I need some help guys qwq

Using it's concept, it is found that there is a 0.3 = 30% probability to select a pupil that uses the swimming pool but not the gym.

A probability is given by the number of desired outcomes divided by the number of total outcomes.

In this problem, we have that 50 - 7 = 43 pupils use at least one of the pool or the gym.

We use the following relation, considering the numbers of each:

Both = Pool + Gym - At least one

Hence:

Both = 31 + 28 - 43 = 16.

From this, we have that out of 50 pupils, there are 31 - 16 = 15 pupils who use the pool but not the gym, hence the probability is:

p = 15/50 = 0.3 = 30%.

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Answer:

Hey! I'm gonna go ahead and give ya the "meaning" of the fundamental theorem of algebra but not the answer to the A. B. and C.

Step-by-step explanation:

fundamental theorem of algebra, theorem of equations proved by Carl Friedrich Gauss in 1799. It states that every polynomial equation of degree n with complex number coefficients has n roots, or solutions, in the complex numbers. The roots can have a multiplicity greater than zero.

Answer:

a

Step-by-step explanation:

the equation of a line in slope- intercept form is

y = mx + c ( m is the slope and c the y- intercept )

calculate m using the slope formula

m = [tex]\frac{y_{2}-y_{1} }{x_{2}-x_{1} }[/tex]

with (x₁, y₁ ) = (4, - 1 ) and (x₂, y₂ ) = (- 2, 3 )

m = [tex]\frac{3-(-1)}{-2-4}[/tex] = [tex]\frac{3+1}{-6}[/tex] = [tex]\frac{4}{-6}[/tex] = - [tex]\frac{2}{3}[/tex] , then

y = - [tex]\frac{2}{3}[/tex] x + c ← is the partial equation

to find c substitute either of the 2 points into the partial equation

using (- 2, 3 ) , then

3 = [tex]\frac{4}{3}[/tex] + c ⇒ c = 3 - [tex]\frac{4}{3}[/tex] = [tex]\frac{5}{3}[/tex]

y = - [tex]\frac{2}{3}[/tex] x + [tex]\frac{5}{3}[/tex] ← in slope- intercept form

multiply through by 3 to clear the fractions

3y = - 2x + 5 ( subtract - 2x + 5 from both sides )

2x + 3y - 5 = 0 ← in general form

Answer:

its log 5 7/8

Step-by-step explanation:

Answer:

The first option

Kindly award branliest

Answer:

x =  -1,  7,  3 + i,  3 - i.

Step-by-step explanation:

(x^2-6x+9)^2-15(x^2-6x+10)=1

(x^2 - 6x + 9)^2  - 15(x^2 - 6x + 9) - 15*1 = 1

(x^2 - 6x + 9)^2  - 15(x^2 - 6x + 9) - 16 = 0

Let Z = x^2 - 6x + 9, then we have:

Z^2 - 15Z - 16 = 0

(Z - 16)(Z + 1) = 0

Z = 16 or Z = -1

so  x^2 - 6x + 9 = -1 or x^2 - 6x + 9 = 16

x^2 - 6x + 9 = -1

---> x^2 - 6x + 10 = 0

Using the Quadratic Formula:

---> x = [6 +/- √((-6)^2 - 4* 1* 10) / 2

---> x = 6/2 +/- √-4/2

---> x = 3 + i , 3 - i.

x^2 - 6x + 9 = 16

---> x^2 - 6x - 7 = 0

---> (x - 7)(x + 1) = 0

---> x = 7, -1.

The producer surplus at the equilibrium point is 1.0146

An equilibrium (or equilibrium point) of a dynamical machine-generated via an autonomous machine of everyday differential equations (ODEs) is a solution that doesn't change with time.

Equilibrium is the country in which the marketplace delivers and calls for stability each different, and as a result, expenses turn out to be solid. usually, an over-deliver of products or services reasons charges to head down, which leads to higher demand—at the same time as a below-supply or scarcity reasons costs to head up resulting in fewer calls.

The equilibrium equations (balance of linear momentum) are given in index form as(1.4)σji,j+bi=ρu¨i, I,j=1,2,3where σij are additives of (Cauchy) stress, ρ is mass density, and bi is body force additives.

Producer Surplus = area of shadestportion

Area of shaded portion = free of rect minus area of OneD

Aves of OABO=Now serving from

"[3√74-3/5]

= √(x+3)" de

=2 (2√6-√5)

(Jon= √46 = Fure=456-255

(x+3)+1

Area of rectangle OA BC =316

= 3√6

producer surplus = 3√6 - (ure -2√5)

= 2√3-66

= 1.0146

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Answer:

B {3/2, -5/3}

Step-by-step explanation:

6x² + x - 15 = 0

6x² + 10x - 9x - 15 = 0

2x(3x + 5) - 3(3x + 5) = 0

(3x + 5)(2x - 3) = 0

3x + 5 = 0  or  2x - 3 = 0

3x = -5  or  2x = 3

x = -5/3  or  x = 3/2

Answer: B {3/2, -5/3}

Answer:

[tex]\left\{\dfrac{3}{2},\ -\dfrac{5}{3}\right\}[/tex]

Step-by-step explanation:

Given quadratic equation: [tex]6x^2+x-15=0[/tex]

[tex]\implies a=\textsf{6},b=\textsf{1},c=\textsf{-15}[/tex]

..................................................................................................................................................

[tex]{\large \textsf{ Quadratic Formula: }}x=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}\ \Bigg{\|}\ \textsf{when }ax^2+bx+c=0[/tex]

..................................................................................................................................................

Step 1: Substitute the given values into the formula and simplify.

[tex]\implies x=\dfrac{-{(\textsf1})\pm \sqrt{\textsf{(1)}^2-4(\textsf{6}){(\textsf{-15})}}}{2{(\textsf{6})}}[/tex]

[tex]\implies x=\dfrac{-1\pm \sqrt{1-4(-90)}}{12}[/tex]

[tex]\implies x=\dfrac{-1\pm \sqrt{1+360}}{12}[/tex]

[tex]\implies x=\dfrac{-1\pm \sqrt{361}}{12}[/tex]

[tex]\implies x=\dfrac{-1\pm 19}{12}[/tex]

Step 2: Separate into two cases.

[tex]x_1=\dfrac{-1+19}{2}[/tex]

[tex]x_2=\dfrac{-1-19}{2}[/tex]

Step 3: Solve both cases.

[tex]\implies x_1=\dfrac{-1+19}{12}=\dfrac{18}{12}=\dfrac{6\times3}{6\times2}=\boxed{\dfrac{3}{2}}[/tex]

[tex]\implies x_2=\dfrac{-1-19}{12}=\dfrac{-20}{12}=\dfrac{-4\times5}{4\times3}=\boxed{-\dfrac{5}{3}}[/tex]

The solutions to this quadratic are: [tex]x_1=\dfrac{3}{2},\ x_2=-\dfrac{5}{3}[/tex]

..................................................................................................................................................

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Answer:

-4

Step-by-step explanation:

Wehn dealing with order of operations, we can remember the expression PEMDAS, which stands for parentheses, exponents, multiplication, division, addition, and subtraction.

So, our first step in the above problem is the parentheses:

[tex]2(3-5)^3+4(-4+7)\\2(-2)^3+4(3)[/tex]

The second step is exponents:

[tex]2(-2)^3+4(3)\\2(-8)+4(3)[/tex]

The third step is multiplication:

[tex]2(-8)+4(3)\\-16+12[/tex]

The fourth and final step is addition:

[tex]-16+12\\-4[/tex]

Answer:

  67 square units

Step-by-step explanation:

The area using the left-hand sum is the sum of products of the function value at the left side of the interval and the width of the interval.

The attachment shows a table of the x-value at the left side of each interval, and the corresponding function value there. The interval width is 1 unit in every case, so the desired area is simply the sum of the function values.

The approximate area is 67 square units.

Split up the interval [0, 6] into 6 equally spaced subintervals of length [tex]\Delta x = \frac{6-0}6 = 1[/tex]. So we have the partition

[0, 1] U [1, 2] U [2, 3] U [3, 4] U [4, 5] U [5, 6]

where the left endpoint of the [tex]i[/tex]-th interval is

[tex]\ell_i = i - 1[/tex]

with [tex]i\in\{1,2,3,4,5,6\}[/tex].

The area under [tex]f(x)=x^2+2[/tex] on the interval [0, 6] is then given by the definite integral and approximated by the Riemann sum,

[tex]\displaystyle \int_0^6 f(x) \, dx \approx \sum_{i=1}^6 f(\ell_i) \Delta x \\\\ ~~~~~~~~ = \sum_{i=1}^6 \bigg((i-1)^2 + 2\bigg) \\\\ ~~~~~~~~ = \sum_{i=1}^6 \bigg(i^2 - 2i + 3\bigg) \\\\ ~~~~~~~~ = \frac{6\cdot7\cdot13}6 - 6\cdot7 + 3\cdot6 = \boxed{67}[/tex]

where we use the well-known sums,

[tex]\displaystyle \sum_{i=1}^n 1 = \underbrace{1 + 1 + \cdots + 1}_{n\,\rm times} = n[/tex]

[tex]\displaystyle \sum_{i=1}^n i = 1 + 2 + \cdots + n = \frac{n(n+1)}2[/tex]

[tex]\displaystyle \sum_{i=1}^n i^2 = 1 + 4 + \cdots + n^2 = \frac{n(n+1)(2n+1)}6[/tex]

The answer is tenths

Step-by-step explanation:

According to the place value chart of decimals,the first number after the decimal point starts from tenths and continues.so the answer is tenths

Hi :)

Remember Place value in decimal numbers

[tex]\large\boldsymbol{\hfill\stackrel{hundreds}{9}~\hfill\stackrel{tenths}0~\hfill\stackrel{ones}6.\hfill\stackrel{tenths}{7}}[/tex]

Then

The place-value of [tex]\boldsymbol{7}[/tex] is [tex]\boldsymbol{tenths}[/tex].

[tex]\tt{Learn~More;Work~Harder}[/tex]

:)

Answer:

put four balls in each been

Step-by-step explanation:

Answer:

PQ = 10 units

Step-by-step explanation:

to find where the lines cross the x- axis let y = 0 and solve for x , that is

2x + 8 = 0 ( subtract 8 from both sides )

2x = - 8 ( divide both sides by 2 )

x = - 4 ← point P

and

2x - 12 = 0 ( add 12 to both sides )

2x = 12 ( divide both sides by 2 )

x = 6 ← point Q

the lines cross the x- axis at x = - 4 and x = 6

using the absolute value of the difference , then

PQ = | - 4 - 6 | = | - 10 | = 10 units

or

PQ = | 6 - (- 4) | = | 6 + 4 | = | 10 | = 10 units

Answer: [tex]\Huge\boxed{Distance=10~units}[/tex]

Step-by-step explanation:

Given expression

y = 2x + 8

Substitute 0 for the y value to find the x value

This is the definition of x-intercepts

(0) = 2x + 8

Subtract 8 on both sides

0 - 8 = 2x + 8 - 8

-8 = 2x

Divide 2 on both sides

-8 / 2 = 2x / 2

x = -4

[tex]\large\boxed{P~(-4,0)}[/tex]

Given expression

y = 2x - 12

Substitute 0 for the y value to find the x value

(0) = 2x - 12

Add 12 on both sides

0 + 12 = 2x - 12 + 12

12 = 2x

Divide 2 on both sides

12 / 2 = 2x / 2

x = 6

[tex]\large\boxed{Q~(6,0)}[/tex]

Given information

[tex](x_1,~y_1)=(-4,~0)[/tex]

[tex](x_2,~y_2)=(6,~0)[/tex]

Substitute values into the distance formula

[tex]Distance=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}[/tex]

[tex]Distance=\sqrt{(6-(-4))^2+(0-0)^2}[/tex]

Simplify values in the parenthesis

[tex]Distance=\sqrt{(10)^2+(0)^2}[/tex]

Simplify values in the radical sign

[tex]Distance=\sqrt{100}[/tex]

[tex]\Huge\boxed{Distance=10~units}[/tex]

Hope this helps!! :)

Please let me know if you have any questions

See below for the values of the probabilities

The complete question is added as an attachment

Write down the probability he gets 'Miss a turn'.

From the spinner, we have

Sections = 6

Miss a turn = 1

So, the probability that he gets 'Miss a turn is

P = 1/6

Write down the probability that he gets 'Go back 1 square'.

Here, we have:

Number of rows = 4

'Go back 1 square' = 4

The probability that he gets 'Go back 1 square' is

P = 'Go back 1 square'/Number of rows

This gives

P = 4/4

Evaluate

P = 1

How many of these spins are expected to be 'Go forward 3 squares'?

Here, we have

Spins = 96

From the spinner, the probability of 'Go forward 3 squares' is

P =1/6

So, the expected number is

E(x) = np

This gives

E(x) = 96 * 1/6

Evaluate

E(x) = 16

Hence, 16 spins are expected to be 'Go forward 3 squares'

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The cost of one yard of black fabric is $4.

The cost of one yard of red fabric is $5.

2r + b = 14 equation 1

2r + 4b = 26 equation 2

Where:

r = cost of one yard of red fabric

b = cost of one yard of black fabric

Subtract equation 1 from equation 2

3b = 12

b = 12 / 3

b = 4

Substitute for b in equation 1

2r + 4 = 14

2r = 14 - 4

2r = 10

r = 10 / 2

r = $5

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Answer:

5x^2-x+6

Step-by-step explanation:

That should work

The given condition on the basis of truth table is true.

According to the statement

we have given that the some condition p and q. and we have to find that  the given condition is true or false on the basis of the truth table.

So, The given conditions are:

If 3 + 2 = 5, Then 5 + 5 = 10

Now,

The Truth table is a tabular representation of all the combinations of values for inputs and their corresponding outputs. It is a mathematical table that shows all possible outcomes that would occur from all possible scenarios that are considered.

So, the given conditions are properly verified with the truth table. So, it is true.

So, The given condition on the basis of truth table is true.

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Answer:8 feet

Step-by-step explanation:

The $315 cost of the couch after tax, gives the cost before tax, C, as the following equation;

[tex]Cost \: before \: tax, \: \: \mathbf {C} = \frac{315}{1 + t} [/tex]

The cost of the couch with tax = $315

Let, t, represent the tax rate, and let C, represent the cost before tax, we have;

315 = C + C × t

Which gives;

Mathematically, we have the following equation;

[tex]Cost \: before \: tax, \: \: \mathbf {C} = \frac{315}{1 + t} [/tex]

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Her profit % was increased by 2.77%

Answer:

Solution Given:

discount =10%

profit =25%

S.P with 13% vat = Rs 5,085

S.P+Vat% of S.P= Rs 5,085

S.P(1+13%)=Rs 5,085

S.P=Rs 5,085/1.13

S.P =Rs 4,500

Again

M.P = S.P+discount% of M.P

M.P- discount% of M.P= Rs 4,500

M.P(1-10%)=Rs 4,500

M.P=Rs 4,500/0.9

M.P = Rs 5,000

Now

C.P= (S.P*100)/(1+profit%)

C.P=(4,500*100)/(100+25%)

C.P= Rs 3,600

Again

profit =S.P- C.P= Rs 4,500-Rs 3,600=Rs 900

Again

Due to the excessive demands of her items, she decreased the discount percent by 2%.

So,

New discount = 10%-2%=8%

New S.P= M.P -discount% of M.P

=M.P(1-discount%)

=Rs 5,000(1-8%)

=Rs 4,600

Again

Profit=S.P- C.P

New profit =Rs 4,600 - Rs 3,600=Rs 1,000

Profit% =profit/C.P*100%= 1000/3600*100=27.77%

Profit % increased =New profit%- profit%

=27.77-25

=2.77%

Answer:

baiBsiasbisvdxhsbhssj

Answer:

-1/7

is the answer of your question

1. Sea level (0 m) - 100%

2. Mid-level elevation (4,500 m) - 57%

3. Peak (8,850 m) - 33%

Identify the elevation and the percentage of oxygen available at each elevation.

Elevation has a significant impact on the amount of oxygen in the atmosphere. The oxygen level increases with decreasing height and decreases with increasing elevation. Thus, we have a situation where the oxygen concentration is highest at sea level or 0 elevations. The oxygen level drops dramatically when we ascend to greater elevations, say 4,500 m, making breathing difficult. Even at higher altitudes, such as 8,850 m, the oxygen level will drop significantly, making it necessary for a person to undergo extensive training and carry oxygen to survive.

1. Sea level (0 m) - 100%

2. Mid-level elevation (4,500 m) - 57%

3. Peak (8,850 m) - 33%

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Answer:

Answer:

  48 minutes

Step-by-step explanation:

The rate of filling from both pipes is the sum of the rates in units of tanks per hour. We want the time for 2/3 of a tank.

The sum of the filling rates is ...

  rate A + rate B = (1/3) tank/hour + (1/2) tank/hour = 5/6 tank/hour

The time to fill a tank or part thereof will be the number of tanks divided by the rate in tanks per hour:

  time = (2/3 tank)/(5/6 tank/hour) = ((4/6)/(5/6)) hour = 4/5 hour

There are 60 minutes in an hour, so the fill time is ...

  (4/5 hour)×(60 min/hour) = 48 min

It will take 48 minutes for both pipes to fill 2/3 of the tank.

The volume of the cylinder that has same height and radius with the given cone is 5,346 cm³

Pyramids and prisms are examples of three-dimensional shapes, as are shapes with curved surfaces. The three-dimensional prism shape has two bases that are parallel and congruent. Any polygon is acceptable for the bases, which are two of the faces. Rectangles make up the remaining faces.

Given that the cylinder and the cone have the same radius and height, the volume of a cylinder equals 3 times the volume of a cone.

According to the given figure:

Given, a cone that has a volume of 1,782 cubic cm, has the same radius and height with a cylinder,

therefore,

Volume of the cylinder = 3 x (1,782)

Volume of the cylinder = 5,346 cubic centimeters

Hence,  the volume of the cylinder is 5,346 cubic centimeter.

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The 8 cm circle radius and the location of the circle X gives the following values;

(a) tan(<XAB) = -(r - 8)/(2•r + 4) = (8 - r)/8

Which gives;

(b) <XAB is approximately 0.644 radians

(c) The area of the shaded region is approximately 3.107 cm²

8 × (8 - r) = 0.5 × 16 × (8 + r) × sin(A)

(8 - r) = (8 + r) × sin(A)

sin(A) = (8 - r)÷(8 + r)

(8 - r)² = 8² + (8 + r)² - 2×8×(8 + r)×cos(A)

16×(8 + r)×cos(A) = (8² + (8 + r)²) - (8 - r)²

Which gives;

cos(A) = ((8² + (8 + r)²) - (8 - r)²) ÷ (16×(8 + r))

cos(A) = (2•r + 4)/(r + 8)

tan(A) = ((8 - r)÷(8 + r))/( (2•r + 4)/(r + 8))

tan(A) = -(r - 8)/(2•r + 4) = (8 - r)/8

(8 - r)/(2•r + 4) = (8 - r)/8

8 = 2•r + 4

2•r + 4 = 8

Therefore;

(b) sin(A) = (8 - r)÷(8 + r)

sin(A) = (8 - 2)÷(8 + 2) = 0.6

Therefore;

<XAB = <A = arcsin(0.6) ≈ 0.644 rad

<XAB in degrees ≈ 36.87°

Angle at sector PXQ = 180° - 2 × (36.87°) ≈ 106.26°

Area of sector PXQ ≈ (106.26°/(360°)) × π × 2²

(106.26°/(360°)) × π × 2² ≈ 3.71

Area of sector APO ≈ (36.87°/(360°)) × π × 8² ≈ 20.59

Area of triangle AXB = 8 × (8-2) = 48

The shaded area is therefore;

48 - (2×20.59 + 3.71) ≈ 3.107

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a) The radius of the small circle is 2 centimeters.

b) The measure of the angle XAB is approximately equal to 51.340°.

c) The area of the shaded region is approximately equal to 2.423 square centimeters.

In this question we must analyze a geometrical system formed by three semicircles and a circle. The small circle touches the two side semicircles tangentially at points P and Q and the uppermost section of it also tangent to the central semicircle (point R).

(a) Therefore, we have to solve the following system of equations:

8² + h² = (8 + r)²      (1)

h + r = 8      (2)

By (2) in (1):

8² + (8 - r)² = (8 + r)²

32 · r = 64

r = 2

The radius of the small circle is 2 centimeters.

(b) The measure of the angle XAB is found by trigonometric functions:

tan m ∠ XAB = OX / AO

tan m ∠ XAB = 10 / 8

m ∠ XAB ≈ 51.340°

The measure of the angle XAB is approximately equal to 51.340°.

(c) The area of the shaded region if the area of the triangle AXB minus the areas of the three circular sections, that is:

A = 0.5 · (16 cm) · (10 cm) · sin 51.340° - (51.340° / 180°) · π · (8 cm)² - 0.5 · (77.320° / 180°) · π · (2 cm)²

A ≈ 2.423 cm²

The area of the shaded region is approximately equal to 2.423 square centimeters.

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Answer:

no solution

Step-by-step explanation:

We can think of an absolute value as the magnitude of the number. We can also define it as how far the number is from 0 on the number line. Thus, absolute value must always be positive (or 0).

Start by adding 1 to both sides:

[tex]\frac{|m|}{2}=-2[/tex]

Multiply both sides by 2:

[tex]|m|=-4[/tex]

The absolute value of a number is always positive (with the exception of 0), therefore there is no solution.

Notice we can only add/multiply (or subtract/divide) an equation because of the Addition/Subtraction/Multiplication/Division Property of Equality which state that if you add/subtract/multiply/divide one side of the equation by a certain value, you must do the same to the other side of the equation.

45 square feet is the area of the base of the pedestal.

what is rectangular prism?

we know that

The volume of the pedestal (rectangular prism) is given by the formula

   V =   B × h

where

B is the area of the rectangular base of pedestal

V = 405 ft³

h = 9 ft

put the given values in the formula and solve for B

405 = B × 9

B = 405/9

B = 45 ft²

Therefore, 45 square feet is the area of the base of the pedestal.

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Answer:

  p(x) = x² +30x -375

Step-by-step explanation:

When a quadratic has real rational coefficients, any irrational or complex roots come in conjugate pairs.

A root of p means (x -p) is a factor of the polynomial. Here, we have roots of -15+10√6 and -15-10√6, so the factored form can be written ...

  p(x) = (x -(-15 +10√6))(x -(-15 -10√6))

Using the factoring of the difference of squares, we can write this as ...

  p(x) = (x +15)² -(10√6)²

Expanding the factored form, we can write the polynomial as ...

  p(x) = x² +30x +225 -600

  p(x) = x² +30x -375

The number of contemporary titles and classic titles in Jarred DVDs collection is 2,921 and 579 respectively.

Simultaneous equation is an equation which involves the solving for two unknown values at the same time.

x + y = 3,500

x – y = 2,342

Add both be equation

x + x = 3,500 + 2,342

2x = 5,842

x = 5,842 ÷ 2

x = 2,921

Substitute x = 2,921 into

x – y = 2,342

2,921 - y = 2, 342

-y = 2,342 - 2,921

-y = -579

y = 579

Therefore, the number of contemporary titles and classic titles in Jarred DVDs collection is 2,921 and 579 respectively.

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The solution of the differential equation is [tex]y = (C_{1} + C_{2} x)e^{-2x} +\frac{1}{2} [x^{3} -\frac{3}{2} x-6][/tex] .

According to the given question.

We have a differential equation

[tex]y^{"} + 4y^{'} + 4y = 2x^{3}[/tex]

The above differerntial equation acn be written as

[tex](D^{2} +4D+ 4)= 2x^{3}[/tex]

Now, the auxillary equation for the above differential equation is given by

[tex]m^{2} + 4m + 4 = 0[/tex]

[tex]\implies m^{2} + 2m + 2m + 4 = 0[/tex]

⇒ m (m + 2) + 2(m + 2) = 0

⇒ m(m + 2)(m + 2) = 0

Therefore,

[tex]C.F = (C_{1} + C_{2}x)e^{-2x}[/tex]

Now,

[tex]PI = \frac{1}{D^{2} +4D+4} 2x^{3}[/tex]

[tex]\implies PI = \frac{1}{4(1+\frac{D^{2}+4D }{4} )} 2x^{3}[/tex]

[tex]\implies PI = \frac{1}{4} [1+(\frac{D^{2}+4D }{4} )]^{-1} 2x^{3}[/tex]

[tex]\implies PI = \frac{1}{4} [ 1 - (\frac{D^{2} +4D}{4} )+(\frac{D^{2} +4D}{4} )^{2} -(\frac{D^{2}+4D }{4}) ^{3} ...]2x^{3}[/tex]

[tex]\implies PI =\frac{1}{2} [ x^{3} -\frac{1}{4} (6x)-3x^{2} +3x^{2} -6][/tex]

[tex]\implies PI = \frac{1}{2} [x^{3} -\frac{3}{2} x-6][/tex]

Therefore, the solution of the differential equation will be

y = CI + PI

[tex]y = (C_{1} + C_{2} x)e^{-2x} +\frac{1}{2} [x^{3} -\frac{3}{2} x-6][/tex]

Hence, the solution of the differential equation is [tex]y = (C_{1} + C_{2} x)e^{-2x} +\frac{1}{2} [x^{3} -\frac{3}{2} x-6][/tex] .

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