So, to factorise something you are taking out the highest common factor.Here, 10 is an example of the highest common factor (I’ll refer to this as HCF)
10(2k+5)
Everything inside the brakets is multiplied by what’s directly next to it (if there is a negative sign, then that is included). So, 10 x 2k = 20k and 10 x 5 = 50
10(2k + 5) is the factorised version of 20k + 50
Given the numbers, let’s set up an equation:
2 + 6 + n = 15
We can solve this using steps:
1. Simplify
8 + n = 15
2. Subtract 8 from both sides
(8 - 8) + n = 15 - 8
3. Simplify
n = 7
Let’s plug in 7 to see if it makes sense:
2 + 6 + 7 = 15
Because this equation is true, we have found our number.
Our unknown number is 7.
Answer:
a) A-B = { 1 , 3, 5 }
b) The remainder is 50
c) f + g)(x) = x⁴+ 5x³
d) 
Step-by-step explanation:
Step(I):-
1) a)
Let A = { 1, 3, 5, 7 } and B = { 7 , 14 , 21 }
A - B = {1,3,5,7} - { 7,14,21}
A - B = { 1 , 3, 5}
b)
Given f (x)=3x³−11x−48
x-1 ) 3 x³ - 11x -48 (- 3x²-3x -2
-3<u>x³ +3x</u>²
3x²-11x-48
<u> -3x²+9x</u>
<u> </u> 2x -48
-2x +2
<u> </u>
50
The remainder is 50
c) Given f(x) = x³ and g(x) = x + 5
(f. g)(x) = f(x) .g(x) = x³ . (x+5) = x³(x) + 5 x³ = x⁴+ 5x³
d)
Given function y = 3 + x²+x³y³ ...(I)
By using derivative formulas
y = xⁿ

Apply d/dx ( UV) = U V¹ + V U¹
Differentiating equation (I) with respective to 'x' , we get


Answer:
you are on the right track
Step-by-step explanation:
the second ones right so you are right :)