K2 + 18k + 81 what’s the factor completely

Mathematics

2 answers:

Vilka [71]1 year ago

8 0

Answer:

(K+9)^2

Step-by-step explanation:

You have to find what two numbers muliply to 81 and add to 18, in this case the numbers that satisfy that are 9 and 9. So you then get (k+9)(k+9) which is the same as (k + 9)^2.

LenaWriter [7]1 year ago

3 0

Answer:

The answer is (0)

Step-by-step explanation:

You might be interested in

Help me pleeeeeeeeeaze

kobusy [5.1K]

$a(x+3) < 5x+15-x\\\\ax+3a < 5x+15-x\\\\ax+3a < 4x+15\ \ \ |-3a\\\\ax < 4x+15-3a\ \ \ \ |-4x\\\\ax-4x < 15-3a\\\\(a-4)x < 15-3a\\\\\text{The solution is all real numbers if}\ a-4=0\ \ \ |+4\\\\\boxed{a=4}\\\\\text{Check:}\\\\4(x+3) < 5x+15-x\\4x+12 < 4x+15\ \ \ \ |-4x\\12 < 15\ TRUE\to x\in\mathbb{R}$

5 0

3 years ago

How many 10-digit ternary strings are there that contain exactly two 0s, three 1s, and five 2s?

Svet_ta [14]

There are $\dbinom{10}2$ ways of picking 2 of the 10 available positions for a 0. 8 positions remain.

There are $\dbinom83$ ways of picking 3 of the 8 available positions for a 1. 5 positions remain, but we're filling all of them with 2s, and there's $\dbinom55=1$ way of doing that.