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lbvjy [14]
3 years ago
10

Frank is trying to factor y^2+6y-27 . He has determined that one factor is (y + 9). What is the other factor? A. y – 27 B. y – 6

C. y – 9 D. y – 3
Mathematics
2 answers:
beks73 [17]3 years ago
5 0
The other factor is (y - 3).  If you multiply the two terms together, you will end up with y^2+6y-27:

(y - 3)(y + 9) \\ y^{2} +9y-3y-27 \\ y^2+6y-27
Doss [256]3 years ago
4 0
Identify b and c in the equation:

b = 6
c = -27

Two factors must add to b, and two factors must multiply to c.

We know that one of the factors is 9, so we can set up the following equations to solve for the missing factor:

9 + y = 6

Subtract 9 from both sides:

y = -3

-

9y = -27

Divide both sides by 9 to get y by itself:

y = -3

The other factor is D. (y - 3).

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2 years ago
Find the general term of sequence defined by these conditions.
disa [49]

Answer:

\displaystyle  a_{n}  =     (2)^{2n -1}   -   (3) ^{n-1 }

Step-by-step explanation:

we want to figure out the general term of the following recurrence relation

\displaystyle \rm a_{n + 2} - 7a_{n + 1} + 12a_n = 0  \:  \: where :  \:  \:a_1 = 1 \: ,a_2 = 5,

we are given a linear homogeneous recurrence relation which degree is 2. In order to find the general term ,we need to make it a characteristic equation i.e

  • {x}^{n}  =  c_{1} {x}^{n - 1}  + c_{2} {x}^{n - 2}  + c_{3} {x}^{n -3 } { \dots} + c_{k} {x}^{n - k}

the steps for solving a linear homogeneous recurrence relation are as follows:

  1. Create the characteristic equation by moving every term to the left-hand side, set equal to zero.
  2. Solve the polynomial by factoring or the quadratic formula.
  3. Determine the form for each solution: distinct roots, repeated roots, or complex roots.
  4. Use initial conditions to find coefficients using systems of equations or matrices.

Step-1:Create the characteristic equation

{x}^{2}  - 7x+ 12= 0

Step-2:Solve the polynomial by factoring

factor the quadratic:

( {x}^{}  - 4)(x - 3) =  0

solve for x:

x =  \rm 4 \:and \: 3

Step-3:Determine the form for each solution

since we've two distinct roots,we'd utilize the following formula:

\displaystyle a_{n}  = c_{1}  {x} _{1} ^{n }  + c_{2}  {x} _{2} ^{n }

so substitute the roots we got:

\displaystyle a_{n}  = c_{1}  (4)^{n }  + c_{2}  (3) ^{n }

Step-4:Use initial conditions to find coefficients using systems of equations

create the system of equation:

\begin{cases}\displaystyle 4c_{1}    +3 c_{2}    = 1  \\ 16c_{1}    + 9c_{2}     =  5\end{cases}

solve the system of equation which yields:

\displaystyle c_{1}  =  \frac{1}{2}     \\  c_{2}   =   - \frac{1}{3}

finally substitute:

\displaystyle  a_{n}  =  \frac{1}{2}   (4)^{n }   -  \frac{1}{3}  (3) ^{n }

\displaystyle \boxed{ a_{n}  =    (2)^{2n-1 }   -   (3) ^{n -1}}

and we're done!

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