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3 years ago

Please show me step by step: x^3+2x^2-5x-6=0

Mathematics

1 answer:

Kazeer [188]3 years ago

3 0

Answer:

x=-1, x=2, x=3

Step-by-step explanation:

Factor the left side of the equation.

(x+1)(x^2+x-6)

(x+1)(x-2)(x-3)

Now set everything equal to 0

x+1=0

x-2=0

x-3=0

Now solve for x

therefore, the answer is x=-1, x=2, x=3

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VMariaS [17]

Answer:

It is the third one

Step-by-step explanation

3 times 3 equals 9.

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2 years ago

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Select a setting so that each given point will lie within the viewing window.

NemiM [27]

Answer:

The given points are

$(-13,3)\\(0,0)\\(-2,-7)$

The setting would have a interval or 2 units above and below the minimum and maximum of each coordinate.

The given maxium horizontal coordinate is 0.

The given minimum horizontal coordinate is -13.

The given maximum vertical coordinate is 3.

The given minimum vertical coordinate is -7.

Now, we extend each maximum and minimum value by 2 units to create the setting.

So, the setting is

$x_{min}=-15\\x_{max}=2\\ y_{min}=-9\\y_{max}=5$

With a scale of 2 units.

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2 years ago

What is 5/8 + 3/4 divided by -2 /3 - 5/6

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$\text{Hey there!}$

$\frac{5}{4}+\frac{3}{4}\div-\frac{2}{3}\ -\ \frac{5}{6}$

$\frac{5}{8} +\frac{3}{4}=\frac{11}{8}$

$\frac{-2}{3}-\frac{5}{6}=\frac{-3}{2}$

$\text{Problem becomes:}\frac{11}{8}\div\frac{-3}{2}$

$\frac{11}{8}\div\frac{-3}{2}=\frac{-11}{12}$

$\boxed{\boxed{\bf{Answer:\frac{-11}{12}}}}\checkmark$

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3 years ago

Find the area of the shape shown below

Arlecino [84]

Answer: 12.5

Step-by-step explanation:

area= 1/2 (2.5+7.5) 2.5

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2 years ago

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Consider the differential equation x2y′′ − 9xy′ + 24y = 0; x4, x6, (0, [infinity]). Verify that the given functions form a funda

pantera1 [17]

Answer:

The functions satisfy the differential equation and linearly independent since W(x)≠0

Therefore the general solution is

$y= c_1x^4+c_2x^6$

Step-by-step explanation:

Given equation is

$x^2y'' - 9xy+24y=0$

This Euler Cauchy type differential equation.

So, we can let

$y=x^m$

Differentiate with respect to x

$y'= mx^{m-1}$

Again differentiate with respect to x

$y''= m(m-1)x^{m-2}$

Putting the value of y, y' and y'' in the differential equation

$x^2m(m-1) x^{m-2} - 9 x m x^{m-1}+24x^m=0$

$\Rightarrow m(m-1)x^m-9mx^m+24x^m=0$

$\Rightarrow m^2-m-9m+24=0$

⇒m²-10m +24=0

⇒m²-6m -4m+24=0

⇒m(m-6)-4(m-6)=0

⇒(m-6)(m-4)=0

⇒m = 6,4

Therefore the auxiliary equation has two distinct and unequal root.

The general solution of this equation is

$y_1(x)=x^4$

and

$y_2(x)=x^6$

First we compute the Wronskian

$W(x)= \left|\begin{array}{cc}y_1(x)&y_2(x)\\y'_1(x)&y'_2(x)\end{array}\right|$

$= \left|\begin{array}{cc}x^4&x^6\\4x^3&6x^5\end{array}\right|$

=x⁴×6x⁵- x⁶×4x³

=6x⁹-4x⁹

=2x⁹

≠0

The functions satisfy the differential equation and linearly independent since W(x)≠0

Therefore the general solution is

$y= c_1x^4+c_2x^6$

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2 years ago