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Mashcka [7]
3 years ago
14

Factorization of 3x2-8x-3

Mathematics
2 answers:
choli [55]3 years ago
8 0
Your answer is <span>(3x + 1) (x - 3)

Hope this helps.

</span>
pychu [463]3 years ago
3 0
<span>3x</span>² <span>- 8x -3  = 3x</span>² - 9x + x - 3 =

= 3x² - 9x + x - 3 = 3x* ( x - 3) + 1* (x - 3) = 

= 3x * (x - 3) + 1 * (x - 3) = (3x + 1) (x - 3) 

Answer (3x + 1) (x - 3)

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Find the least common multiple (LCM) of 8y^6+ 144y^5+ 640y^4 and 2y^4 + 40y^3 + 200y^2.
Mice21 [21]

Answer:

<em>LCM</em> = 8y^{4}(y+ 10)^{2}(y + 8)

Step-by-step explanation:

Making factors of 8y^{6}+ 144y^{5}+ 640y^{4}

Taking 8y^{4} common:

\Rightarrow 8y^{4} (y^{2}+ 18y+ 80)

Using <em>factorization</em> method:

\Rightarrow 8y^{4} (y^{2}+ 10y + 8y + 80)\\\Rightarrow 8y^{4} (y (y+ 10) + 8(y + 10))\\\Rightarrow 8y^{4} (y+ 10)(y + 8))\\\Rightarrow \underline{2y^{2}} \times  4y^{2} \underline{(y+ 10)}(y + 8)) ..... (1)

Now, Making factors of 2y^{4} + 40y^{3} + 200y^{2}

Taking 2y^{2} common:

\Rightarrow 2y^{2} (y^{2}+ 20y+ 100)

Using <em>factorization</em> method:

\Rightarrow 2y^{2} (y^{2}+ 10y+ 10y+ 100)\\\Rightarrow 2y^{2} (y (y+ 10) + 10(y + 10))\\\Rightarrow \underline {2y^{2} (y+ 10)}(y + 10)        ............ (2)

The underlined parts show the Highest Common Factor(HCF).

i.e. <em>HCF</em> is 2y^{2} (y+ 10).

We know the relation between <em>LCM, HCF</em> of the two numbers <em>'p' , 'q'</em> and the <em>numbers</em> themselves as:

HCF \times LCM = p \times q

Using equations <em>(1)</em> and <em>(2)</em>: \Rightarrow 2y^{2} (y+ 10) \times LCM = 2y^{2} \times  4y^{2}(y+ 10)(y + 8) \times 2y^{2} (y+ 10)(y + 10)\\\Rightarrow LCM = 2y^{2} \times  4y^{2}(y+ 10)(y + 8) \times (y + 10)\\\Rightarrow LCM = 8y^{4}(y+ 10)^{2}(y + 8)

Hence, <em>LCM</em> = 8y^{4}(y+ 10)^{2}(y + 8)

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Answer:

Theres more orange juice than pineapple juice.

Step-by-step explanation:

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Answer:

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8 0
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german tiene 13 billetes en su bolsillo , algunos son de $2000 y otros de $5000 si en total German tiene $40.000 ¿Cuantos billet
Rufina [12.5K]

Answer:

hay 10 billetes de 2000

hay 4 billetes de 5000

Step-by-step explanation:

Planteamos las siguientes ecuaciones...:

x= billetes de 2000

y= billetes de 5000

ecuaciones :

1)    x+y = 14

2)  2000x + 5000y =40000

despejamos primera ecuacion y remplazamos en segunda  :

1) x= 14-y

2)  2000(14-y)  +  5000y  =  40000

    28000- 2000y+5000y  = 40000

                           3000y=  40000 - 28000

                                  y= 4

reemplazamos este valor en primera ecuacion:

x+y = 14

 x+ 4= 14

    x= 10

ahora reemplazamo valor de X       y      valor de Y

2000(10)   + 5000(4)  =40000

                  40000    =40000                        

hay 10 billetes de 2000

hay 4 billetes de 5000

Espero que esto te ayude

8 0
3 years ago
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