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Llana [10]
3 years ago
11

Please help me!! thank you

Mathematics
1 answer:
Ierofanga [76]3 years ago
8 0
A and c are the answers because if x,y and z are whole number they are positive numbers
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HELP ASAP!! Identify the least common multiple of x^2 + 7x + 6 and x^2 − 3x − 4.
vesna_86 [32]

(x+1)(x+6)(x-4)

The least common multiple (LCM) of two or more non-zero whole numbers is the smallest whole number that is divisible by each of those numbers. In other words, the LCM is the smallest number that all of the numbers divide into evenly.

Hope This Helps & Good Luck

3 0
4 years ago
Read 2 more answers
– 18q+10q+16q–3= – 19
sergeinik [125]

Step-by-step explanation:

  • – 18q+10q+16q–3= – 19
  • 8q=-19
  • q=-19/8
  • q=-2.375

8 0
3 years ago
Read 2 more answers
Write a rule in words and as an algebraic expression to model the relationship in each table
goblinko [34]
The cost is the number of videos (V) times the cost of a video, plus the monthly fee.

C=2.25V+5
6 0
3 years ago
A parallelogram with an equal base and height and an area greater than 64 square meters can you give me the answer please
Roman55 [17]
8
if the base and height are equal
and the formula equals b x h = A^2
then u can make a formula b x b = 64
then u take the square root of 64

5 0
3 years ago
Divide 6k2-15-5 by 3k
TEA [102]
Based on the question above, I believe that '6k2' would be (6k^2). So by this, I would continue to with the following expression as: 6k^2-15-5\div3k

Simplify: \frac{5}{3} \\ \\  ((6 * (k2)) -  15) - ( \frac{5}{3} *k) \\ \\  ((2*3k^2) -  15) -  \frac{5k}{3} \\ \\ \boxed{\boxed{6k^2 - 15 = \frac{6k^2-15}{1} =  \frac{(6k^2-15)*3}{3} }} \\ \\ (pull \ out \ like \ terms) \\ \\  6k^2 - 15  =   3 * (2k^2 - 5) \\ \\ Factoring:  2k^2 - 5

And by this, in order for us to fully understand that this would be true, the following would enable us to understand.

\Rightarrow  \left[\begin{array}{ccc}A+B) * (A-B) =\\
         A2 - AB + BA - B2 =\\
         A2 - AB + AB - B2 = \\
         A2 - B2\end{array}\right] \Leftarrow

\boxed{\bf{ \frac{3*(2k^2-5)*3-(5k)}{3} =  \frac{18k^2-5k-45}{3} }}

Factoring : 18k^2 - 5k - 45 \\ \\   \left[\begin{array}{ccc}   -810    +    1    =    -809 \\
      -405    +    2    =    -403 \\
      -270    +    3    =    -267\\  -162    +    5    =    -157 \\
      -135    +    6    =    -129 \\
      -90    +    9    =    -81 
 
  
\end{array}\right]

(Your \ Answer) : \boxed{\boxed{\bf{ \frac{18k^2-5k-45}{3} }}}
4 0
4 years ago
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