PLEASE HELP! (3/4) - 50 POINTS -

Answer:

1/3(a+b)

hope it helps :>

Answer:

Perimeter = 42 cm

Step-by-step explanation:

A square has all equal sides so you would just add 10.5 + 10.5 + 10.5 + 10.5 to get 42 cm.

Answer:

42 cm

Step-by-step explanation:

Side of square = 10.5 cm (given)

Perimeter of square = Side X 4

                                  = 10.5 X 4

                                  = 42 cm

HOPE THIS HELPED YOU !

:)

Complete question is;

A machine used to fill gallon-sized paint cans is regulated so that the amount of paint dispensed has a mean of 128 ounces and a standard deviation of 0.20 ounce. You randomly select 35 cans and carefully measure the contents. The sample mean of the cans is 127.9 ounces. Does the machine need to be? reset? Explain your reasoning.

(yes/no)?, it is (very unlikely/ likely) that you would have randomly sampled 35 cans with a mean equal to 127.9 ?ounces, because it (lies/ does not lie) within the range of a usual? event, namely within (1 standard deviation, 2 standard deviations 3 standard deviations) of the mean of the sample means.

Answer:

Yes, we should reset the machine because it is unusual to have a mean equal to 127.9 from a random sample of 35 as the mean of 127.9 doesn't fall within range of a usual event with 2 standard deviations of the mean of the sample means.

Step-by-step explanation:

We are given;

Mean: μ = 128

Standard deviation; σ = 0.2

n = 35

Now, formula for standard error of mean is given as;

se = σ/√n

se = 0.2/√35

se = 0.0338

Normally, the range of values should be within 2 standard deviations of mean. In this case, normal range of values will be;

μ ± 2se = 128 ± 0.0338

This gives; 127.9662, 128.0338

So, Yes, we should reset the machine because it is unusual to have a mean equal to 127.9 from a random sample of 35 as the mean of 127.9 doesn't fall within range of a usual event with 2 standard deviations of the mean of the sample means.

Answer:

16.50

Step-by-step explanation:

Constant of proportionality = no of gallons of water per 1 minute.

In the first row, we have 16.50 gallons of water per 1 minute.

In the 2nd row, we have 24.75 gallons of water in 1.5 minutes. In 1 minute, we will have 24.75 ÷ 1.5 = 16.50 gallons

In the 3rd row, we have 33 gallons in 2 minutes. In 1 minute, we will have 33 ÷ 2 = 16.50 gallons.

We can see that there seems to be the same constant of proportionality for the 2nd and 3rd row, which is 16.50.

Thus, a relationship between gallons of water (w) and time (t), considering the constant, 16.50, can be written as: [tex] w = 16.50t [/tex]

This means the constant of proportionality, 16.50, is same for all rows.

Find the powers [tex]a=\sqrt{2}+\sqrt{3}[/tex]

$a^{2}=5+2 \sqrt{6}$

$a^{3}=11 \sqrt{2}+9 \sqrt{3}$

The cubic term gives us a clue, we can use a linear combination to eliminate the root 3 term $a^{3}-9 a=2 \sqrt{2}$ Square $\left(a^{3}-9 a\right)^{2}=8$ which gives one solution. Expand we have $a^{6}-18 a^{4}-81 a^{2}=8$ Hence the polynomial $x^{6}-18 x^{4}-81 x^{2}-8$ will have a as a solution.

Note this is not the simplest solution as $x^{6}-18 x^{4}-81 x^{2}-8=\left(x^{2}-8\right)\left(x^{4}-10 x^{2}+1\right)$

so fits with the other answers.

Answer:

[tex]y^3 -6y-6[/tex]

Step-by-step explanation:

utilise the formula a+(n-1)d

a is the first number while d is common difference

Answer:

22

Step-by-step explanation:

Using the formular, Un = a + (n - 1)d

Where n = 10; a = -23; d = 5

U10 = -23 + (9)* 5

U10 = -23 + 45 = 22

Answer:

X is uniformly distributed.

Step-by-step explanation:

Uniform Distribution:

This is the type of distribution where all outcome of a certain event have equal likeliness of occurrence.

Example of Uniform Distribution is - tossing a coin. The probability of getting a head is the same as the probability of getting a tail. The have equal likeliness of occurrence.

Answer:

The answer is D)90

Hope I helped

Answer:

The upper bound of the confidence interval is 0.34

Step-by-step explanation:

Here in this question, we want to calculate the upper bound of the confidence interval.

We start by calculating the margin of error.

Mathematically, the margin of error = 0.29 -0.24 = 0.05

So to get the upper bound of the confidence interval, we simply add this margin of error to the mean

That would be 0.05 + 0.29 = 0.34

[tex]\displaystyle\\(a+b)^n\\T_{r+1}=\binom{n}{r}a^{n-r}b^r\\\\\\(x+2)^7\\a=2x\\b=3\\r+1=4\Rightarrow r=3\\n=5\\T_4=\binom{5}{3}\cdot (2x)^{5-3}\cdot3^3\\T_4=\dfrac{5!}{3!2!}\cdot 4x^2\cdot27\\T_4=\dfrac{4\cdot5}{2}\cdot 4x^2\cdot27\\\\T_4=1080x^2[/tex]

Answer:

300.05 miles

Step-by-step explanation:

initial fee= $39.99

final bill = $ 100

cost =$ 0.20 per mile

remaining amount = $ 60.01

solution,

she drive = remaining amount / cost

=60.01/0.20

=300.05 miles

Answer:

500 miles

Step-by-step explanation:

Let us use cross multiplication to find the unknown amount.

Given:

1) Cost for 1 mile=$0.20

2)Cost for x miles=$100

Solution:

No of miles                             Cost

1) 1                                             $0.20

2)x                                             $100

By cross multiplying,

100 x 1= 0.20x

x=100/0.20

x=500 miles

Thank you!

Answer: y=(4/3)x+2/3

Step-by-step explanation:

Slope-intercept form is expressed as y=mx+b

First, find the slope (m):

m= rise/run or vertical/horizontal or y/x (found between the highlighted points)

m = 4/3

Second, find b:

Use one of the highlighted points for (x, y)

2=4/3(1)+b

6/3=4/3+b

2/3=b

b=2/3

Plug it into the equation:

You get y=(4/3)x+2/3 :)

Answer:

n + 1

Step-by-step explanation:

Starting with the integer 'n,' we represent the "next integer" by n + 1.

Answer:

Question 1 = D) Acute

Question 2 = C)3 feet

Question 3 = D) Obtuse

Question 4 = C)27.73 ft.

Step-by-step explanation:

Question 1: A triangle has sides with lengths 5, 6, and 7. Is the triangle right, acute, or obtuse?

In order to be able to accurately classify that a triangle with 3 given sides is either a right , acute or obtuse angle, we use the Pythagoras Theorem

Where:

If a² + b² = c² = Right angle triangle

If a² +b² > c² = Acute triangle.

If a² +b² < c² = Obtuse triangle.

It is important to note that the length ‘‘c′′ is always the longest.

Therefore, for the above question, we have lengths

5 = a, 6 = b and c = 7

a² + b² = c²

5² + 6² = 7²

25 + 36 = 49

61 = 49

61 ≠ 49, Hence 61 > 49

Therefore, this is an Acute Triangle

Question 2: A 15-foot statue casts a 20-foot shadow. How tall is a person who casts a 4-foot-long shadow?

This is question that deals with proportion.

The formula to solve for this:

Height of the statue/ Length of the shadow of the person = Height of the person/ Length of the shadow of the person

Height of the statue = 15 feet

Length of the shadow of the person = 20 feet

Height of the person = unknown

Length of the shadow of the person = 4

15/ 20 = Height of the person/4

Cross Multiply

15 × 4 = 20 × Height of the person

Height of the person = 15 × 4/20

= 60/20

Height of the person = 3 feet

Therefore, the person is 3 feet tall.

Question 3: A triangle has sides with lengths 17, 12, and 9. Is the triangle right, acute, or obtuse?

In order to be able to accurately classify that a triangle with 3 given sides is either a right , acute or obtuse angle, we use the Pythagoras Theorem

Where:

If a² + b² = c² = Right angle triangle

If a² +b² > c² = Acute triangle.

If a² +b² < c² = Obtuse triangle.

It is important to note that the length ‘‘c′′ is always the longest.

Therefore, for the above question, we have lengths 17, 12, 9

9 = a, 12 = b and c = 17

a² + b² = c²

9² + 12² = 17²

81 + 144 = 289

225 = 289

225 ≠ 289

225 < 289

Hence, This is an Obtuse Triangle.

Question 4: Two friends are standing at opposite corners of a rectangular courtyard. The dimensions of the courtyard are 12 ft. by 25 ft. How far apart are the friends?

To calculate how far apart the two friends are we use the formula

Distance = √ ( Length² + Breadth²)

We are given dimensions: 12ft by 25ft

Length = 12ft

Breadth = 25ft

Distance = √(12ft)² + (25ft)²

Distance = √144ft²+ 625ft²

Distance = √769ft²

Distance = 27.730849248ft

Approximately ≈27.73ft

Therefore, the friends are 27.73ft apart.

Answer:

Solution : Option B

Step-by-Step Explanation:

We have the following system of equations at hand here.

{ x = 5 cot(t), y = - 3csc(t) + 4 }

Now instead of isolating the t from either equation, let's isolate cot(t) and csc(t) --- Step #1,

x = 5 cot(t) ⇒ x - 5 = cot(t),

y = - 3csc(t) + 4 ⇒ y - 4 = - 3csc(t) ⇒ y - 4 / - 3 = csc(t)

Now let's square these two equations. We know that csc²θ - cot²θ = 1, so let's subtract the equations  as well. --- Step #2

( y - 4 / - 3 )² = (csc(t))²

- ( x - 5 / 1 )² = (cot(t))²  

___________________

(y - 4)² / 9 - x² / 25 = 1

And as we are subtracting the two expressions, this is an example of a hyperbola. Therefore your solution is option b.

Answer:

$3

Step-by-step explanation:

Given

Total Cost Price: $12,000

Unit Cost Price= $6

Total Selling Price = $18,000

Required

Determine the profit on each share

First, we need to determine the units of share bought;

Units = Total cost price / Unit Cost Price

[tex]Units = \frac{\$12000}{\$6}[/tex]

[tex]Units = 2000[/tex]

Next is to determine the selling price of each share; This is calculated as follows;

Unit Selling Price = Total Selling Price / Units Sold

[tex]Unit\ Selling\ Price = \frac{\$18000}{\$2000}[/tex]

[tex]Unit\ Selling\ Price = \$9[/tex]

The profit is the difference between the unit cost price and unit selling price

[tex]Profit = Unit\ Selling\ Price - Unit\ Cost\ Price[/tex]

[tex]Profit = \$9 - \$6[/tex]

[tex]Profit = \$3[/tex]

Answer:

A.the type 1 error probability is [tex]\mathbf{\alpha = 0.0244 }[/tex]

B. β  = 0.0122

C. β  = 0.0000

Step-by-step explanation:

Given that:

Mean = 100

standard deviation = 2

sample size = 9

The null and the alternative hypothesis can be computed as follows:

[tex]\mathtt{H_o: \mu = 100}[/tex]

[tex]\mathtt{H_1: \mu \neq 100}[/tex]

A. If the acceptance region is defined as [tex]98.5 < \overline x > 101.5[/tex] , find the type I error probability [tex]\alpha[/tex] .

Assuming the critical region lies within [tex]\overline x < 98.5[/tex] or [tex]\overline x > 101.5[/tex], for a type 1 error to take place, then the sample average x will be within the critical region when the true mean heat evolved is [tex]\mu = 100[/tex]

∴

[tex]\mathtt{\alpha = P( type \ 1 \ error ) = P( reject \ H_o)}[/tex]

[tex]\mathtt{\alpha = P( \overline x < 98.5 ) + P( \overline x > 101.5 )}[/tex]

when  [tex]\mu = 100[/tex]

[tex]\mathtt{\alpha = P \begin {pmatrix} \dfrac{\overline X - \mu}{\dfrac{\sigma}{\sqrt{n}}} < \dfrac{\overline 98.5 - 100}{\dfrac{2}{\sqrt{9}}} \end {pmatrix} + \begin {pmatrix}P(\dfrac{\overline X - \mu}{\dfrac{\sigma}{\sqrt{n}}} > \dfrac{101.5 - 100}{\dfrac{2}{\sqrt{9}}} \end {pmatrix} }[/tex]

[tex]\mathtt{\alpha = P ( Z < \dfrac{-1.5}{\dfrac{2}{3}} ) + P(Z > \dfrac{1.5}{\dfrac{2}{3}}) }[/tex]

[tex]\mathtt{\alpha = P ( Z <-2.25 ) + P(Z > 2.25) }[/tex]

[tex]\mathtt{\alpha = P ( Z <-2.25 ) +( 1- P(Z < 2.25) })[/tex]

From the standard normal distribution tables

[tex]\mathtt{\alpha = 0.0122+( 1- 0.9878) })[/tex]

[tex]\mathtt{\alpha = 0.0122+( 0.0122) })[/tex]

[tex]\mathbf{\alpha = 0.0244 }[/tex]

Thus, the type 1 error probability is [tex]\mathbf{\alpha = 0.0244 }[/tex]

B. Find beta for the case where the true mean heat evolved is 103.

The probability of type II error is represented by β. Type II error implies that we fail to reject null hypothesis [tex]\mathtt{H_o}[/tex]

Thus;

β = P( type II error) - P( fail to reject [tex]\mathtt{H_o}[/tex] )

∴

[tex]\mathtt{\beta = P(98.5 \leq \overline x \leq 101.5) }[/tex]

Given that [tex]\mu = 103[/tex]

[tex]\mathtt{\beta = P( \dfrac{98.5 -103}{\dfrac{2}{\sqrt{9}}} \leq \dfrac{\overline X - \mu}{\dfrac{\sigma}{n}} \leq \dfrac{101.5-103}{\dfrac{2}{\sqrt{9}}}) }[/tex]

[tex]\mathtt{\beta = P( \dfrac{-4.5}{\dfrac{2}{3}} \leq Z \leq \dfrac{-1.5}{\dfrac{2}{3}}) }[/tex]

[tex]\mathtt{\beta = P(-6.75 \leq Z \leq -2.25) }[/tex]

[tex]\mathtt{\beta = P(z< -2.25) - P(z < -6.75 )}[/tex]

From standard normal distribution table

β  = 0.0122 - 0.0000

β  = 0.0122

C. Find beta for the case where the true mean heat evolved is 105. This value of beta is smaller than the one found in part (b) above. Why?

[tex]\mathtt{\beta = P(98.5 \leq \overline x \leq 101.5) }[/tex]

Given that [tex]\mu = 105[/tex]

[tex]\mathtt{\beta = P( \dfrac{98.5 -105}{\dfrac{2}{\sqrt{9}}} \leq \dfrac{\overline X - \mu}{\dfrac{\sigma}{n}} \leq \dfrac{101.5-105}{\dfrac{2}{\sqrt{9}}}) }[/tex]

[tex]\mathtt{\beta = P( \dfrac{-6.5}{\dfrac{2}{3}} \leq Z \leq \dfrac{-3.5}{\dfrac{2}{3}}) }[/tex]

[tex]\mathtt{\beta = P(-9.75 \leq Z \leq -5.25) }[/tex]

[tex]\mathtt{\beta = P(z< -5.25) - P(z < -9.75 )}[/tex]

From standard normal distribution table

β  = 0.0000 - 0.0000

β  = 0.0000

The reason why the value of beta is smaller here is that since the difference between the value for the true mean and the hypothesized value increases, the probability of type II error decreases.

Answer:

x=9/2

Step-by-step explanation:

Let's solve your equation step-by-step.

5x+4(−x−2)=−5x+2(x−1)+12

Step 1: Simplify both sides of the equation.

5x+4(−x−2)=−5x+2(x−1)+12

5x+(4)(−x)+(4)(−2)=−5x+(2)(x)+(2)(−1)+12 (Distribute)

5x+−4x+−8=−5x+2x+−2+12

(5x+−4x)+(−8)=(−5x+2x)+(−2+12) (Combine Like Terms)

x+−8=−3x+10

x−8=−3x+10

Step 2: Add 3x to both sides.

x−8+3x=−3x+10+3x

4x−8=10

Step 3: Add 8 to both sides.

4x−8+8=10+8

4x=18

Step 4: Divide both sides by 4.

4x/4=18/4

x=9/2

Answer:

A. 10%

B. Lower limit= 21

Upper limit = 39

Step-by-step explanation:

Mean = 30

SD = 3

a. COV = SD/|x| × 100

= 3/30 × 100

= 10%

= 0.1

B. For 88.9 chevbychev interval:

= (1 - 1/K²) = 0.889

= 1/K² = 1 - 0.889

= 1/K² = 0.111

= K² = 1/0.111

= K² = 9

= K = √9

K = 3

Lower limit = 30 - 3(3)

Lower limit = 21

Upper limit = 30 + 3(3)

Upper limit = 39

Therefore lower limit is 21 and upper limit is 39

Answer:

  x = 1.00995066776

  x = 2.52925492433

Step-by-step explanation:

This sort of equation is best solved using a graphing calculator. For that purpose, I like to rewrite the equation as a function whose zeros we're seeking. Here, that becomes ...

  [tex]f(x)=\log{(x)}-\log{(x-1)}^2-2\log{(x-1)}[/tex]

The attached graph shows zeros at

  x = 1.00995066776 and 2.52925492433

_____

Comment on the equation

Note that we have taken the middle term to be the square of the log, rather than the log of a square. For the latter interpretation, see mberisso's answer at https://brainly.com/question/17210068

Comment on the answer refinement

We have used Newton's method iteration to refine the solutions to this equation. The solution near 1.00995 requires the initial guess be very close for that method to work properly. Fortunately, the 1.01 value shown on the graph is sufficient for the purpose.

Answer:

(a). x = 80°

(b). x = 7.2 units

Step-by-step explanation:

Angle formed between the tangents from a point outside the circle measure the half of the difference of intercepted arcs.

(a). Here the intercepted arcs are,

    Measure of major arc = 360° - 100°

                                        = 260°

    Measure of minor arc = 100°

   x° = [tex]\frac{1}{2}[m(\text{Major arc})-m(\text{Minor arc})][/tex]

       = [tex]\frac{1}{2}(260-100)[/tex]

    x = 80°

(b). If a secant and tangent are drawn form a point outside the circle, then square of the measure of tangent is equal to the product of the measures of the secant segment and and its external segment.

x² = 4(4 + 9)

x² = 4 × 13

x² = 52

x = √52

x = 7.211 ≈ 7.2 units

Complete Question

On the uploaded image is a similar question that will explain the given question

Answer:

The value of k is  [tex]k = 214285.7[/tex]

The percentage  of the oil that will be cleaned is [tex]x = 80.77\%[/tex]

Step-by-step explanation:

From the question we are told that

   The  cost of cleaning up the spillage is  [tex]C = \frac{ k x }{100 - x }[/tex]  [tex]x \le x \le 100[/tex]

     The  cost of cleaning x =  70% of the oil is  [tex]C = \$500,000[/tex]

Now at  [tex]C = \$500,000[/tex] we have  

       [tex]\$ 500000 = \frac{ k * 70 }{100 - 70 }[/tex]

       [tex]\$ 500000 = \frac{ k * 70 }{30 }[/tex]

      [tex]\$ 500000 = \frac{ k * 70 }{30 }[/tex]

      [tex]k = 214285.7[/tex]

Now  When  [tex]C = \$900,000[/tex]

       [tex]x = 80.77\%[/tex]

De acordo com a disponibilidade da unidade, há apenas a seguinte dosagem: 1g/2mL - ou seja, uma grama de dipirona a cada 2mL

O enunciado está meio mal formulado, pois é dito que foram prescritos 500mg de dipirona e é essa quantidade de farmaco que a criança tem que tomar. Deseja-se saber quantos mL deverao ser administrados.

Fazendo a classica regra de 3, podemos chegar no volume desejado:

(atentar que 500mg = 0,5g)

     g               mL

     1    ---------   2

    0,5  ---------  X    

1 . X = 0,5 . 2

There will be a total of 4 test scores including the final exam. To get a 70, the 4 tests need to equal 4 x 70 = 280 points , to be 79, they have to equal 4 x 79 = 316 points.

The 3 already done = 61 + 62 + 86 = 209 points.

The final exam needs to be between :

280 -209 = 71

316 -209 = 107. The answer would be between 71 and 100%

Complete Question

The complete question is shown on the first uploaded image

Answer:

Option C is the correct option

Step-by-step explanation:

From the question we are told that

   The equation is  [tex]f (x, y , z ) = x^2 +y^2 + z^2[/tex]

    The constraint is  [tex]P(x, y , z) = x + y + z - 24 = 0[/tex]

Now using Lagrange multipliers  we have that  

   [tex]\lambda = \frac{ \delta f }{ \delta x } = 2 x[/tex]  

   [tex]\lambda = \frac{ \delta f }{ \delta y } = y[/tex]  

   [tex]\lambda = \frac{ \delta f }{ \delta z } = 2 z[/tex]

=>       [tex]x = \frac{ \lambda }{2}[/tex]

          [tex]y = \frac{ \lambda }{2}[/tex]

         [tex]z = \frac{ \lambda }{2}[/tex]

From the constraint  we have

      [tex]\frac{\lambda }{2} + \frac{\lambda }{2} + \frac{\lambda }{2} = 24[/tex]

=>   [tex]\frac{3 \lambda }{2} = 24[/tex]

=>   [tex]\lambda = 16[/tex]

substituting for x, y, z

=>   x =  8

=>  y =  8

=>   z =  8        

Hence

    [tex]f (8, 8 , 8 ) = 8^2 +8^2 + 8^2[/tex]

    [tex]f (8, 8 , 8 ) = 192[/tex]

Answer:

Step-by-step explanation:

[tex]-3\left(x+1\right)=-3\left(x+1\right)-5\\\\\mathrm{Add\:}3\left(x+1\right)\mathrm{\:to\:both\:sides}\\\\-3\left(x+1\right)+3\left(x+1\right)=-3\left(x+1\right)-5+3\left(x+1\right)\\\\\mathrm{Simplify}\\\\0=-5\\\\\mathrm{The\:sides\:are\:not\:equal}\\\\\mathrm{No\:Solution}[/tex]

Answer:

The 99% confidence interval for estimating the population mean μ is ($60,112.60, $68087.40).

Step-by-step explanation:

The complete question is:

Salaries of 42 college graduates who took a statistics course in college have a​ mean, [tex]\bar x[/tex] of, $64, 100. Assuming a standard​ deviation, σ of ​$10​,016 construct a ​99% confidence interval for estimating the population mean μ.

Solution:

The (1 - α)% confidence interval for estimating the population mean μ is:

[tex]CI=\bar x\pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}[/tex]

The critical value of z for 99% confidence interval is:

[tex]z_{\alpha/2}=z_{0.01/2}=z_{0.005}=2.57[/tex]

Compute the 99% confidence interval for estimating the population mean μ as follows:

[tex]CI=\bar x\pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}[/tex]

     [tex]=64100\pm 2.58\times\frac{10016}{\sqrt{42}}\\\\=64100+3987.3961\\\\=(60112.6039, 68087.3961)\\\\\approx (60112.60, 68087.40)[/tex]

Thus, the 99% confidence interval for estimating the population mean μ is ($60,112.60, $68087.40).

Answer:

It sold 14 cans boxes of food and 12 cans of food.

Step-by-step explanation:

The factor for the food cans depend upon every seven food boxes .So, the same no. of sets of food cans will be sold.

Let the no. of sets of food boxes be x.

According to the question,

6x+7x=26

13x=26

x=26/13

x=2

No. of food cans =6x=6×2=12 cans

No. of food boxes=7x=7×2=14 boxes

Please mark brainliest ,if it is truly the best ! Thank you!

Answer:

[tex]L(f(t)) = \dfrac{6}{S^2+1} [\sqrt{3} \ S +1 ][/tex]

Step-by-step explanation:

Given that:

[tex]f(t) = 12 cos (t- \dfrac{\pi}{6})[/tex]

recall that:

cos (A-B) = cos AcosB + sin A sin B

∴

[tex]f(t) = 12 [cos\ t \ cos \dfrac{\pi}{6}+ sin \ t \ sin \dfrac{\pi}{6}][/tex]

[tex]f(t) = 12 [cos \ t \ \dfrac{3}{2}+ sin \ t \ sin \dfrac{1}{2}][/tex]

[tex]f(t) = 6 \sqrt{3} \ cos \ (t) + 6 \ sin \ (t)[/tex]

[tex]L(f(t)) = L ( 6 \sqrt{3} \ cos \ (t) + 6 \ sin \ (t) ][/tex]

[tex]L(f(t)) = 6 \sqrt{3} \ L [cos \ (t) ] + 6\ L [ sin \ (t) ][/tex]

[tex]L(f(t)) = 6 \sqrt{3} \dfrac{S}{S^2 + 1^2}+ 6 \dfrac{1}{S^2 +1^2}[/tex]

[tex]L(f(t)) = \dfrac{6 \sqrt{3} +6 }{S^2+1}[/tex]

[tex]L(f(t)) = \dfrac{6( \sqrt{3} \ S +1 }{S^2+1}[/tex]

[tex]L(f(t)) = \dfrac{6}{S^2+1} [\sqrt{3} \ S +1 ][/tex]