One week, the music store sold 2 trumpets, 3 clarinets, and 5 violins for $1,240. The next
week, they sold 3 trumpets, 1 clarinet, and 4 violins for $1,027. The following week, they sold
5 7 trumpets, 7 clarinets, and 2 violins for $2,091. Find the cost of a trumpet, clarinet, and
violin

The greatest number of binders Eula can buy is 5

The greatest number of notebooks Eula can buy is 10

The greatest number of binders she can buy  if Eula buys 7 notebooks is 3/2

It is an order relationship that is greater than, greater than, or equal to, less than, or less than or equal to—between two numbers or algebraic expressions.

We have,

4x + 2y ≤ 20

x = number of binders

y = number of notebooks

The greatest number of binders Eula can buy:

Put y = 0.

4x + 2 x 0 ≤ 20

4x ≤ 20

x ≤ 20/4

x ≤ 5

The greatest number of notebooks Eula can buy:

Put x = 0.

4x + 2y ≤ 20

4 x 0 + 2y ≤ 20

2y ≤ 20

y ≤ 10

Eula buys 7 notebooks then, the greatest number of binders she can buy:

4x + 2y ≤ 20

4x + 2 x 7 ≤ 20

4x + 14 ≤ 20

4x ≤ 20 - 14

4x ≤ 6

x ≤ 6/4

x ≤ 3/2

Thus,

The greatest number of binders Eula can buy is 5

The greatest number of notebooks Eula can buy is 10

The greatest number of binders she can buy  if Eula buys 7 notebooks is 3/2

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Answer:

Eula needs to buy binders that cost $4 each and notebooks that cost $2 each. She has $20. The graph of the inequality 4x + 2y ≤ 20, which represents the situation, is shown.What is the greatest number of binders Eula can buy? What is the greatest number of notebooks Eula can buy? If Eula buys 7 notebooks, what is the greatest number of binders she can buy?  ⇒ 1

Step-by-step explanation:

Answer:

between 4 and 5

Step-by-step explanation:

The slope of line AB with points A(4, 5) and B(9, 7) is 2/5.

The slope or gradient of a line is a number that describes both the direction and the steepness of the line.

We have,

A(4, 5) and B(9, 7)

The slope of a line with points A and B is given by:

= d - b / c - a

Where A(a, b) and B(c, d) are the coordinates of the points.

We have the points:

A(4, 5) = (a, b)

B(9, 7) = (c, d)

The slope of the line AB:

= (7 - 5) / (9 - 4) = 2 / 5

Therefore the slope of line AB with points A(4, 5) and B(9, 7) is 2/5.

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The ratio of high school students who are in a music class to students who are not is 9 : 41.

So, 18% of high school students are in a music class which means the rest 82% are not in a music class.

Then the ratio:

Simplest form:

Therefore, the ratio is 9 : 41.

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Using prime factors the root of given variety of 777924 is 882 .

The image of prime factorization of 777924 is given below

By prime factorization of 777924 we have a tendency to follow five easy steps:

1. we have a tendency to write variety 777924 higher than a 2-column table

2. we have a tendency to divide 777924 by the tiniest attainable factor

3. we have a tendency to write down on the left facet of the table the factor and next variety to factorize on the ride facet

4. we have a tendency to still consider this fashion (we subsume odd numbers by making an attempt little prime factors)

5. we have a tendency to continue till we have a tendency to reach one on the ride facet of the table

Prime factorization of 777924 = 1×2×2×3×3×3×3×7×7×7×7

On doing square root , we get

[tex]\sqrt{777924} =\sqrt{ 1\times2\times2\times3\times3\times3\times3\times7\times7\times7\times7}[/tex]

              = 882

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9.9 and 9.95 are the result of rounding off 9.949 to the nearest tenth and hundredth.

So, rounding off:

Therefore, rounding off of 9.949 to the nearest tenth and hundredth is 9.9 and 9.95.

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Step-by-step explanation:

So, the expression “three times the variable x” can be written in a number of ways: 3x, 3(x), or 3 · x. Use the distributive property to expand the expression 9(4 + x).

Answer:

[tex]2(35)+4=74[/tex]

∠SQR = 74

Step-by-step explanation:

[tex](2m+4)+(3m+1)=180\\5m+5=180\\ -5\\5m=175\\/5\\175/5=35[/tex]

Answer:

SQR = 74

Step-by-step explanation:

2m + 4 + 3m + 1 = 180

5m + 5 = 180

5m = 175

m = 35

35 * 2 + 4

70 + 4 = 74

Angle SQR = 74

Answer:

11

Step-by-step explanation:

Answer:

[tex]\textsf{Rewrite the original equation as $x^2+\dfrac{1}{3}x=\boxed{\dfrac{2}{9}}$}[/tex]

[tex]\textsf{Add appropriate number to make the left side a perfect square trinomial}[/tex]

[tex]x^2+\dfrac{1}{3}x+\boxed{\dfrac{1}{36}}=\dfrac{2}{9}+\boxed{\dfrac{1}{36}}[/tex]

[tex]\textsf{Factor the left side as a perfect square and combine the right hand side into one number}[/tex][tex]\left(x+\boxed{\dfrac{1}{6}}\:\right)^2=\boxed{\dfrac{1}{4}}[/tex]

[tex]\textsf{Final answers $x=\boxed{\dfrac{1}{3}, - \dfrac{2}{3}}$}[/tex]

Step-by-step explanation:

Given equation:

[tex]18x^2+6x-4=0[/tex]

Add 4 to both sides:

[tex]\implies 18x^2+6x-4+4=0+4[/tex]

[tex]\implies 18x^2+6x=4[/tex]

Divide both sides by 18:

[tex]\implies \dfrac{18x^2}{18}+\dfrac{6x}{18}=\dfrac{4}{18}[/tex]

[tex]\implies x^2+\dfrac{1}{3}x=\dfrac{2}{9}[/tex]

Add the square of half the coefficient of x to both sides:

[tex]\implies x^2+\dfrac{1}{3}x+\left(\dfrac{\frac{1}{3}}{2}\right)^2=\dfrac{2}{9}+\left(\dfrac{\frac{1}{3}}{2}\right)^2[/tex]

[tex]\implies x^2+\dfrac{1}{3}x+\left(\dfrac{1}{6}}\right)^2=\dfrac{2}{9}+\left(\dfrac{1}{6}\right)^2[/tex]

[tex]\implies x^2+\dfrac{1}{3}x+\dfrac{1}{36}=\dfrac{2}{9}+\dfrac{1}{36}[/tex]

Factor the perfect square trinomial on the left side and combine the numbers on the right side:

[tex]\implies \left(x+\dfrac{1}{6}\right)^2=\dfrac{1}{4}[/tex]

Square root both sides:

[tex]\implies \sqrt{\left(x+\dfrac{1}{6}\right)^2}=\sqrt{\dfrac{1}{4}}[/tex]

[tex]\implies x+\dfrac{1}{6}=\pm \dfrac{\sqrt{1}}{\sqrt{4}}[/tex]

[tex]\implies x+\dfrac{1}{6}=\pm \dfrac{1}{2}[/tex]

Subtract 1/6 from both sides:

[tex]\implies x+\dfrac{1}{6}-\dfrac{1}{6}=\pm\dfrac{1}{2}-\dfrac{1}{6}[/tex]

[tex]\implies x=-\dfrac{1}{6}\pm\dfrac{1}{2}[/tex]

Therefore:

[tex]\implies x=-\dfrac{1}{6}+\dfrac{1}{2}=\dfrac{1}{3}[/tex]

[tex]\implies x=-\dfrac{1}{6}-\dfrac{1}{2}=-\dfrac{2}{3}[/tex]

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Answer:

108

Step-by-step explanation:

3²(2³+4) calculate to the power of 2 and get 9.

9 (2³+4) calculate 2 to the power of 3 and get 8.

9 (8+4) add 8 and 4 to get 12.

9 · 12 = 108 multiply 9 and 12 to get 108.

Answer:

3 < x < 5

Step-by-step explanation:

Expression is |2x − 8| < 2

The absolute rule says that if |u| < a then -a < u < a

Here u = 2x-8

So we get -2 < 2x - 8 < 2

This means

2x - 8 > -2            

==> 2x > -2 + 8       (add 8 to both sides)

==> 2x > 6               (simplify)

==> x > 3                  (divide by 2 both sides)

and,

2x - 8 < 2 gives

==> 2x < 2 + 8          (add 8 to both sides)

==> 2x < 10              (simplify)

==> x < 5                  (divide by 2 both sides)

So the solution to |2x − 8| < 2

is 3 < x < 5

Answer:

Step-by-step explanation:

According to the question, the cupcake cafe makes 1/2 times as much revenue on doughnuts as muffins and the total revenue is total sales were $44,000. Let us assume that the revenue on muffins is X and then the revenue on doughnuts is 412X. 4 1 2 X . The amount of money earned from muffins is 20000 dollars

The length and width of the yard with perimeter of 400 feet are 136 ft and 64 ft respectively.

The yard has a perimeter of 400 feet.

Eight times the length of the yard equals seventeen times the width.

Therefore,

perimeter of a rectangular yard = 2(l + w)

where

Therefore,

8l = 17w

Hence,

l = 17 / 8 w

perimeter of a rectangular yard = 2(17 / 8 w + w)

perimeter of a rectangular yard = 2( 25  /8 w)

perimeter of a rectangular yard = 50/ 8 w

400 = 50 / 8 w

cross multiply

3200 = 50w

divide both sides by 50

w = 3200 / 50

w = 64 ft

l = 17 / 8 × 64 = 136 ft

Therefore, the length and width of the yard with perimeter of 400 feet are 136 ft and 64 ft respectively.

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[tex] \large \bf \implies \angle{BAC} = 40 \degree[/tex]

[tex] \bf \implies \angle{ABC} + \angle{BAC} = 90\degree[/tex] [The acute angles of a right triangle are complementary]

[tex] \bf \implies 50x + 40x = 90\degree[/tex]

Substitute :

[tex]\angle{BAC} = 40x \: \: , \: \: \angle{ABC} = 50x \: into \: \angle{ABC} + \angle{BAC} = 90\degree[/tex]

Calculate 50x + 40x = 90° ↑

Substitute x = 1 into [tex]\bf{\angle{BAC} = 40x}[/tex] ↑

[tex] \boxed{ \bold{\angle{BAC} = 40} }\: \mathfrak{ans.}[/tex]

The garden has a length of 37.5 yards and width of 37.5 yards as well as an opening of 12.5 yards.

An equation is an expression that shows the relationship between two or more numbers and variables.

Let x represent the length and y represent the width.

150 yards of fencing is available, hence:

2(x + y) = 150

x + y = 75

y = 75 - x     (1)

The area (A) of the garden is given as:

A = xy

A = x(75 - x)

A = 75x - x²

The maximum area is at A' = 0

A' = 75 - 2x

75 - 2x = 0

x = 37.5 yards

y = 75 - x = 75 - 37.5 = 37.5

Opening of the garden = 1/3 * 37.5 = 12.5 yards

The garden has a length of 37.5 yards and width of 37.5 yards as well as an opening of 12.5 yards.

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Answer: 6.12315 × 105

Step-by-step explanation: No worry's just here for the fun of it

There are 40 rows with 53 seats in each row.

This auditorium is divided into rows of seats, and because it is rectangular, each row has the same number of seats.

So,

Therefore, there are 40 rows with 53 seats in each row because we can't have a negative number of rows.

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In the differential equation

[tex]\dfrac{dy}{dx} + 12y = \cos\left(\dfrac{\pi x}{12}\right)[/tex]

multiply on both sides by the integrating factor

[tex]\mu = \exp\left(\displaystyle\int12\,dx\right) = e^{12x}[/tex]

Then the left side condenses to the derivative of a product.

[tex]e^{12x} \dfrac{dy}{dx} + 12 e^{12x} y = e^{12x} \cos\left(\dfrac{\pi x}{12}\right)[/tex]

[tex]\dfrac{d}{dx}\left[e^{12x}y\right] = e^{12x}\cos\left(\dfrac{\pi x}{12}\right)[/tex]

Integrate both sides with respect to [tex]x[/tex], and use the initial condition [tex]y(0)=0[/tex] to solve for the constant [tex]C[/tex].

[tex]\displaystyle \int \frac{d}{dx} \left[e^{12x}y\right] \, dx = \int e^{12x} \cos\left(\dfrac{\pi x}{12}\right) \, dx[/tex]

As an alternative to integration by parts, recall

[tex]e^{ix} = \cos(x) + i \sin(x)[/tex]

Now

[tex]e^{12x} \cos\left(\dfrac{\pi x}{12}\right) = e^{12x} \mathrm{Re}\left(e^{i\pi x/12}\right) = \mathrm{Re}\left(e^{(12+i\pi/12)x}\right)[/tex]

[tex]\displaystyle \int \mathrm{Re}\left(e^{(12+i\pi/12)x}\right) \, dx = \mathrm{Re}\left(\int e^{(12+i\pi/12)x} \, dx\right)[/tex]

[tex]\displaystyle. ~~~~~~~~ = \mathrm{Re}\left(\frac1{12+i\frac\pi{12}} e^{(12+i\pi/12)x}\right) + C[/tex]

[tex]\displaystyle. ~~~~~~~~ = \mathrm{Re}\left(\frac{12 - i\frac\pi{12}}{12^2 + \frac{\pi^2}{12^2}} e^{12x} \left(\cos\left(\frac{\pi x}{12}\right) + i \sin\left(\frac{\pi x}{12}\right)\right)\right) + C[/tex]

[tex]\displaystyle. ~~~~~~~~ = \frac{12}{12^2 + \frac{\pi^2}{12^2}} e^{12x} \cos\left(\frac{\pi x}{12}\right) + \frac\pi{12} e^{12x} \sin\left(\frac{\pi x}{12}\right) + C[/tex]

[tex]\displaystyle. ~~~~~~~~ = \frac1{12(12^4+\pi^2)} e^{12x} \left(12^4 \cos\left(\frac{\pi x}{12}\right) + \pi (12^4+\pi^2) \sin\left(\frac{\pi x}{12}\right)\right) + C[/tex]

Solve for [tex]y[/tex].

[tex]\displaystyle e^{12x} y = \frac1{12(12^4+\pi^2)} e^{12x} \left(12^4 \cos\left(\frac{\pi x}{12}\right) + \pi (12^4+\pi^2) \sin\left(\frac{\pi x}{12}\right)\right) + C[/tex]

[tex]\displaystyle y = \frac1{12(12^4+\pi^2)} \left(12^4 \cos\left(\frac{\pi x}{12}\right) + \pi (12^4+\pi^2) \sin\left(\frac{\pi x}{12}\right)\right) + C[/tex]

Solve for [tex]C[/tex].

[tex]y(0)=0 \implies 0 = \dfrac1{12(12^4+\pi^2)} \left(12^4 + 0\right) + C \implies C = -\dfrac{12^3}{12^4+\pi^2}[/tex]

So, the particular solution to the initial value problem is

[tex]\displaystyle y = \frac1{12(12^4+\pi^2)} \left(12^4 \cos\left(\frac{\pi x}{12}\right) + \pi (12^4+\pi^2) \sin\left(\frac{\pi x}{12}\right)\right) - \frac{12^3}{12^4+\pi^2}[/tex]

Recall that

[tex]R\cos(\alpha-\beta) = R\cos(\alpha)\cos(\beta) + R\sin(\alpha)\sin(\beta)[/tex]

Let [tex]\alpha=\frac{\pi x}{12}[/tex]. Then

[tex]\begin{cases} R\cos(\beta) = 12^4 \\ R\sin(\beta) = 12^4\pi+\pi^3 \end{cases} \\\\ \implies \begin{cases} (R\cos(\beta))^2 + (R\sin(\beta))^2 = R^2 = 12^8 + (12^4\pi + \pi^3)^2 \\ \frac{R\sin(\beta)}{R\cos(\beta)}=\tan(\beta)=\pi+\frac{\pi^3}{12^4}\end{cases}[/tex]

Whatever [tex]R[/tex] and [tex]\beta[/tex] may actually be, the point here is that we can condense [tex]y[/tex] into a single cosine expression, so choice (D) is correct, since [tex]\cos(x)[/tex] is periodic. This also means choice (C) is also correct, since [tex]\beta=\cos(x)\implies\beta=\cos(x+2n\pi)[/tex] for infinitely many integers [tex]n[/tex]. This simultaneously eliminates (A) and (B).

The system of inequalities for the given situation is x+y≤11 and 2x+4y≥40.

Given that, the cost of each drink = $2 and the cost of each taco = $4.

A system of inequalities is a set of two or more inequalities in one or more variables. Systems of inequalities are used when a problem requires a range of solutions, and there is more than one constraint on those solutions.

x represents the number of drinks purchased and y represents the number of tacos purchased.

Juan has $40 to spend and must buy a minimum of 11 drinks and tacos altogether.

So inequalities are x+y≤11 and 2x+4y≥40

Therefore, the system of inequalities for the given situation is x+y≤11 and 2x+4y≥40.

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The dimensions of the rug would be 27 feet long and 19 feet wide with a 6 foot space around the carpet uniformly.

She wants to buy a rug for a room that is 25 ft wide and 33 ft long.

She wants  to leave a uniform strip of floor around the rug.

she can afford to buy 513 square feet of carpeting.

Cynthia room area = lw

where

Therefore,

Cynthia room area = 25 × 33

Cynthia room area = 825 ft²

She wants to leave a uniform strip of floor all around the rug and has affordability to buy 513 square feet of carpet.

Therefore, the dimension the rug should have is as follows:

(25 - 6) (33 - 6) = 513

19 × 27 = 513 ft²

Therefore, the dimensions of the rug would be 27 feet long and 19 feet wide with a 6 foot space around the carpet uniformly.

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If the school earns $264 for 8 senior tickets and 12 child tickets and $237 for 11 senior tickets and 6 child tickets then the price of 1 senior ticket be $15 and the price of child ticket be $12.

Given that on the first day of ticket sales, the school sold 8 senior citizen tickets and 12 child tickets for a total of $264 and the school earns $237 on IInd day by selling 11 senior citizen tickets and 6 child tickets.

We are required to define the variables and solve the system of the equations.

Suppose the price of 1 ticket of senior citizen be x.

Suppose the price of 1 ticket of child be y.

The equations will be:

8x+12y=264--------1

11x+6y=237--------2

Multiply equation 1 by 11 and multiply equation 2 by 8 and then subtract equation 2 from equation 1.

88x+132y-88x-48y=2904-1896

84y=1008

y=1008/84

y=12

Use the value of y in equation 1 to get the value of x.

8x+12y=264

8x+12*12=264

8x+144=264

8x=264-144

8x=120

x=120/8

x=15

Hence if the school earns $264 for 8 senior tickets and 12 child tickets and $237 for 11 senior tickets and 6 child tickets then the price of 1 senior ticket be $15 and the price of child ticket be $12.

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The cost of a Senior citizen ticket is $15, while a children's ticket cost $12.

On the first day:

The school sold 12 child tickets and 8 senior citizen tickets for a total amount of $264.

On the second day:

The school sold 11 senior citizen tickets and 6 child tickets for a total of $237.

Let A be the price of a senior citizen ticket and B be the price of a child ticket.

So, the equation for the first day:

8A + 12B = 264

The equation for the second day,

11A + 6B = 237

Multiplying the equation for the second day by 2 and subtracting the equation for the first day.

We get,

22A + 12B - 8A - 12B = 474 - 264

14A = 210

A = 15

Substituting the value of A in the equation 8A + 12B = 264,

8A + 12B = 264

8(15) + 12B = 264

120 + 12B = 264

12B = 144

B = 12

Therefore, the price of a senior citizen ticket is $15 and the price of a child ticket is $12.

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Answer:Domar range x

Step-by-step explanation:

The solution to the equation 1/3|x - 8| = 10 is x = 38

Expressions are mathematical statements that are represented by variables, coefficients and operators

The equation is given as

1/3|x - 8| = 10

Multiply through by 3

So, we have

|x - 8| = 30

Remove the absolute bracket

So, we have

x - 8 = 30

Add 8 to both sides of the equation

So, we have

x = 38

Hence, the solution to the equation is x = 38

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Answer:

The reference angle is π/6.

Cosine is -(√3)/2.

Step-by-step explanation:

To find the reference angle, find the acute angle in quadrant I and use it as a reference for the given expression.

For the cosine, the cosine is the sine of the complementary angle. The complementary angle is the given angle beside it minus a right angle, which is exactly 90 degrees. If the angle is 25 degrees, its complementary angle will be double its amount, 50 degrees. Then, for angle angle measured "theta", the cosine is equal to the sine's right-angle subtracted by theta.

The base 10 representation of the number 142 base 5 as required to be determined in the task content is; 47.

Since, the conversion of numbers from any base to base 10 requires imaginary exponents to represent place value in such number as follows;

142₅ can therefore be written as follows; 1²4¹2⁰₅.

On this note the evaluation is carried out by summing the product of each digits and 5 to the corresponding power as follows;

1²4¹2⁰₅ = (2 × 5⁰) + (4 × 5¹) + (1 × 5²)

= (2 × 1) + (4 × 5) + (1 × 25)

= 2 + 20 + 25.

= 45.

Ultimately, the base 10 representation of the number 142 base 5 as required to be determined in the task content is; 47.

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Answer:

The circumference is 153.93804... or 153.94...

Step-by-step explanation:

Hope it helps! =D

Answer:

153.94 meters.

Step-by-step explanation:

The formula for finding the circumference of a circle is C = πd, where d is the diameter of the circle. With a diameter of 49m, the circumference can be calculated as C = π(49) ≈ 153.94m. Therefore, the circumference of the circle is approximately 153.94 meters.

Based on the graph, which statement could describe Janelle’s trip home from school is that D. Janelle rode the bus to the bus stop, talked with a friend, and then walked home.

It should be noted that a graph is a diagram that is used to represent a system of connections or interrelations that is among two or more things by a number of distinctive dots, bars, etc

Therefore, based on the graph, which statement could describe Janelle’s trip home from school is that Janelle rode the bus to the bus stop, talked with a friend, and then walked home.

In conclusion, the correct option is D.

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Based on the graph, which statement could describe Janelle's trip home from school?

Janelle waited for the bus, rode the bus, and then walked home.

Janelle walked the opposite direction from home to the library, rode the bus, and then walked to a friend's house.

Janelle walked home at a constant speed.

Janelle rode the bus to the bus stop, talked with a friend, and then walked home

If Leslie is about to type 56 words per minute, she would be able to type 8 pages in one hour

How many words can Leslie type in one hour?

The fact that Leslie can type 56 words per minute means that he is able, means that the number of words she is able to type in one hour is determined as 56 words multiplied 60 minutes which make an hour

number of words in one hour=56*60

number of words in one hour=3360

The number of pages typed is determined as the 3360 words typed in one hour divided by the number of words in a page

number of pages type=3360/420

number of pages typed=8 pages

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Step-by-step explanation:

We have,

[tex]\begin{gathered} \bullet \: \bold{ - \dfrac{\pi}{4} < \beta < 0 < \alpha < \dfrac{\pi}{4} } \\ \\ \implies \: - \dfrac{\pi}{4} < \alpha + \beta < \dfrac{\pi}{4} \end{gathered}[/tex]

[tex]\begin{gathered} \rm\bullet \: \: \: \sin( \alpha + \beta ) = \dfrac{1}{3} \: \: \: \: \: and \: \: \: \: \: \cos( \alpha - \beta ) = \dfrac{2}{3} \\ \end{gathered} [/tex]

Now,

[tex]\begin{gathered} y= \bigg( \dfrac{ \sin( \alpha ) }{ \cos( \beta ) } + \dfrac{ \cos( \beta ) }{ \sin( \alpha ) } + \dfrac{ \cos( \alpha ) }{ \sin( \beta ) } + \dfrac{ \sin( \beta ) }{ \cos( \alpha ) } \bigg)^{2} \\\end{gathered} [/tex]

[tex]\begin{gathered} \implies y= \bigg( \dfrac{ \sin( \alpha ) }{ \cos( \beta ) } + \dfrac{ \sin( \beta ) }{ \cos( \alpha ) } + \dfrac{ \cos( \beta ) }{ \sin( \alpha ) } + \dfrac{ \cos( \alpha ) }{ \sin( \beta ) } \bigg)^{2} \\\end{gathered} [/tex]

[tex]\begin{gathered} \implies y= \bigg( \dfrac{ \sin( \alpha ) \cos( \alpha ) + \sin( \beta \cos( \beta ) ) }{ \cos( \beta ) \cos( \alpha) } + \dfrac{ \sin( \alpha) \cos( \alpha ) + \sin( \beta ) \cos( \beta ) }{ \sin( \alpha ) \sin( \beta ) } \bigg)^{2} \\\end{gathered}[/tex]

[tex]\begin{gathered} \implies y= \bigg( \dfrac{ \sin( \alpha + \beta ) }{ \cos( \beta ) \cos( \alpha) } + \dfrac{ \sin( \alpha + \beta ) }{ \sin( \alpha ) \sin( \beta ) } \bigg)^{2} \\\end{gathered} [/tex]

[tex]\begin{gathered} \implies y= \sin^{2} ( \alpha + \beta ) \bigg( \dfrac{ 1 }{ \cos( \beta ) \cos( \alpha) } + \dfrac{ 1 }{ \sin( \alpha ) \sin( \beta ) } \bigg)^{2} \\\end{gathered} [/tex]

[tex]\begin{gathered} \implies y= \sin^{2} ( \alpha + \beta ) \bigg( \dfrac{\cos( \beta ) \cos( \alpha) + \sin( \alpha ) \sin( \beta ) }{ \cos( \beta ) \cos( \alpha) \sin( \alpha ) \sin( \beta )} \bigg)^{2} \\\end{gathered} [/tex]

[tex]\begin{gathered} \implies y= \sin^{2} ( \alpha + \beta ) \bigg( \dfrac{\cos( \alpha - \beta ) }{ \cos( \beta ) \cos( \alpha) \sin( \alpha ) \sin( \beta )} \bigg)^{2} \\\end{gathered} [/tex]

[tex]\begin{gathered} \implies y= \dfrac{4\sin^{2} ( \alpha + \beta ) \cdot\cos^{2} ( \alpha - \beta ) }{ \left \{2\cos( \alpha ) \cos( \beta ) \cdot 2\sin( \alpha ) \sin( \beta ) \right \}^{2} } \\\end{gathered} [/tex]

[tex]\begin{gathered} \implies y= \dfrac{4\sin^{2} ( \alpha + \beta ) \cdot\cos^{2} ( \alpha - \beta ) }{ \left \{\cos( \alpha + \beta ) + \cos( \alpha + \beta ) \right \} ^{2} \left \{ \cos( \alpha - \beta ) - \cos( \alpha + \beta ) \right \}^{2} } \\\end{gathered} [/tex]

[tex]\begin{gathered} \implies y= \dfrac{4\sin^{2} ( \alpha + \beta ) \cdot\cos^{2} ( \alpha - \beta ) }{ \left \{ \cos^{2} ( \alpha - \beta ) - \cos^{2} ( \alpha + \beta ) \right \}^{2} } \\\end{gathered}[/tex]

[tex]\begin{gathered} \implies y= \dfrac{4\sin^{2} ( \alpha + \beta ) \cdot\cos^{2} ( \alpha - \beta ) }{ \left \{ \cos^{2} ( \alpha - \beta ) - 1 + \sin^{2} ( \alpha + \beta ) \right \}^{2} } \\\end{gathered}[/tex]

Putting the values given above, we get,

[tex]\begin{gathered} \implies y= \dfrac{4 \cdot \dfrac{1}{9} \cdot\dfrac{4}{9} }{ \left \{ \dfrac{4}{9} - 1 + \dfrac{1}{9} \right \}^{2} } \\\end{gathered} [/tex]

[tex]\begin{gathered} \implies y= \dfrac{\dfrac{16}{81} }{ \left \{ \dfrac{5}{9} - 1\right \}^{2} } \\\end{gathered}[/tex]

[tex]\begin{gathered} \implies y= \dfrac{\dfrac{16}{81} }{ \left \{ \dfrac{5 - 9}{9}\right \}^{2} } \\\end{gathered} [/tex]

[tex]\begin{gathered} \implies y= \dfrac{\dfrac{16}{81} }{ \left \{ \dfrac{- 4}{9}\right \}^{2} } \\\end{gathered}[/tex]

[tex]\begin{gathered} \implies y= \dfrac{\dfrac{16}{81} }{ \dfrac{16}{81}} \\\end{gathered} [/tex]

[tex]⟹y=1[/tex]