Let S={2sin(2x):−π/2​≤x≤π/2​} find supremum and infrimum for S

The future value of the ordinary annuity with a payment of $1200, earning 8% compounded quarterly for 10 years is $72,483 (Option d).

To find the future value of an ordinary annuity, we can use the formula:

[tex]FV = PMT * [(1 + r)^n - 1] / r[/tex]

Where:

FV is the future value of the annuity,

PMT is the payment amount,

r is the interest rate per period, and

n is the number of periods.

In this case, the payment amount is $1200, the interest rate is 8% (or 0.08), and the annuity is compounded quarterly, so the interest rate per quarter is 0.08/4 = 0.02. The number of periods is 10 years * 4 quarters per year = 40 quarters.

Plugging these values into the formula, we get:

FV = $1200 * [(1 + 0.02)⁴⁰ - 1] / 0.02

  = $1200 * [(1.02)⁴⁰  - 1] / 0.02

  ≈ $72,483

Therefore, the future value of the ordinary annuity is approximately $72,483.

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Answer:

169 / 14

Step-by-step explanation:

7/1 + 71/14 = 7/1 * 14/14 + 71/14

= 98/14 + 71/14

= (98 + 71) / 14

= 169 / 14

So, the answer is 169 / 14

i) The polar form of Z is:[tex]Z = 3\sqrt 2 \left( {\cos \frac{\pi }{4} + i\sin \frac{\pi }{4}} \right),[/tex]

ii) [tex]Z^7 = -2187 - 2187i[/tex] and is expressed in the form a + bi

Polar Form of Z = -3 -3i.

In order to express the complex number -3-3i in polar form, we use the formula:

r = \sqrt {a^2 + b^2 }

where a = -3 and b = -3,

hence;[tex]r &= \sqrt {a^2 + b^2 } \\&= \sqrt {{\left( { - 3} \right)^2} + {\left( { - 3} \right)^2}} \\&= \sqrt {18} \\&= 3\sqrt 2 \[/tex]

We can calculate the argument [tex]\theta of Z as:\theta = \tan ^{ - 1} \left( {\frac{b}{a}} \right)[/tex]

where a = -3 and b = -3,

hence;

  [tex]\theta &= \tan ^{ - 1} \left( {\frac{b}{a}} \right) \\&= \tan ^{ - 1} \left( {\frac{{ - 3}}{{ - 3}}} \right) \\&= \tan ^{ - 1} \left( 1 \right) \\&= \frac{\pi }{4} \[/tex]

Therefore, the polar form of Z is:

Z = [tex]3\sqrt 2 \left( {\cos \frac{\pi }{4} + i\sin \frac{\pi }{4}} \right)[/tex]

ii)  Z^7 = -2187 - 2187i and is expressed in the form a + bi

Since we already have Z in polar form we can now easily find

Z^7.Z^7 = [tex]{\left( {3\sqrt 2 } \right)^7}{\left( {\cos \frac{\pi }{4} + i\sin \frac{\pi }{4}} \right)^7}[/tex]

We can expand [tex]\left( {\cos \frac{\pi }{4} + i\sin \frac{\pi }{4}} \right)^7[/tex] using De Moivre's theorem:

[tex]\left( {\cos \theta + i\sin \theta } \right)^n = \cos n\theta + i\sin n\ \\theta\\Therefore; \\Z^7 &= {\left( {3\sqrt 2 } \right)^7}\left( {\cos \frac{{7\pi }}{4} + i\sin \frac{{7\pi }}{4}} \right) \\&= 3^7\left( {2\sqrt 2 } \right)\left( {\cos \left( {\frac{{6\pi }}{4} + \frac{\pi }{4}} \right) + i\sin \left( {\frac{{6\pi }}{4} + \frac{\pi }{4}} \right)} \right) \\&= 2187\sqrt 2 \left( { - \frac{1}{{\sqrt 2 }}} \right) + 2187i\left( { - \frac{1}{{\sqrt 2 }}} \right) \\&=  - 2187 - 2187i \[/tex]

Thus, Z^7 = -2187 - 2187i and is expressed in the form a + bi

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18. there are two solutions in the interval 0 ≤ x ≤ 2π.

19. the current altitude of the plane is approximately 226.406 meters.

21. Since cos 20 is not given, we cannot find the exact value of sin 20 without additional information or a trigonometric table.

18. The number of solutions to the equation 2sin²x - sin x - 1 = 0 in the interval 0 ≤ x ≤ 2π is:

C. 2

To solve this quadratic equation, we can factor it as follows:

2sin²x - sin x - 1 = 0

(2sin x + 1)(sin x - 1) = 0

Setting each factor equal to zero:

2sin x + 1 = 0 or sin x - 1 = 0

Solving for sin x in each equation:

2sin x = -1 or sin x = 1

sin x = -1/2 or sin x = 1

The solutions for sin x = -1/2 in the interval 0 ≤ x ≤ 2π are π/6 and 5π/6.

The solution for sin x = 1 in the interval 0 ≤ x ≤ 2π is π/2.

As a result, the range 0 x 2 contains two solutions.

19. The current altitude of the plane with a 6.5-degree angle of elevation, when it is 2 kilometers from the airport, can be calculated using trigonometry.

We can use the tangent function:

tan(angle) = opposite/adjacent

In this case, the opposite side is the altitude of the plane and the adjacent side is the distance from the airport.

tan(6.5 degrees) = altitude/2 kilometers

Using a calculator to find the tangent of 6.5 degrees, we have:

tan(6.5 degrees) ≈ 0.113203

altitude/2 = 0.113203

altitude = 0.113203 * 2

altitude ≈ 0.226406 kilometers

Converting the altitude to meters:

altitude ≈ 0.226406 * 1000

altitude ≈ 226.406 meters

As a result, the aircraft is currently flying at a height of about 226.406 metres.

21. To find the exact value of sin 20, we will use the trigonometric identity:

sin²θ + cos²θ = 1

Given that cos 0 = 4/5 and 0 is a first-quadrant angle, we can find sin 0 using the identity:

cos²θ + sin²θ = 1

Since θ is a first-quadrant angle, cos 0 = 4/5 implies sin 0 = √(1 - cos²0):

sin 0 = √(1 - (4/5)²)

sin 0 = √(1 - 16/25)

sin 0 = √(9/25)

sin 0 = 3/5

Now, we can find sin 20 using the half-angle formula for sin:

sin (20/2) = √((1 - cos 20)/2)

We cannot determine the precise value of sin 20 without additional information or a trigonometric table because cos 20 is not given.

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A: ZABC is a right angle. (Given)

B: DB bisects ZABC. (Given)

C: m/ABD = m/CBD. (Definition of angle bisector)

D: m/ABD + m/CBD = 90°. (Sum of angles in a right triangle)

By substitution property, m/CBD + m/CBD = 90° should be m/ABD + m/CBD = 90°.

A: Given: ZABC is a right angle.

B: Given: DB bisects ZABC.

C: To prove: m/CBD = 45°

D: Proof:

ZABC is a right angle. (Given)

DB bisects ZABC. (Given)

m/ABD = m/CBD. (Definition of angle bisector)

m/ABD + m/CBD = 90°. (Sum of angles in a right triangle)

Substitute m/CBD with m/ABD in equation (4).

m/ABD + m/ABD = 90°.

2 [tex]\times[/tex] m/ABD = 90°. (Simplify equation (5))

Divide both sides of equation (6) by 2.

m/ABD = 45°.

Therefore, m/CBD = 45°. (Substitute m/ABD with 45°)

Thus, we have proved that m/CBD is equal to 45° based on the given statements and the reasoning provided.

Please note that in step 5, the substitution of m/CBD with m/ABD is valid because DB bisects ZABC. By definition, an angle bisector divides an angle into two congruent angles.

Therefore, m/ABD and m/CBD are equal.

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The Taylor series expansion of ln(1+x) at x=2.

To find the Taylor series expansion of ln(1+x) at x=2, we can start by finding the derivatives of ln(1+x) with respect to x and evaluating them at x=2.

The derivatives of ln(1+x) are:

f(x) = ln(1+x)

f'(x) = 1/(1+x)

f''(x) = -1/(1+x)^2

f'''(x) = 2/(1+x)^3

f''''(x) = -6/(1+x)^4

...

Evaluating these derivatives at x=2, we get:

f(2) = ln(1+2) = ln(3)

f'(2) = 1/(1+2) = 1/3

f''(2) = -1/(1+2)^2 = -1/9

f'''(2) = 2/(1+2)^3 = 2/27

f''''(2) = -6/(1+2)^4 = -6/81

The Taylor series expansion of ln(1+x) centered at x=2 is given by:

ln(1+x) = f(2) + f'(2)(x-2) + f''(2)(x-2)^2/2! + f'''(2)(x-2)^3/3! + f''''(2)(x-2)^4/4! + ...

Substituting the values we calculated earlier, the Taylor series expansion becomes:

ln(1+x) = ln(3) + (1/3)(x-2) - (1/9)(x-2)^2/2 + (2/27)(x-2)^3/3 - (6/81)(x-2)^4/4 + ...

This is the Taylor series expansion of ln(1+x) at x=2.

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(a) The partial derivatives D1f and D2f of the function f(x, y) are bounded in R2. Moreover, f is continuous.

(b) The directional derivative (Dvf)(0, 0) exists for a unit vector v, and its absolute value is at most 1. Additionally, f is Gâteaux differentiable at the origin (0, 0).

(c) The function g(t) = f(γ(t)) is differentiable at every t ∈ R, and if γ ∈ C1(R, R2), then g ∈ C1(R, R).

(d) Despite the aforementioned properties, f is not Fréchet differentiable at the origin (0, 0).

(a) To prove that the partial derivatives ∂f/∂x and ∂f/∂y are bounded in R², we need to show that there exists a constant M such that |∂f/∂x| ≤ M and |∂f/∂y| ≤ M for all (x, y) in R².

Calculating the partial derivatives:

∂f/∂x = [tex](0 - 2xy^2)/(x^4 + y^4)[/tex]= [tex]-2xy^2/(x^4 + y^4)[/tex]

∂f/∂y = [tex]2yx^2/(x^4 + y^4)[/tex]

Since[tex]x^4 + y^4[/tex] > 0 for all (x, y) ≠ (0, 0), we can bound the partial derivatives as follows:

|∂f/∂x| =[tex]2|xy^2|/(x^4 + y^4) ≤ 2|x|/(x^4 + y^4) \leq 2(|x| + |y|)/(x^4 + y^4)[/tex]

|∂f/∂y| = [tex]2|yx^2|/(x^4 + y^4) ≤ 2|y|/(x^4 + y^4) \leq 2(|x| + |y|)/(x^4 + y^4)[/tex]

Letting M = 2(|x| + |y|)/[tex](x^4 + y^4)[/tex], we can see that |∂f/∂x| ≤ M and |∂f/∂y| ≤ M for all (x, y) in R². Hence, the partial derivatives are bounded.

Furthermore, f is continuous since it can be expressed as a composition of elementary functions (polynomials, division) which are known to be continuous.

(b) To show the existence and bound of the directional derivative (Dvf)(0,0), we use the limit definition of the directional derivative. Let v = (v1, v2) be a unit vector.

(Dvf)(0,0) = lim(h→0) [f((0,0) + hv) - f(0,0)]/h

           = lim(h→0) [f(hv) - f(0,0)]/h

Expanding f(hv) using the given formula: f(hv) = 0(hv²)/(h³) = v²/h

(Dvf)(0,0) = lim(h→0) [v²/h - 0]/h

           = lim(h→0) v²/h²

           = |v²| = 1

Therefore, the absolute value of the directional derivative (Dvf)(0,0) is at most 1.

(c) Let γ: R → R² be a differentiable curve such that γ(0) = (0,0), and γ'(t) ≠ (0,0) whenever γ(t) = (0,0) for some t ∈ R. We define g(t) = f(γ(t)).

To prove that g is differentiable at every t ∈ R, we can use the chain rule of differentiation. Since γ is differentiable, g(t) = f(γ(t)) is a composition of differentiable functions and is therefore differentiable at every t ∈ R.

If γ ∈ [tex]C^1(R, R^2)[/tex], which means γ is continuously differentiable, then g ∈ [tex]C^1(R, R)[/tex] as the composition of two continuous functions.

(d) To show that f is

not Fréchet differentiable at the origin (0,0), we need to demonstrate that the formula (Dvf)(0,0) = ⟨∇f(0,0), v⟩ fails for some direction(s) v, where ⟨⋅,⋅⟩ denotes the standard dot product in R².

The gradient of f is given by ∇f = (∂f/∂x, ∂f/∂y). Using the previously derived expressions for the partial derivatives, we have:

∇f(0,0) = (∂f/∂x, ∂f/∂y) = (0, 0)

However, if we take v = (1, 1), the formula (Dvf)(0,0) = ⟨∇f(0,0), v⟩ becomes:

(Dvf)(0,0) = ⟨(0, 0), (1, 1)⟩ = 0

But from part (b), we know that the absolute value of the directional derivative is at most 1. Since (Dvf)(0,0) ≠ 0, the formula fails for the direction v = (1, 1).

Therefore, f is not Fréchet differentiable at the origin (0,0).

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Correct answer is "None of the mentioned".

The given linear ODE is:y' - y = x

We want to find the one-parameter solution of the above linear ODE.For the linear ODE:y' + p(t)y = g(t), the solution is given byy = (1/u) [ ∫u g(t) dt + C ], where u is the integrating factor, which is given by u(t) = e^∫p(t)dt.

In our case,p(t) = -1, so we haveu(t) = e^∫-1dt= e^-t.The integrating factor isu(t) = e^-t.Multiplying both sides of the linear ODE by the integrating factor, we get:e^-ty' - e^-ty = xe^-t

Now, we have:(e^-ty)' = xe^-t∫(e^-ty)' dt = ∫xe^-t dtIntegrating both sides, we get:-e^-ty = -xe^-t - e^-t + C1

Multiplying both sides by -1, we get:e^-ty = xe^-t + e^-t + C2

Taking exponential on both sides, we get:e^(-t) * e^y = e^(-t) * (x + 1 + C2)or e^y = x + 1 + C2or y = ln(x + 1 + C2)

Therefore, the one-parameter solution of the given linear ODE is y = ln(x + 1 + C2), where C2 is an arbitrary constant. None of the options given in the question matches with the solution.

Hence, the correct answer is "None of the mentioned".

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Answer: 3 sec

Step-by-step explanation:

They want to know how long? That is time, which is the x-axis.  How long is your curve, it goes til 3 so the ball was in the air for 3 sec.

Finding the midpoint of a line segment is easy.

In a two-dimensional Cartesian plane with known endpoints, the abscissa value of the midpoint is half the sum of the abscissa values of the endpoints, and the ordinate value is half the sum of the ordinate values of the endpoints.

Based on this information, we can comfortably say that the midpoint of this line segment is as follows;

Let the midpoint of this segment is [tex]M(x_{1},y_{1})[/tex].

Hence, the midpoint of this segment is [tex](6,4)[/tex].

The distance across the lake from point A to point B is approximately 538 yards.

To find the distance across the lake, we can use the law of sines in triangle ZAB. The law of sines states that the ratio of the length of a side of a triangle to the sine of its opposite angle is constant. In this case, we have the angle ZAB (43.6 degrees) and the lengths ZC (325 yards) and AC (unknown).

Using the law of sines, we can set up the following equation:

sin(ZAB) / ZC = sin(ZCA) / AC

Substituting the known values, we have:

sin(43.6°) / 325 = sin(ZCA) / AC

Solving for sin(ZCA), we get:

sin(ZCA) = (sin(43.6°) / 325) * AC

To find the length of AC, we need to rearrange the equation:

AC = (325 * sin(ZCA)) / sin(43.6°)

Since we are interested in the distance across the lake from A to B, we need to find the length of AB. We know that AB = AC + BC, where BC is the distance from C to B.

To find BC, we can use the law of sines again in triangle ZCB:

sin(ZCB) / ZC = sin(ZCA) / BC

Substituting the known values, we have:

sin(ZCB) / 325 = sin(ZCA) / BC

Solving for BC, we get:

BC = (325 * sin(ZCB)) / sin(ZCA)

Finally, we can calculate AB by adding AC and BC:

AB = AC + BC

Plugging in the values we know, we have:

AB = ((325 * sin(ZCA)) / sin(43.6°)) + ((325 * sin(ZCB)) / sin(ZCA))

Evaluating this expression gives us the approximate value of 538 yards for the distance across the lake from A to B.

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The solution to the system of equations is:

x = -2, y = -4, z = -3

So, the correct option is:

e. x = -2, y = -4, z = -3; (-2, -4, -3)

To solve the given system of equations:

1) x + 2y + 2z = -16

2) 4y + 5z = -31

3) z = -3

We can substitute the value of z from equation 3 into equations 1 and 2 to solve for x and y.

Substituting z = -3 into equation 1:

x + 2y + 2(-3) = -16

x + 2y - 6 = -16

x + 2y = -16 + 6

x + 2y = -10

Substituting z = -3 into equation 2:

4y + 5(-3) = -31

4y - 15 = -31

4y = -31 + 15

4y = -16

y = -16/4

y = -4

Now, substituting y = -4 back into equation 1:

x + 2(-4) = -10

x - 8 = -10

x = -10 + 8

x = -2

Therefore, the solution to the system of equations is:

x = -2, y = -4, z = -3

So, the correct option is:

e. x = -2, y = -4, z = -3; (-2, -4, -3)

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In a linear regression model, the linearity assumption refers to the relationship between the independent variables and the dependent variable.

It assumes that the dependent variable is a linear combination of the independent variables, with the coefficients representing the effect of each independent variable on the dependent variable.

In the given model, y^2 = ax_1 + bx_2 + u, the dependent variable y is squared, which introduces a non-linearity to the model. The presence of y^2 in the equation makes the model non-linear, as it cannot be expressed as a linear combination of the independent variables.

If you want to include quadratic or cubic terms for the dependent variable y, you would need to transform the model accordingly. For example, you could use a quadratic or cubic transformation of y, such as y^2, y^3, or even log(y), and include those transformed variables in the linear regression model along with the independent variables. This would allow you to capture non-linear relationships between the dependent variable and the independent variables in the model.

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Answer:

  C.  75°

Step-by-step explanation:

You want the angle marked ∠1 in the trapezoid shown.

Where a transversal crosses parallel lines, same-side interior angles are supplementary. In this trapezoid, this means the angles at the right side of the figure are supplementary:

  ∠1 + 105° = 180°

  ∠1 = 75° . . . . . . . . . . . . subtract 105°

__

Additional comment

The given relation also means that the unmarked angle is supplementary to the one marked 50°. The unmarked angle will be 130°.

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The property that isn't satisfied for this relation to be an equivalence relation is transitivity.

To be an equivalence relation, a relation must satisfy three properties: reflexivity, symmetry, and transitivity. Reflexivity means that every element is related to itself. Symmetry means that if a is related to b, then b is related to a. Transitivity means that if a is related to b and b is related to c, then a must be related to c.

In this case, we have a relation R defined on the set P = {Amy, Bob, John, Steve}. The relation R is defined as aRb if and only if a and b are classmates in the courses CSE116 and CSE191.

Reflexivity is satisfied because each student is a classmate of themselves. Symmetry is satisfied because if a is a classmate of b, then b is also a classmate of a. However, transitivity is not satisfied.

To demonstrate the lack of transitivity, let's consider the students' enrollment in the courses. Bob and John are taking CSE116, and John and Steve are taking CSE191. Based on the definition of R, we can say that Bob is a classmate of John and John is a classmate of Steve.

However, this does not imply that Bob is a classmate of Steve. Transitivity would require that if Bob is a classmate of John and John is a classmate of Steve, then Bob must also be a classmate of Steve. But this is not the case here.

In conclusion, the relation R defined as aRb if and only if a and b are classmates does not satisfy the property of transitivity, which is necessary for it to be an equivalence relation.

The lack of transitivity in this relation can be illustrated by the enrollment of the students in specific courses. Transitivity would require that if a is related to b and b is related to c, then a must be related to c. In this case, Bob is related to John because they are classmates in CSE116, and John is related to Steve because they are classmates in CSE191.

However, Bob is not related to Steve because they are not classmates in any of the specified courses. This violates the transitivity property and prevents the relation from being an equivalence relation.

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12. The number of ways to line up five dogs is calculated using permutations, resulting in 120 different arrangements.

13. Cynthia can choose three flavors out of 15 options, and the number of different selections is calculated using combinations, resulting in 455 possibilities.

14. There are 56 different arrangements of president and vice-president from a club consisting of eight members, calculated using permutations.

12. 1: Identify that we need to find the number of arrangements (permutations) of the five dogs.

2: Use the formula for permutations: P(n, r) = n! / (n - r)!

3: Substitute the values: P(5, 5) = 5! / (5 - 5)!

4: Simplify the expression: P(5, 5) = 5! / 0! = 5! / 1 = 5 x 4 x 3 x 2 x 1 = 120

Therefore, there are 120 different ways the five dogs can be lined up for the dog show.

13. 1: Recognize that we need to find the number of combinations of three flavors from 15 options.

2: Use the formula for combinations: C(n, r) = n! / (r! * (n - r)!)

3: Substitute the values: C(15, 3) = 15! / (3! * (15 - 3)!)

4: Simplify the expression: C(15, 3) = 15! / (3! * 12!)

5: Calculate the factorial values: 15! = 15 x 14 x 13 x 12!, 3! = 3 x 2 x 1, 12! = 12 x 11 x 10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1

6: Substitute the factorial values: C(15, 3) = (15 x 14 x 13) / (3 x 2 x 1) = 455

Therefore, there are 455 different selections of three flavors possible for Cynthia's banana split.

14. 1: Recognize that we need to find the number of arrangements (permutations) of two positions (president and vice-president) from eight club members.

2: Use the formula for permutations: P(n, r) = n! / (n - r)!

3: Substitute the values: P(8, 2) = 8! / (8 - 2)!

4: Simplify the expression: P(8, 2) = 8! / 6!

5: Calculate the factorial values: 8! = 8 x 7 x 6!, 6! = 6 x 5 x 4 x 3 x 2 x 1

6: Substitute the factorial values: P(8, 2) = (8 x 7) / (6 x 5 x 4 x 3 x 2 x 1) = 56

Therefore, there are 56 different arrangements of president and vice-president possible from the eight club members.

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Since Jeffrey deposits the $450 at the end of every quarter, the type of annuity is an Ordinary Annuity.

An ordinary annuity is a type of annuity where the payment occurs at the end of the period and not at the beginning like Annuity Due.

The ordinary annuity can be computed as follows using an online finance calculator.

Quarterly deposits = $450

Investment period = 4 years and 6 months (4.5 years)

Compounding period = semi-annually

N (# of periods) = 18 (4.5 years x 4)

I/Y (Interest per year) = 5.3%

PV (Present Value) = $0

PMT (Periodic Payment) = $450

P/Y (# of periods per year) = 4

C/Y (# of times interest compound per year) = 2

PMT made = at the of each period

Results:

FV = $9,073.18

Sum of all periodic payments = $8,100 ($450 x 4.5 x 4)

Total Interest = $973.18

Thus, the annuity is not an Annuity Due but an Ordinary Annuity.

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In the case of the given 1x4 matrix [-4 3 2 0], since it does not meet the requirement of being a square matrix, it does not have a determinant. The determinant is only applicable to matrices with dimensions of n x n, where n is a positive integer and hence, the determinant of the given matrix is undefined.

The given matrix is a 1x4 matrix, which means it has only one row and four columns. Determinants are defined for square matrices, so a 1x4 matrix does not have a determinant.

The determinant is a scalar value that represents certain properties of a square matrix, such as invertibility and the scaling factor of the linear transformation it represents. It is only defined for square matrices, which have an equal number of rows and columns.

In the case of the given 1x4 matrix [-4 3 2 0], since it does not meet the requirement of being a square matrix, it does not have a determinant. The determinant is only applicable to matrices with dimensions of n x n, where n is a positive integer.

Therefore, the determinant of the given matrix is undefined.

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Answer:  21 + 8√5

Step-by-step explanation:

(4+√5) (4+√5)                             >FOIL

16 + 4√5 + 4√5 + √5√5            >combine like terms

16 + 8√5 + 5

21 + 8√5

Answer:

8√5+21

Step-by-step explanation:

Simplify the given expression.

(4+√5) (4+√5)

Start by distributing, using F.O.I.L. (First, outer, inner, last).

(4+√5) (4+√5)

=> 4(4)+4(√5)+√5(4)+√5(√5)

Simplify what's above.

4(4)+4(√5)+√5(4)+√5(√5)

=> 16+4√5+4√5+5

=> 8√5+21

Thus, the given expression has been simplified.

Answer:

x - intercepts are x = - 3, x = 2 , y- intercept = - 6

Step-by-step explanation:

the x- intercepts are the points on the x- axis where the graph of f(x) crosses the x- axis.

any point on the x- axis has a y- coordinate of zero.

let y = f(x) = 0 and solve for x, that is

x² + x - 6 = 0

consider the factors of the constant term (- 6) which sum to give the coefficient of the x- term (+ 1)

the factors are + 3 and - 2 , since

3 × - 2 = - 6 and 3 - 2 = - 1 , then

(x + 3)(x - 2) = 0 ← in factored form

equate each factor to zero and solve for x

x + 3 = 0 ( subtract 3 from both sides )

x = - 3

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

x = 2

the x- intercepts are x = - 3 and x = 2

the y- intercept is the point on the y- axis where the graph of f(x) crosses the y- axis.

any point on the y- axis has an x- coordinate of zero

let x = 0 in y = f(x)

f(0) = 0² + 0 - 6 = 0 + 0 - 6 = - 6

the y- intercept is y = - 6

Answer: Choice B

Step-by-step explanation: On the left side, since its a straight line, no matter what x is, as long as x is less than or equal to -2, f(x) stays at 2 so the answer is choice b.

a) AD can be expressed as AD = 6a - 4b.

b) ABCD is a parallelogram.

a) To express AD in terms of 'a' and/or 'b', we can observe that AD is the difference between AB and BC. Using the given values, we have:

AD = AB - BC

= (8a + 12b) - (2a + 16b)

= 8a + 12b - 2a - 16b

= 6a - 4b

Therefore, AD can be expressed as 6a - 4b.

b) Based on the given information, the shape ABCD is a parallelogram. This is because a parallelogram has opposite sides that are parallel and equal in length, which is satisfied by the given sides AB and DC.

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(a) The statement assumes that the population mean of radon in New Brunswick houses (β) is equal to the EPA cutoff of 4.

The null hypothesis can be written as:

H0: β = 4

(b) The alternative hypothesis can be written as:

Ha: β ≠ 4

This statement suggests that the population mean of radon in New Brunswick houses (β) is not equal to the EPA cutoff of 4.

(c) When testing this null hypothesis, a two-tailed test is used. This is because the alternative hypothesis does not specify a direction (greater than or less than), but instead allows for the possibility that the population mean can differ from the EPA cutoff in either direction.

(d) To test the null hypothesis, we need to use an estimator of β. In this case, the sample mean (x) will serve as the estimator of β. The formula for the variance of this estimator, assuming simple random sampling, is:

Var(x) = σ²/n

Here, σ represents the population standard deviation and n is the sample size. To estimate this variance formula, we need the sample standard deviation (s). The estimated variance formula becomes:

Var(x)≈ s²/n

To obtain a standard error for the estimator of β, we take the square root of the estimated variance:

SE(x) ≈ √(s²/n)

4. To test the null hypothesis using a t-test, we will compute the t-statistic using the formula:

t = (x-β) / (SE(x))

In this case, since β is known (4), the formula simplifies to:

t = (x- 4) / (SE(x))

To carry out the test at the 5% significance level (α = 0.05), we will compare the computed t-statistic to the critical value(s) from the t-distribution with appropriate degrees of freedom. The rejection rule is as follows: If the absolute value of the computed t-statistic is greater than the critical value(s), we reject the null hypothesis; otherwise, we fail to reject the null hypothesis.

The result of the test will indicate whether there is sufficient evidence to reject the null hypothesis or not.

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Answer: The original price is $460.

Step-by-step explanation: Since the sofa is sold at a 68% discount (0.68) from the original price, the sofa during the sale cost 32% (0.32) of the original price. Therefore, $147.20 =  (0.32)* original price and dividing both sides by 0.32, the original price is $460.

To find the vector x determined by the set of vectors B and the given vector [x], we need to solve the system of linear equations formed by equating the linear combination of vectors in B to the given vector [x].  the vector x determined by B is:

x = [ -7.5 ]

        [ 1.5 ]

        [ -5 ]

The step-by-step process of finding the vector x determined by B.

We are given the set of vectors B:

B = {[ 1  1  -1 ],

    [-1 -2  3 ],

    [-2  0  3 ]}

And the vector [x] = [ -5  1  -9 ].

1. Write the vectors in B as column vectors:

v₁ = [ 1 ]

       [ 1 ]

       [ -1 ]

v₂ = [ -1 ]

       [ -2 ]

       [ 3 ]

v₃ = [ -2 ]

       [ 0 ]

       [ 3 ]

2. We want to find the coefficients c₁, c₂, and c₃ such that:

c₁ * v₁ + c₂ * v₂ + c₃ * v₃ = [ -5 ]

                                             [ 1 ]

                                             [ -9 ]

3. Set up the system of equations using the coefficients:

c₁ * [ 1 ] + c₂ * [ -1 ] + c₃ * [ -2 ] = [ -5 ]

          [ 1 ]              [ -2 ]                     [  1  ]

          [ -1 ]             [ 3 ]                       [ -9 ]

4. Write the system of equations in matrix form:

A * c = b

where A is the coefficient matrix, c is the column vector of coefficients c₁, c₂, and c₃, and b is the given vector [ -5, 1, -9 ].

The matrix A is:

A = [  1   -1   -2 ]

      [  1   -2    0 ]

      [ -1    3    3 ]

The column vector b is:

b = [ -5 ]

      [  1 ]

      [ -9 ]

5. Calculate the inverse of matrix A:

[tex]A^(-1)[/tex] = [  -3/2   -1/2    1/2 ]

             [ -1/2   -1/2    1/2 ]

             [  1/2    1/2   -1/2 ]

6. Multiply A^(-1) with b to find the vector c:

c =[tex]A^(-1)[/tex]* b

c = [  -3/2   -1/2    1/2 ] * [ -5 ] = [ -9 ]

       [ -1/2   -1/2    1/2 ]        [  1 ]        [  1 ]

       [  1/2    1/2   -1/2 ]        [ -9 ]       [ -5 ]

7. Finally, calculate the vector x using the coefficients c and the vectors in B:

x = c₁ * v₁ + c₂ * v₂ + c₃ * v₃

 = [  -3/2   -1/2    1/2 ] * [  1 ] + [ -1/2   -1/2    1/2 ] * [ -1 ] + [  1/2    1/2   -1/2 ] * [ -2 ]

x = [ -9 ] + [ 1/2 ] + [ 2/2 ]

         [ 1 ]        [ 1/2 ]       [ 1/2 ]

         [ -5 ]       [ -1/2 ]     [ 3/2 ]

Simplifying the expression, we get:

x = [ -7.5 ]

         [ 1.5 ]

         [ -5 ]

Therefore, the vector x determined by B is:

x = [ -7.5 ]

        [ 1.5 ]

        [ -5 ]

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An equation for the function graphed above is g(x) = |x - 1| - 2.

In Mathematics and Geometry, the translation of a graph to the right means adding a digit to the numerical value on the x-coordinate of the pre-image;

g(x) = f(x - N)

By critically observing the graph of this absolute value function, we can reasonably infer and logically deduce that the parent absolute value function f(x) = |x| was vertically translated to the right by 1 unit and 2 units down, in order to produce the transformed absolute value function g(x) as follows;

f(x) = |x|

g(x) = f(x - 1)

g(x) = |x - 1| - 2

In conclusion, the value of the variables A, B, and C are 4, 2, and 8 respectively.

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The result of entering "2/3 4" and using the Preview button is 8/3.

The order of operations, also known as precedence rules, is crucial in mathematics to ensure consistent and accurate calculations. These rules dictate the order in which different mathematical operations should be performed when evaluating an expression.

The standard order of operations, often remembered using the acronym PEMDAS (Parentheses, Exponents, Multiplication and Division from left to right, Addition and Subtraction from left to right), helps us determine which operations to prioritize.

When evaluating expressions, it is important to consider the order of operations. In this case, the expression "2/3 4" consists of a division operation followed by a multiplication operation. According to the rules of precedence, multiplication and division have the same level of precedence and should be evaluated from left to right.

Therefore, we first perform the division operation: 2 divided by 3, which gives us the fraction 2/3. Then, we proceed to the multiplication operation: multiplying the fraction 2/3 by 4. This yields a result of 8/3.

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To find the highest common factor (HCF) of 540 and 629 using the continued division method, we will perform a series of divisions until we reach a remainder of 0.The HCF of 540 and 629 is 1.

Step 1: Divide 629 by 540.

The quotient is 1, and the remainder is 89.

Step 2: Divide 540 by 89.

The quotient is 6, and the remainder is 54.

Step 3: Divide 89 by 54.

The quotient is 1, and the remainder is 35.

Step 4: Divide 54 by 35.

The quotient is 1, and the remainder is 19.

Step 5: Divide 35 by 19.

The quotient is 1, and the remainder is 16.

Step 6: Divide 19 by 16.

The quotient is 1, and the remainder is 3.

Step 7: Divide 16 by 3.

The quotient is 5, and the remainder is 1.

Step 8: Divide 3 by 1.

The quotient is 3, and the remainder is 0.

Since we have reached a remainder of 0, the last divisor used (in this case, 1) is the HCF of 540 and 629.

Therefore, the HCF of 540 and 629 is 1.

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The graph of the linear function y = -x - 4 will look like a straight line that passes through the points (-3, -1), (-2, -2), (0, -4), (1, -5), and (2, -6).

To graph the linear function y = -x - 4, we can start by plotting a few points and then connecting them with a straight line.

We'll choose some x-values and substitute them into the equation to find the corresponding y-values. Let's choose x = -3, -2, 0, 1, and 2.

When x = -3:

y = -(-3) - 4 = 3 - 4 = -1

So, we have the point (-3, -1).

When x = -2:

y = -(-2) - 4 = 2 - 4 = -2

So, we have the point (-2, -2).

When x = 0:

y = -(0) - 4 = 0 - 4 = -4

So, we have the point (0, -4).

When x = 1:

y = -(1) - 4 = -1 - 4 = -5

So, we have the point (1, -5).

When x = 2:

y = -(2) - 4 = -2 - 4 = -6

So, we have the point (2, -6).

Now, let's plot these points on a coordinate plane.

The x-axis represents the values of x, and the y-axis represents the values of y. We can plot the points (-3, -1), (-2, -2), (0, -4), (1, -5), and (2, -6).

After plotting the points, we can connect them with a straight line. Since the equation is y = -x - 4, the line will have a negative slope and will be sloping downward from left to right.

The graph of the linear function y = -x - 4 will look like a straight line that passes through the points (-3, -1), (-2, -2), (0, -4), (1, -5), and (2, -6).

Please note that without an actual graphing tool, I can only describe the process of graphing the function. The actual graph would be a line passing through the mentioned points.

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