Select the correct answer. which expression is equivalent to 2x√44-2√11x³, if x > 0? OA 2x√11x OB. 6x√22x OC. 21 O D. 8x²√11 Reset Nex​

Answer:

  279/1640

Step-by-step explanation:

The relations between the various trig functions can be used to find the necessary function values.

Solving for cos(θ), we have ...

  41 cos(θ) = 9

  cos(θ) = 9/41 . . . . . divide by 41

The Pythagorean relation tells you ...

  sin²(θ) +cos²(θ) = 1   ⇒   sin(θ) = √(1 -cos²(θ))

  sin(θ) = √(1 -(9/41)²) = (√(41² -9²))/41 = 40/41

The tangent relation is ...

  tan(θ) = sin(θ)/cos(θ) = (40/41)/(9/41) = 40/9

The value of the given expression is ...

  (sin(θ) -cos(θ))/tan(θ) = (40/41 -9/41)/(40/9) = (31/41)(9/40) = 279/1640

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Additional comment

This can also be figured by a suitable calculator or spreadsheet. Output formatted as a fraction is often an option. Here, the result is rational.

Answer:

4x + 3 just take away rhe values

The value of the probability is 0.30

Using the table of values, we have:

P(x <= -3) = P(x = -5) + P(x = -3)

From the table of values, we have:

P(x = -5) = 0.17

P(x = -3) = 0.13

Substitute the known values in the above equation

P(x <= -3) = 0.17 + 0.13

Evaluate the sum

P(x <= -3) = 0.30

Hence, the value of the probability is 0.30

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Answer:

Most you can do is factor out the x and turn it into   x (3x + 7)

roots would be x = 0 and x = -7/3

Answer:  x(3x+7)

Step-by-step explanation:

You would factor the x out of both of your values and put on the outside of the parenthesis. And you put the two numbers that you have left inside of the parenthesis. And that is as far down as this function can be factored.

Given the width and length of Teresa's new desktop computer, the length of the diagonal of the desktop screen is approximately 22.8 inches.

If a diagonal line cuts through a rectangle, it forms two equal right triangles. the side lengths of this triangle can be easily determined using Pythagoras theorem. Pythagoras theorem is expressed as;

c² = a² + b²

Where c is the hypotenuse or diagonal, a is base length and b is perpendicular height.

Given the data in the question;

We substitute into the equation above.

c² = a² + b²

c² = (18in)² + (14in)²

c² = 324in² + 196in²

c² = 520in²

c = √( 520in² )

c = 22.8in

Given the width and length of Teresa's new desktop computer, the length of the diagonal of the desktop screen is approximately 22.8 inches.

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Answer:

z = [tex]\sqrt{30}[/tex]

Step-by-step explanation:

using the Altitude- on- Hypotenuse theorem

(leg of big Δ )² = (part of hypotenuse below it ) × (whole hypotenuse)

z² = 3 × (3 + 7) = 3 × 10 = 30 ( take square root of both sides )

z = [tex]\sqrt{30}[/tex]

The relationship between all pairs of special angles 1, 2, 3, and 4 created by transversal line b and parallel lines d and e include:

These are usually formed when two straight lines meet at a common endpoint or vertex.

Angles 3 and 4 are equal due to them being alternative interior angles and other relationships are mentioned above.

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the coordinates of the vertices of D2(WXYZ) are;

W' ( 2, 3)

X' ( 3, -1)

Y' ( -2, -2)

Z'(−1, −2)

It is important to note that the general formula for finding the transformation of the coordinates is;

M ( h, k) = M' ( k, -h)

Where

For the quadrilateral WXYZ we have to find the transformed coordinates

For W, we have

W(−3, 2)

h = -3

k = 2

Substitute the values

W' ( 2, 3)

For X,

X(1, 3)

h = 1

k = 3

Substitute the values

X' ( 3, -1)

For Y

Y(2, −2)

h = 2

k = -2

Substitute the values

Y' ( -2, -2)

For Z

Z(−1, −2)

h = -1

k = -2

Z'(−1, −2)

Thus, the coordinates of the vertices of D2(WXYZ) are;

W' ( 2, 3)

X' ( 3, -1)

Y' ( -2, -2)

Z'(−1, −2)

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Answer:

Step-by-step explanation:

8 adults means 48 children

11 adults means 66 children

15 adults means 120 children

[tex] \qquad \qquad \bf \huge\star \: \: \large{ \underline{Answer} } \huge \: \: \star[/tex]

Combining the above inequalities, we get :

[tex]\qquad❖ \: \sf \:0.2 < 0.8 < 0.801 < 0.9[/tex]

[tex] \qquad \large \sf {Conclusion} : [/tex]

0.1 or 10%

The numbers between 1 and 50 that end in 6 are as follows:

6, 16, 26, 36, 46

There are therefore 5 numbers between 1 and 50 that end in 6.

This means there are 5 out of 50, or 5/50.

5/50 is 0.1 as a decimal, or 10% in percentage form.

The area below the mean compares to the area above the mean in a normal distribution as the areas are always equal regardless of the mean. Option A This is further explained below.

Generally, The normal distribution, also known as the Gaussian distribution, is a kind of probability distribution that is symmetric around the mean. This means that it demonstrates that data that are closer to the mean are more likely to occur than data that are farther away from the mean. When represented graphically, the normal distribution takes the shape of a "bell curve."

In conclusion, In a normal distribution, the area below the mean is compared to the area above the mean since the areas are always equal regardless of the mean. This is true even if the mean is different.

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Complete Question

How does the area below the mean compare to the area above the mean in a normal distribution?

A. the areas are always equal regardless of the mean

B. the areas are sometimes equal depending upon the standard deviation of the distribution

C. the area above the mean is larger since the values are larger as you move above the mean

D. the areas are sometimes equal depending upon the value of the mean

In matrix form, the system is given by

[tex]\begin{bmatrix} -1 & 1 & -1 \\ 2 & -1 & 1 \\ 3 & 2 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} -20 \\ 29 \\ 29 \end{bmatrix}[/tex]

I'll use G-J elimination. Consider the augmented matrix

[tex]\left[ \begin{array}{ccc|c} -1 & 1 & -1 & -20 \\ 2 & -1 & 1 & 29 \\ 3 & 2 & 1 & 29 \end{array} \right][/tex]

• Multiply through row 1 by -1.

[tex]\left[ \begin{array}{ccc|c} 1 & -1 & 1 & 20 \\ 2 & -1 & 1 & 29 \\ 3 & 2 & 1 & 29 \end{array} \right][/tex]

• Eliminate the entries in the first column of the second and third rows. Combine -2 (row 1) with row 2, and -3 (row 1) with row 3.

[tex]\left[ \begin{array}{ccc|c} 1 & -1 & 1 & 20 \\ 0 & 1 & -1 & -11 \\ 0 & 5 & -2 & -31 \end{array} \right][/tex]

• Eliminate the entry in the second column of the third row. Combine -5 (row 2) with row 3.

[tex]\left[ \begin{array}{ccc|c} 1 & -1 & 1 & 20 \\ 0 & 1 & -1 & -11 \\ 0 & 0 & 3 & 24 \end{array} \right][/tex]

• Multiply row 3 by 1/3.

[tex]\left[ \begin{array}{ccc|c} 1 & -1 & 1 & 20 \\ 0 & 1 & -1 & -11 \\ 0 & 0 & 1 & 8 \end{array} \right][/tex]

• Eliminate the entry in the third column of the second row. Combine row 2 with row 3.

[tex]\left[ \begin{array}{ccc|c} 1 & -1 & 1 & 20 \\ 0 & 1 & 0 & -3 \\ 0 & 0 & 1 & 8 \end{array} \right][/tex]

• Eliminate the entries in the second and third columns of the first row. Combine row 1 with row 2 and -1 (row 3).

[tex]\left[ \begin{array}{ccc|c} 1 & 0 & 0 & 9 \\ 0 & 1 & 0 & -3 \\ 0 & 0 & 1 & 8 \end{array} \right][/tex]

Then the solution to the system is

[tex]\boxed{x=9, y=-3, z=8}[/tex]

If you want to use G elimination and substitution, you'd stop at the step with the augmented matrix

[tex]\left[ \begin{array}{ccc|c} 1 & -1 & 1 & 20 \\ 0 & 1 & -1 & -11 \\ 0 & 0 & 1 & 8 \end{array} \right][/tex]

The third row tells us that [tex]z=8[/tex]. Then in the second row,

[tex]y-z = -11 \implies y=-11 + 8 = -3[/tex]

and in the first row,

[tex]x-y+z=20 \implies x=20 + (-3) - 8 = 9[/tex]

Answer:

  (f•g)(x) = 4x² -7x -11

Step-by-step explanation:

The product of the two functions is the product of their respective definitions.

(f•g)(x) = f(x)•g(x) = (x+1)•(4x -11)

  = x(4x -11) +1(4x -11) . . . . . use the distributive property

  = 4x² -11x +4x -11 . . . . . . . and again

(f•g)(x) = 4x² -7x -11 . . . . . collect terms

The length of the rectangular section of the roof is 25 m

From the question, we are to determine the length of the section of the roof

From the given information,

The area of the rectangular section = 180 m²

The width of the rectangular section = 7 1/5 m = 7.2 m

Using the formula for area of a rectangle

A = l × w

Where A is the area

l is the length

and w is the width

Then,

180 = l × 7.2

l = 180/7.2

l = 25 m

Hence, the length of the rectangular section of the roof is 25 m

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Answer:

25

Step-by-step explanation:

Just did it on Khan Academy.

Answer:

(4,8)

hi hi hi hi hi hi hi hi

Using the quadratic formula, we solve for [tex]z^2[/tex].

[tex]z^4 - 5(1+2i) z^2 + 24 - 10i = 0 \implies z^2 = \dfrac{5+10i \pm \sqrt{-171+140i}}2[/tex]

Taking square roots on both sides, we end up with

[tex]z = \pm \sqrt{\dfrac{5+10i \pm \sqrt{-171+140i}}2}[/tex]

Compute the square roots of -171 + 140i.

[tex]|-171+140i| = \sqrt{(-171)^2 + 140^2} = 221[/tex]

[tex]\arg(-171+140i) = \pi - \tan^{-1}\left(\dfrac{140}{171}\right)[/tex]

By de Moivre's theorem,

[tex]\sqrt{-171 + 140i} = \sqrt{221} \exp\left(i \left(\dfrac\pi2 - \dfrac12 \tan^{-1}\left(\dfrac{140}{171}\right)\right)\right) \\\\ ~~~~~~~~~~~~~~~~~~~~= \sqrt{221} i \left(\dfrac{14}{\sqrt{221}} + \dfrac5{\sqrt{221}}i\right) \\\\ ~~~~~~~~~~~~~~~~~~~~= 5+14i[/tex]

and the other root is its negative, -5 - 14i. We use the fact that (140, 171, 221) is a Pythagorean triple to quickly find

[tex]t = \tan^{-1}\left(\dfrac{140}{171}\right) \implies \cos(t) = \dfrac{171}{221}[/tex]

as well as the fact that

[tex]0<\tan(t)<1 \implies 0along with the half-angle identities to find

[tex]\cos\left(\dfrac t2\right) = \sqrt{\dfrac{1+\cos(t)}2} = \dfrac{14}{\sqrt{221}}[/tex]

[tex]\sin\left(\dfrac t2\right) = \sqrt{\dfrac{1-\cos(t)}2} = \dfrac5{\sqrt{221}}[/tex]

(whose signs are positive because of the domain of [tex]\frac t2[/tex]).

This leaves us with

[tex]z = \pm \sqrt{\dfrac{5+10i \pm (5 + 14i)}2} \implies z = \pm \sqrt{5 + 12i} \text{ or } z = \pm \sqrt{-2i}[/tex]

Compute the square roots of 5 + 12i.

[tex]|5 + 12i| = \sqrt{5^2 + 12^2} = 13[/tex]

[tex]\arg(5+12i) = \tan^{-1}\left(\dfrac{12}5\right)[/tex]

By de Moivre,

[tex]\sqrt{5 + 12i} = \sqrt{13} \exp\left(i \dfrac12 \tan^{-1}\left(\dfrac{12}5\right)\right) \\\\ ~~~~~~~~~~~~~= \sqrt{13} \left(\dfrac3{\sqrt{13}} + \dfrac2{\sqrt{13}}i\right) \\\\ ~~~~~~~~~~~~~= 3+2i[/tex]

and its negative, -3 - 2i. We use similar reasoning as before:

[tex]t = \tan^{-1}\left(\dfrac{12}5\right) \implies \cos(t) = \dfrac5{13}[/tex]

[tex]1 < \tan(t) < \infty \implies \dfrac\pi4 < t < \dfrac\pi2 \implies \dfrac\pi8 < \dfrac t2 < \dfrac\pi4[/tex]

[tex]\cos\left(\dfrac t2\right) = \dfrac3{\sqrt{13}}[/tex]

[tex]\sin\left(\dfrac t2\right) = \dfrac2{\sqrt{13}}[/tex]

Lastly, compute the roots of -2i.

[tex]|-2i| = 2[/tex]

[tex]\arg(-2i) = -\dfrac\pi2[/tex]

[tex]\implies \sqrt{-2i} = \sqrt2 \, \exp\left(-i\dfrac\pi4\right) = \sqrt2 \left(\dfrac1{\sqrt2} - \dfrac1{\sqrt2}i\right) = 1 - i[/tex]

as well as -1 + i.

So our simplified solutions to the quartic are

[tex]\boxed{z = 3+2i} \text{ or } \boxed{z = -3-2i} \text{ or } \boxed{z = 1-i} \text{ or } \boxed{z = -1+i}[/tex]

The two triangles exist congruent if they contain two congruent corresponding sides and their contained angles exist congruent.

Let [tex]$&\overline{A B} \cong \overline{D E} \\[/tex] and [tex]$&\overline{A C} \cong \overline{D F}[/tex]

Angle between [tex]$\overline{A B}$[/tex] and [tex]$\overline{A C}$[/tex] exists [tex]$\angle A$[/tex].

Angle between [tex]$\overline{D E}$[/tex] and [tex]$\overline{D F}$[/tex] exists [tex]$\angle D$[/tex].

Therefore, [tex]$\triangle A B C \cong \triangle D E F$[/tex] by SAS, if [tex]$\angle A \cong \angle D$$[/tex].

Given:

[tex]$&\overline{A B} \cong \overline{D E} \\[/tex] and

[tex]$&\overline{A C} \cong \overline{D F}[/tex]

According to the SAS congruence property, two triangles exist congruent if they contain two congruent corresponding sides and their contained angles exist congruent.

Let [tex]$&\overline{A B} \cong \overline{D E} \\[/tex] and [tex]$&\overline{A C} \cong \overline{D F}[/tex]

Angle between [tex]$\overline{A B}$[/tex] and [tex]$\overline{A C}$[/tex] exists [tex]$\angle A$[/tex].

Angle between [tex]$\overline{D E}$[/tex] and [tex]$\overline{D F}$[/tex] exists [tex]$\angle D$[/tex].

Therefore, [tex]$\triangle A B C \cong \triangle D E F$[/tex] by SAS, if [tex]$\angle A \cong \angle D$$[/tex].

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Answer:

A = $11; C = $7 tickets

Step-by-step explanation:

Using the given info, we can create a system of equations to find the price of adult and children tickets.

Thus, we have 5A + 7C = 104 and 7A + 3C = 98

The easiest method to solve would be elimination:

Answer:

Step-by-step explanation:

The sales on the two days can be expressed using equations that can be solved for ticket prices.

Let x and y represent the prices of adult and children's tickets, respectively. Then the sales revenue for the two days can be expressed in the equations ...

One of the easiest solution methods is to use a graphing calculator. The first attachment shows the prices are $11 for an adult ticket; $7 for children tickets.

Using the matrix functions of a calculator, the augmented matrix of the equation coefficients can be reduced to row-echelon form. This, too, shows the solution to be (adult price, children price) = ($11, $7). See the second attachment. (The solution is the right-most column of the reduced matrix.)

Yet another solution method can use the coefficients from the equations written in general form:

In this form, we define three products of "cross multiplication":

  Δ1 = (5)(3) -(7)(7) = -34

  Δ2 = (7)(-98) -(3)(-104) = -374

  Δ3 = (-104)(7) -(-98)(5) = -238

Using those, we find the variable values to be ...

  x = Δ2/Δ1 = -374/-34 = 11

  y = Δ3/Δ1 = -238/-34 = 7

(adult price, children price) = ($11, $7)

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Answer:

98-54+12-36

110-54-36

56-36

20Ans

The height of the container be so as to minimize cost will be 1.20. inches.

The volume of the box will be:

= 2x × 3x × h

= 6x²h

Volume = 6x²h

12 = 6x²h

h = 2x²

The cost function will be:

C = 2.60(2)(6x²) + 4.30(12x)h

C = 31.2x² + 51.6xh

Taking the derivative

62.4x + 51.6h

h = 1.20

Therefore, the height of the container be so as to minimize cost will be 1.20 inches.

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Answer:

7/18

Step-by-step explanation:

P(sophomore girl) is 7 out of a total 30.

P(girl) is 18 out of 30.

(7/30) / (18/30) = 7/18

If you just look at the girls column you can see this immediately. A given is that we're looking for a girl (18 girls), then what is the chance that it's a sophomore. That is 7 out of 18.

Answer:

The answer would be 7 over 18 in fraction form fill in 7 then 18

Step-by-step explanation:

7/18

Answer:

1 +-3i

Step-by-step explanation:

Answer:

answer is 1) x=±3i

Step-by-step explanation:

The left hand derivative of the given function comes out to be 3a² + 3ah + h².

Deducing the Left Derivative:

The given function is,

f(x) = x³ + 2

⇒ f(a) = a³ + 2

The left hand limit is the definition of the left-hand derivative of f: f′⁻(x) = [tex]lim_{h- > 0}[/tex]f(x+h)f(x)h. F is said to be left-hand differentiable at x if the left-hand derivative exists.

Now, the formula for the left derivative of a function is given as,

f'(a)⁻ = [ f(a+h) - f(a) ] / [ (a+h) - a]

f'(a)⁻ = [ ((a+h)³ + 2) - (a³+2) ] / h

f'(a)⁻ = (a³ + 3a²h + 3ah² + h³ + 2 - a³ - 2) / h

f'(a)⁻ = (3a²h + 3ah² + h³) / h

f'(a)⁻ = h(3a² + 3ah + h²) / h

f'(a)⁻ = 3a² + 3ah + h²

Hence, the left derivative is 3a² + 3ah + h².

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Answer: 16

Step-by-step explanation:

1.5x = 24

24/1.5 = 16 students.

The answers to the questions are:

1. The confidence level is 99 percent.

2. We have to conclude that there is no sufficient evidence available to support this claim because the Confidence interval contains 75 sec.

1. The confidence level here should be

1- 0.01 = 0.99

= 99 percent

Given that, 99% confidence interval for population mean (μ) is (-34.1 sec u< u < 264.1 ) seconds.

We are to test  the claim that the population mean is greater than 75 sec.

2.

The given confidence interval contains the value of 75 sec, so there is not sufficient evidence to support the claim that the mean is greater than 75 sec.

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Answer:

To know the final result we have to make 30% of 56 dollars and add the result to the 56 dollars to know the original price.

Operations:

30% de 56 = (30x56)/100= 16,8 dollars.

56+16,8= 72,8 dollars (final result)

A pair of shoes is reduced by 30% down to a price of £56 then the original price of the pair of shoes was £80.

To find the original price of a pair of shoes using a mathematical approach. We are given that the shoes were reduced by 30% and their final price is £56. We will use algebraic equations to determine the original price of the shoes.

Let's denote the original price of the shoes as "P". Since the shoes were reduced by 30%, we can express this reduction as a decimal, which is 0.30 (30% = 30/100 = 0.30).

The amount of reduction is then given by 0.30 multiplied by the original price, which is 0.30P.

The final price of the shoes after the reduction is £56. We can set up an equation to represent this:

P - 0.30P = 56

Now, let's simplify the equation:

0.70P = 56

Next, we need to isolate "P" on one side of the equation. To do this, we divide both sides by 0.70:

[tex]\[ \frac{{0.70P}}{{0.70}} = \frac{{56}}{{0.70}} \][/tex]

This simplifies to:[tex]\[ P = \frac{{56}}{{0.70}} \][/tex]

Now, let's calculate the value of "P":

[tex]\[ P = \frac{{56}}{{0.70}} = 80 \][/tex]

So, the original price of the pair of shoes was £80.

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We can write the integration domain as

[tex]D = \left\{(x,y) \mid -1 \le y \le 1 \text{ and } 2y-2 \le x \le -y+1\right\}[/tex]

so that the integral is

[tex]\displaystyle \iint_D -\sin(y+x) \, dA = \int_{-1}^1 \int_{2y-2}^{-y+1} -\sin(y+x) \, dx \, dy[/tex]

Compute the integral with respect to [tex]x[/tex].

[tex]\displaystyle \int_{2y-2}^{-y+1} -\sin(y+x) \, dx = \cos(y+x)\bigg|_{x=2y-2}^{x=-y+1} \\\\ ~~~~~~~~ = \cos(y+(2y-2)) - \cos(y+(-y+1)) \\\\ ~~~~~~~~ = \cos(3y-2) - \cos(1)[/tex]

Compute the remaining integral.

[tex]\displaystyle \int_{-1}^1 (\cos(3y-2) - \cos(1)) \, dy = \left(\frac13 \sin(3y-2) - \cos(1) y\right) \bigg|_{y=-1}^{y=1} \\\\ ~~~~~~~~ = \left(\frac13 \sin(3-2) - \cos(1)\right) - \left(\frac13 \sin(-3-2) + \cos(1)\right) \\\\ ~~~~~~~~ = \boxed{\frac13 \sin(1) - 2 \cos(1) + \frac13 \sin(5)}[/tex]

The option that is a proportion is: B. 4/6 = 2/3.

A proportion can be defined as an equation whereby two ratios are set equal to each other, in such a way that the ratio on one side equals the ratio on the other side of the equation when simplified.

First Option is not a proportion because:

5/7 ≠ 10/12 (10/12 can be simplified further as 5/6, which is not equal to 5/7).

Second option is a proportion because:

4/6 = 2/3

8/12 = 2/3

Thus, 4/6 = 8/12.

Third Option is not a proportion because:

14/21 = 2/3

9/12 = 3/4

Therefore, 14/21 ≠ 9/12.

Fourth Option is not a proportion because:

9/15 = 3/5

12/18 = 2/3

Therefore, 9/15 ≠ 12/18.

In conclusion, the option that is a proportion is: B. 4/6 = 2/3.

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The following statements are considered to be propositions:

Deductive reasoning can be defined as a type of logical reasoning that typically involves drawing conclusions based on a given set of rules and conditions or from one or more premises (factual statements) that are assumed to be generally (universally) true.

A proposition can be defined as a type of statement (assertion) that is typically used to express an opinion or a judgement, with either a true or false answer.

This ultimately implies that, a proposition refers to a type of statement (assertion) that is either a true or false.

In this context, we can infer and logically deduce that the following statements are considered to be propositions:

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