 × 3 = 3², 27 = 3 × 3 × 3 = 3³, 81 = 3 × 3 × 3 × 3 = .
- = (3^2)^4, 27³ = (3^3)^3, 81^2 = (3^4)^2.
- According to the power rule, (a^x)^y = a^xy.
- So the given numbers can be written as follows = 3^8, 27³ = 3^9 and 81² = 3^8.
- According to the product rule, (a^x) × (a^y) = a^xy.
- So the numerator can be written as × 27³ × 81² = (3^8) × (3^9) × (3^8) = 3^(8+9+8) = 3^25.
- So the fraction becomes 3^25 / 3^24 = 3 × 3^24 / 3^24 = 3.