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Akimi4 [234]
3 years ago
11

Factor completely: x^3 - x^2 - 6x

Mathematics
1 answer:
PSYCHO15rus [73]3 years ago
4 0

Answer:

x(x-3)(x+2)

Step-by-step explanation:

1. Find the GCF (Greatest Common Factor)

GCF=x

2. Factor out the GCF and simplify.

x(x^{2} -x-6)

3. Factor x^{2} -x-6.

Which two numbers add up to -1 and multiply to -6?

<u>-3 and 2</u>

Rewrite the expression using the numbers above.

(x-3)(x+2)

4. Done!

x(x-3)(x+2)

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Answer:

–9xy2 – 6x2y + 5x3

–9xy2 – 6x2y – 5x3

9xy2 + 6x2y + 5x3

9xy2 – 6x2y + 5x3

the answer is 9xy2 – 6x2y + 5x3

Step-by-step explanation:

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I will offer brainalist + 10 points for correct answer (:
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Answer:

D) complimentary

Step-by-step explanation:

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Answer:

4.62910049578

Step-by-step explanation:

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The areas of the squares adjacent to two sides of a right triangle are shown below.
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3 years ago
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Find a formula for Y(t) with Y(0)=1 and draw its graph. What is Y\infty?
tino4ka555 [31]

Answer:

(a)\ y(t)\ =\ -2e^{-2t}+3

(b)\ y(t)\ =\ 4e^{-2t}-3

Step-by-step explanation:

(a) Given differential equation is

   Y'+2Y=6

=>(D+2)y = 6

To find the complementary function, we will write

D+2=0

=> D = -2

So, the complementary function can be given by

y_c(t)\ =\ C.e^{-2t}

To find the particular integral, we will write

y_p(t)\ =\ \dfrac{6}{D+2}

          =\ \dfrac{6.e^{0.t}}{D+2}

           =\ \dfrac{6}{0+2}

           = 3

so, the total solution can be given by

y_(t)\ =\ C.F+P.I

         =\ C.e^{-2t}\ +\ 3

y_(0)=C.e^{-2.0}\ +\ 3

but according to question

1 = C +3

=> C = -2

So, the complete solution can be given by

y_(t)\ =\ -2.e^{-2.t}\ +\ 3

(b) Given differential equation is

   Y'+2Y=-6

=>(D+2)y = -6

To find the complementary function, we will write

D+2=0

=> D = -2

So, the complementary function can be given by

y_c(t)\ =\ C.e^{-2t}

To find the particular integral, we will write

y_p(t)\ =\ \dfrac{-6}{D+2}

           =\ \dfrac{-6.e^{0.t}}{D+2}

           =\ \dfrac{-6}{0+2}

           = -3

so, the total solution can be given by

y_(t)\ =\ C.F+P.I

         =\ C.e^{-2t}\ -\ 3

y_(0)\ =C.e^{-2.0}\ -\ 3

but according to question

1 = C -3

=> C = 4

So, the complete solution can be given by

y_(t)\ =\ 4.e^{-2.t}\ -3

4 0
3 years ago
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