I love prime factorizations! 50= 2 x 5 x 5 or 2 x 5 at the power of 2
Answer:
<h2>
</h2>
Solution,
Hope this helps..
Good luck on your assignment...
Answer:
C is correct
Step-by-step explanation:
To factor:
4x^2 + 12x + 5
(4x^2 + 2x)(10x + 5)
2x(2x + 1) + 5(2x + 1)
Then, take the outside numbers make them their own group:
2x + 5
(2x + 5)(2x + 1)
Therefore, it’s (2x + 1)(2x + 5)
Answer:
The 29th term is
<h2>243</h2>
Step-by-step explanation:
The above sequence is an arithmetic sequence
For an nth term in an arithmetic sequence
U(n) = a + ( n - 1)d
where n is the number of terms
a is the first term
d is the common difference
From the question
a = - 121
d = -108 -- 121 = - 108 + 121 = 13
Since we are finding the 29th term
n = 29
The 29th term of the sequence is
U(29) = - 121 + ( 29 - 1) 13
= -121 + 28(13)
= -121 + 364
The final answer is
<h2>243</h2>
Hope this helps you
Answer:
(k·j)(x) = -12x^2 -x +6
Step-by-step explanation:
(k·j)(x) = k(x)·j(x) = (4x+3)(-3x+2) . . . . . substitute function definitions
= 4·(-3)x^2 +(4·2+3(-3))x +3·2 . . . . . multiply using the distributive property
= -12x^2 -x +6