Step-by-step explanation:





<u>Let us assume that:</u>

<u>Therefore, the equation becomes:</u>






<u>Now substitute the value of u. We get:</u>


<u>Therefore:</u>


★ <u>Which is our required answer.</u>

(a + b)² = a² + 2ab + b²
(a - b)² = a² - 2ab + b²
a² - b² = (a + b)(a - b)
(a + b)³ = a³ + 3ab(a + b) + b³
(a - b)³ = a³ - 3ab(a - b) - b³
a³ + b³ = (a + b)(a² - ab + b²)
a³ - b³ = (a - b)(a² + ab + b²)
(x + a)(x + b) = x² + (a + b)x + ab
(x + a)(x - b) = x² + (a - b)x - ab
(x - a)(x + b) = x² - (a - b)x - ab
(x - a)(x - b) = x² - (a + b)x + ab
Answer:
B and D
Step-by-step explanation:
A
× 13 =
=
= 10
< 13
B
4 × 13 = 52 > 13
C
× 13 =
=
= 4
< 13
D
× 13 = 1 × 13 = 13
Band D are equal to or greater than 13
5. . 1
---- - -----
6. . 2
5. . 3
----- - -----
6. . 6
5-3
-------
6
2
-----
6
= 1/3
B
Answer:
D
Step-by-step explanation:
Yes it's D. Answered by Gauthmath