Find the tangential and normal components of the acceleration vector. r(t) = ti + t^2 j + 3tK a_T = a_N =

Answer:

28

Step-by-step explanation:

You must divide 20 by 100 to get 1 percent, then multiply it by 140.

Randy ate 28 pop-tarts this month.

To find out how many pop-tarts Randy ate this month, we first need to calculate 40% of the pop-tarts he ate last month.

To do this, we can multiply 20 by 0.4 to get 8.

Next, we can add this amount to the original number of pop-tarts Randy ate last month (20), giving us a total of 28 pop-tarts for this month.

Therefore, Randy ate 28 pop-tarts this month, which is 40% more than he ate last month.  

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это вопрос 10 класса? Вопрос старшеклассника??

Answer: I think its reflected

Step-by-step explanation:

The largest level of confidence for which the confidence interval will not contain 2 is "A 99% confidence level". Therefore the answer is option (E).

Since the p-value of 0.67 is quite large, we fail to reject the null hypothesis at any reasonable level of significance, and we cannot conclude that the population mean is different from 2.

As we cannot reject the null hypothesis at any reasonable level of significance, we cannot conclude that the population mean is different from 2. Therefore, the confidence interval for µ will always include 2, regardless of the level of confidence.

Therefore, the confidence interval for µ will always contain 2, regardless of the level of confidence. This means that the answer is (E) A 99% confidence level.

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The amount in the annuity after 3 years is $33,371.92.

What is an annuity?

An annuity is a financial product that provides a series of regular payments to the holder for a specified period of time, usually until the end of their life or a predetermined number of years.

This problem involves finding the future value of an annuity, which can be calculated using the formula:

FV = PMT x [[tex](1 + r)^{n}[/tex] - 1] / r

where FV is the future value of the annuity, PMT is the monthly payment, r is the monthly interest rate, and n is the number of payments.

In this case, PMT = $765.13, r = 0.07/12 = 0.00583 (since the interest rate is given as an annual rate and compounded monthly), and n = 3 years x 12 months/year = 36.

Substituting these values into the formula, we get:

FV = $765.13 x [[tex](1 + 0.00583)^{36}[/tex] - 1] / 0.00583

= $765.13 x [1.249542 - 1] / 0.00583

= $765.13 x 43.679

= $33,371.92

Therefore, the amount in the annuity after 3 years is $33,371.92.

An annuity is typically purchased by making a lump sum payment or a series of payments over time, which is then invested to generate a stream of income payments. The payments can be made at a fixed interval, such as monthly or annually, and may be guaranteed for a specific period or for the life of the holder.

There are different types of annuities, including fixed, variable, immediate, and deferred annuities, each with their own unique features and benefits. Annuities are often used for retirement planning or to provide a steady income stream to cover ongoing expenses.

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The amount in the annuity after 3 years is $33,371.92.

What is an annuity?

An annuity is a financial product that provides a series of regular payments to the holder for a specified period of time, usually until the end of their life or a predetermined number of years.

This problem involves finding the future value of an annuity, which can be calculated using the formula:

FV = PMT x [[tex](1 + r)^{n}[/tex] - 1] / r

where FV is the future value of the annuity, PMT is the monthly payment, r is the monthly interest rate, and n is the number of payments.

In this case, PMT = $765.13, r = 0.07/12 = 0.00583 (since the interest rate is given as an annual rate and compounded monthly), and n = 3 years x 12 months/year = 36.

Substituting these values into the formula, we get:

FV = $765.13 x [[tex](1 + 0.00583)^{36}[/tex] - 1] / 0.00583

= $765.13 x [1.249542 - 1] / 0.00583

= $765.13 x 43.679

= $33,371.92

Therefore, the amount in the annuity after 3 years is $33,371.92.

An annuity is typically purchased by making a lump sum payment or a series of payments over time, which is then invested to generate a stream of income payments. The payments can be made at a fixed interval, such as monthly or annually, and may be guaranteed for a specific period or for the life of the holder.

There are different types of annuities, including fixed, variable, immediate, and deferred annuities, each with their own unique features and benefits. Annuities are often used for retirement planning or to provide a steady income stream to cover ongoing expenses.

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The box-and-whisker plot for the data set, 18, 30, 24, 19, 22, 34, 13, 12, 20, 25, 28, 17, is shown in the diagram attached below.

In order to make a box-and-whisker plot for the data given, we have to find the five-number summary of the data, which would be displayed on the box-and-whisker plot.

Given the data as: 18, 30, 24, 19, 22, 34, 13, 12, 20, 25, 28, 17

Minimum: 12 (smallest data value)

First Quartile: 17.5 (middle of the first half of the data when ordered)

Median: 21 (center of the data set)

Third Quartile: 26.5 (middle of the second half)

Maximum: 34 (largest data value)

The box-and-whisker plot is shown below in the attachment.

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There are 3003 ways to select a group consisting of any 6 members from the math club at Utopia University.

To select a group consisting of any 6 members from a total of 14 members, we need to use the combination formula.

The combination formula tells us the number of ways to select k objects from a total of n distinct objects without regard to order. It is given by:

C(n,k) = n! / (k!(n-k)!)

where n! (n factorial) represents the product of all positive integers up to n.

In our case, we want to select a group of 6 members from a total of 14 members, so we can use the combination formula as follows:

C(14,6) = 14! / (6!(14-6)!)

Simplifying the formula using factorials, we get:

C(14,6) = (14 × 13 × 12 × 11 × 10 × 9) / (6 × 5 × 4 × 3 × 2 × 1)

Cancelling out the common factors, we get:

C(14,6) = 3003

Therefore, there are 3003 ways to select a group consisting of any 6 members from the math club at Utopia University.

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There are 3003 ways to select a group consisting of any 6 members from the math club at Utopia University.

To select a group consisting of any 6 members from a total of 14 members, we need to use the combination formula.

The combination formula tells us the number of ways to select k objects from a total of n distinct objects without regard to order. It is given by:

C(n,k) = n! / (k!(n-k)!)

where n! (n factorial) represents the product of all positive integers up to n.

In our case, we want to select a group of 6 members from a total of 14 members, so we can use the combination formula as follows:

C(14,6) = 14! / (6!(14-6)!)

Simplifying the formula using factorials, we get:

C(14,6) = (14 × 13 × 12 × 11 × 10 × 9) / (6 × 5 × 4 × 3 × 2 × 1)

Cancelling out the common factors, we get:

C(14,6) = 3003

Therefore, there are 3003 ways to select a group consisting of any 6 members from the math club at Utopia University.

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it will take the band 50 minutes to walk 10 blocks.

So the answer is (B) 50 minutes.

In general, the term "proportion" refers to a part, share, or amount that is compared to a whole. According to the definition of proportion, two ratios are in proportion when they are equal.

The marching band is marching at a rate of 3 blocks in 15 minutes. To find how long it will take them to walk 10 blocks, we can set up a proportion:

3 blocks / 15 minutes = 10 blocks / x minutes

where x is the time it will take to walk 10 blocks.

Simplifying the proportion:

3/15 = 10/x

Cross-multiplying:

3x = 150

x = 50

Therefore, it will take the band 50 minutes to walk 10 blocks.

So the answer is (B) 50 minutes.

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If the vertices of triangle PQR, are rotated 90 degree in counter-clockwise direction, then the new vertices of triangle P'Q'R' is P'(1,-3), Q'(3,-3) and R'(2,-6), the correct option is (b).

To rotate a point (x,y) 90 degrees counterclockwise about the origin, we can use the following formula:

(x', y') = (-y, x)

that means if the coordinate of triangle PQR is (x,y) , then after 90 degree counter clockwise rotation , the coordinate will be = (-y,x);

Using this formula for each vertex, we get:

⇒ P' = (-(-1), -3)) = (1, -3),

⇒ Q' = (-(-3), -3)) = (3, -3),

⇒ R' = (-(-2), -6)) = (2, -6),

Therefore, the coordinates of the triangle after the rotation are P'(1,3), Q'(3,3), and R'(2,6), Option(b) is correct.

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The given question is incomplete, the complete question is

Triangle PQR has vertices P(-3, -1), Q(-3, -3), and R(-6, -2). The triangle is rotated 90° counterclockwise using the origin as the center of rotation. Which graph shows the image, triangle P’Q’R’?

Group of answer choices

(a) On a coordinate plane, triangle P'Q'R' has points (-3, -1), (-3, -3), (-6, -2).

(b ) On a coordinate plane, triangle P'Q'R' has points (1, -3), (3, -3), (2, -6).

(c) On a coordinate plane, triangle P'Q'R' has points (3, -1), (3, -3), (6, -2).

(d) On a coordinate plane, triangle P'Q'R' has points (-1, 3), (-3, 3), (-2, 6).

The order of[tex]$gN$[/tex] divides [tex]$k$[/tex], as required.

Let [tex]$g\in G$[/tex] be an arbitrary element and let [tex]$k$[/tex] be the order of [tex]$g$[/tex], that is, [tex]$g^k=e_G$[/tex], the identity element of [tex]$G$[/tex]. We want to show that the order of [tex]$gN$[/tex] in [tex]$G/N$[/tex] divides [tex]$k$[/tex].

Consider the coset [tex]$g^iN\in G/N$[/tex] for some positive integer [tex]$i$[/tex]. We want to find the smallest positive integer [tex]$j$[/tex] such that [tex]$(gN)^j = g^jN = g^iN$[/tex]. Since [tex]$g^iN = g^k(g^i)^kN = g^kN$[/tex], we have [tex]$(gN)^j = g^{jk}N = g^iN$[/tex], which implies [tex]$g^{jk-i}\in N$[/tex].

Since [tex]$N$[/tex] is a normal subgroup of[tex]$G$[/tex], we have [tex]$g^{jk-i}\in N$[/tex] if and only if [tex]$g^{jk-i}g^j=g^{jk}\in N$[/tex]. This shows that [tex]$g^{jk}$[/tex] is in the kernel of the canonical homomorphism [tex]$\pi:G\to G/N$[/tex], so[tex]$k$[/tex] is a multiple of the order of[tex]$gN$[/tex] in [tex]$G/N$[/tex].

Therefore, the order of [tex]$gN$[/tex] divides [tex]$k$[/tex], as required.

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(a) A basis for the subspace { (w, x, y) ∈ R³ : 2w + x - y = 0 } is {(1, 0, 2), (0, 1, 1)}.

(b) A basis for the subspace { (x, y, z) ∈ R³ : x - 2y + z = 0, x + 2z = 0 } is {(-2, 1, 1)}.

(a) To find a basis for the subspace { (w, x, y) ∈ R³ : 2w + x - y = 0 }, we can first rewrite the equation as y = 2w + x. Then any vector (w, x, y) in the subspace can be written as (w, x, 2w + x) = w(1, 0, 2) + x(0, 1, 1). Therefore, a basis for the subspace is {(1, 0, 2), (0, 1, 1)}.

(b) To find a basis for the subspace { (x, y, z) ∈ R³ : x - 2y + z = 0, x + 2z = 0 }, we can use the equations to solve for x, y, and z in terms of a free variable. Using z as the free variable, we get x = -2z and y = z. Therefore, any vector (x, y, z) in the subspace can be written as (-2z, z, z) = z(-2, 1, 1). Since there is only one free variable, z, we have a one-dimensional subspace. Therefore, a basis for the subspace is {(-2, 1, 1)}.

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Answer:

B

Step-by-step explanation:

The method that assures random selection is:

The supervisor should select one grade level and survey randomly selected students from that grade.

Option A is the correct answer.

It is the way of choosing a number of required items from a number of populations given.

Each item has an equal of being chosen.

We have,

Option A would be the best method to assure the random selection of a sample population because it involves selecting a specific group (one grade level) and then randomly selecting individuals from that group.

This ensures that all individuals within the chosen group have an equal chance of being selected for the survey.

Option B may not provide an accurate representation of the entire student population as some grades may have a higher or lower percentage of students bringing their lunch.

Option C is not feasible as it would be time-consuming and resource-intensive to survey all students enrolled in the school.
Option D is not a random selection method as it only involves selecting one grade level and surveying all students in that grade, which may not be representative of the entire student population.

Thus,

The supervisor should select one grade level and survey randomly selected students from that grade.

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The correct answer to the following question on regression model are;

1. True 2. False 3. False  4 True

1.  In a regression model with an interaction term, the effect of the dummy variable (d) depends on the value of the numeric variable (x). To determine the partial effect of the dummy variable, the value of x needs to be known.

2.  In a regression model with two dummy variables and an interaction term, the interaction effect (β3d1d2) can be difficult to estimate because of potential multicollinearity between the main effects (d1 and d2) and the interaction term (d1d2).

3. In a quadratic regression model (y = β0 + β1x + β2x² + ε), if β2 > 0, the relationship between x and y is a U-shape, not an inverted U-shape. If β2 < 0, the relationship would be an inverted U-shape.

4. The equation ln(y) = β0 + β1 ln(x) + ε represents an exponential regression model. By exponentiating both sides of the equation, you obtain y = e^(β0) * (x^β1) * e^(ε), which is an exponential relationship between x and y.

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Required length of the ST is 35 feet.

What is circle?

A circle is a geometrical shape in which all points on its boundary or circumference are equidistant from a fixed point called the center. It can also be defined as the locus of all points that are at a fixed distance from the center. The distance from the center to any point on the circle is called the radius, and the distance across the circle through the center is called the diameter.

First, we notice that since PS is a diameter, angle PXS is a right angle (90 degrees) since it subtends the diameter. Therefore, angle QXT = 180 - angle PXT - angle RXS = 180 - 125 - 97 = 38 degrees.

Since X is the center of the circle, PX = RX = SX = TX (the radius of the circle), and so triangle PXS is an isosceles triangle with PS = 28 feet as its base. We can find PX as follows:

cos(125/2) = (PS/2) / PX

PX = (PS/2) / cos(125/2) = 28 / cos(62.5) = 60.1 feet (rounded to one decimal place)

Now we can use the law of cosines to find ST:

ST² = PS² + PT² - 2(PS)(PT)cos(QXT)

ST² = 28² + 2(PX² - 14²) - 2(28)(PX)cos(38)

ST² = 784 + 2(3602 - 196) - 2(28)(3602 - 196)cos(38)

ST² = 784 + 6806 - 2(28)(3406)cos(38)

ST²= 784 + 6806 - 64687.5

ST² = 1245.5

ST ≈ 35.3 feet (rounded to the nearest ten)

Therefore, the length of ST is approximately 35 feet.

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The linear Taylor polynomial is l(x,y) = x-1, and the quadratic Taylor polynomial is q(x,y) = x-1.

To find the linear and quadratic Taylor polynomials for f(x, y) = sin(x-1)cos(y) near (1, 0), we need to find the partial derivatives of f with respect to x and y, evaluated at (1,0):

f(x, y) = sin(x-1)cos(y)

[tex]\dfrac{\partial f}{\partial x}[/tex] = cos(x-1)cos(y)

[tex]\dfrac{\partial f}{\partial y}[/tex] = -sin(x-1)sin(y)

Evaluated at (1,0), we get:

f(1,0) = sin(0)cos(0) = 0

[tex]\dfrac{\partial f}{\partial x}(1,0)[/tex] = cos(0)cos(0) = 1

[tex]\dfrac{\partial f}{\partial y}(1,0)[/tex] = -sin(0)sin(0) = 0

The linear Taylor polynomial is:

l(x,y) = f(1,0) + [tex]\dfrac{\partial f}{\partial x}(1,0)[/tex](x-1) + [tex]\dfrac{\partial f}{\partial y}(1,0)[/tex](y-0)

l(x,y) = 0 + 1(x-1) + 0(y-0)

l(x,y) = x-1

The quadratic Taylor polynomial is:

[tex]q(x,y) = l(x,y) + \dfrac{1}{2} \dfrac{\partial^2f}{\partial x^2}(1,0)(x-1)^2 + \dfrac{\partial^2f}{\partial y^2}(1,0)(y-0)^2 + \dfrac{\partial^2f}{\partial x \partialy}(1,0)(x-1)(y-0)[/tex]

We need to find the second-order partial derivatives:

[tex]\dfrac{\partial^2f}{\partial x^2}[/tex] = -sin(x-1)cos(y)

[tex]\dfrac{\partial^2f}{\partial y^2}[/tex] = -sin(x-1)cos(y)

[tex]\dfrac{\partial^2f}{\partial x \partial y}[/tex] = -cos(x-1)sin(y)

Evaluated at (1,0), we get:

[tex]\dfrac{\partial^2f}{\partial x^2}(1,0)[/tex]= -sin(0)cos(0) = 0

[tex]\dfrac{\partial^2f}{\partial y^2}(1,0)[/tex] = -sin(0)cos(0) = 0

[tex]\dfrac{\partial^2f}{\partial x \partialy}(1,0)[/tex] = -cos(0)sin(0) = 0

Substituting into the quadratic Taylor polynomial formula, we get:

q(x,y) = (x-1) + (1/2)(0)(x-1)² + (1/2)(0)(y-0)² + (0)(x-1)(y-0)

q(x,y) = x-1

Therefore, the linear Taylor polynomial is l(x,y) = x-1, and the quadratic Taylor polynomial is q(x,y) = x-1.

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An operating system such as Windows or Linux is an example of the software component of an information system. So, the correct option is A. Software.

An operating system such as Windows or Linux is an example of the software component of an information system. Software is a set of instructions or programs that tell the hardware what to do and how to do it. In the case of an operating system, it is the software that manages the computer's hardware resources, provides services for applications, and enables users to interact with the computer. An operating system acts as an intermediary between the hardware and other software programs that run on the computer.

In addition to managing hardware resources, an operating system provides several other key functions, such as memory management, file management, security, and network connectivity. These functions are essential for the effective and efficient operation of an information system. Therefore, an operating system plays a crucial role in the overall functioning of an information system.

Therefore, Option A. Software is the correct answer.

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The following parts can be answered by the concept of Sets.

a.  f is not injective.
b. f is surjective.

(a) To prove or disprove that f is injective, we need to determine whether for every x1, x2 in R such that f(x1) = f(x2), it must be the case that x1 = x2.

Assume f(x1) = f(x2). Then, for any Y in R, we have y^2 < x1 if and only if y² < x2. However, this does not guarantee that x1 = x2. For example, let x1 = 2 and x2 = 3. Both f(x1) and f(x2) include all Y such that y² < 2 and y^2 < 3, respectively, but x1 ≠ x2.

Therefore, f is not injective.

(b) To prove or disprove that f is surjective, we need to determine whether for every set S in P(R), there exists an x in R such that f(x) = S.

Consider an arbitrary set S in P(R). If S is the empty set, then we can choose x = 0, since there are no Y in R such that y² < 0, and f(x) = S. If S is non-empty, let m = sup{y² | y in S}. Then for all Y in R, y² < m if and only if Y in S. Thus, we can choose x = m and f(x) = S.

Therefore, f is surjective.

Therefore,

a.  f is not injective.
b. f is surjective.

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Answer:3.5/1     14/4     21/6

Step-by-step explanation:

1. When lab = 0:  f_0 = f(lab = 0, cde, ac, ad) Here, we substitute lab with 0 in the function.
2. When lab = 1: f_1 = f(lab = 1, cde, ac, ad) Here, we substitute lab with 1 in the function.
So, the cofactors of the given function f with respect to lab are f_0 and f_1.

To find the cofactors of f with respect to lab using Shannon's Expansion Theorem, we need to consider two cases:

1. When lab = 0:

In this case, we need to remove the term that contains lab. So we can rewrite f as follows:
f = (ab ac ad) + (cde)

To find the cofactor of f with respect to lab = 0, we need to remove the terms that contain both lab and its complement (la!b):
Cofactor of f with respect to lab = 0 = (ac ad) + (cde) = acd + cde + ace + ade

2. When lab = 1:

In this case, we need to set lab to 1 and remove the term that contains its complement (la!b).

So we can rewrite f as follows: f = (ab ac ad) + (cde)

Setting lab to 1 gives us: f|lab=1 = ac ad cde

To find the cofactor of f with respect to lab = 1, we need to remove the terms that contain both lab and its complement (la!b):
Cofactor of f with respect to lab = 1 = ad cde

Therefore, the cofactors of f with respect to lab are acd + cde + ace + ade and ad cde.

Using Shannon's Expansion Theorem, we can determine the cofactors of the given function f with respect to the variable lab.

The theorem states that any function can be expressed as the sum of its cofactors.

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Answer: At least 96 OR 340 students.

Explanation:

1) Since we know that at least 96 students of the survey have at least one sibling, it is safe to assume that 96 students of the total 425 students have at least one sibling.

2) Since we know that 80% of respondents to the survey have at least one sibling, we could apply that percentage to the whole school:

80% × 425 students = 340 students.

Hence, we could conclude that 340 students have at least one sibling.

The solution to the given initial value problem is[tex]y(t) = -4e^t - e^{(2t)} + e^{(-2t)}[/tex], with initial conditions y(0) = -4, y'(0) = -1, and y"(0) = -19.

To solve the initial value problem y′′′-y′′-4y′ + 4y = 0; y(0) = -4, y'(0) = -1, y"(0) = -19, we can use the characteristic equation method.

The characteristic equation is[tex]r^3 - r^2 - 4r + 4 = 0.[/tex]

Factoring the equation by grouping, we get:

[tex]r^2(r - 1) - 4(r - 1) = 0[/tex]

[tex](r - 1)(r^2 - 4) = 0[/tex]

(r - 1)(r - 2)(r + 2) = 0

Therefore, the roots are r = 1, r = 2, and r = -2.

The general solution of the differential equation is:

[tex]y(t) = c1 e^t + c2 e^{(2t)} + c3 e^{(-2t)}[/tex]

Using the initial conditions, we have:

y(0) = c1 + c2 + c3 = -4

[tex]y'(t) = c1 e^t + 2c2 e^{(2t)} - 2c3 e^{(-2t)}[/tex]

y'(0) = c1 + 2c2 - 2c3 = -1

[tex]y''(t) = c1 e^t + 4c2 e^{(2t)} + 4c3 e^{(-2t)}[/tex]

y''(0) = c1 + 4c2 + 4c3 = -19

Solving the system of equations:

c1 = -4

c2 = -1

c3 = 1

Therefore, the particular solution to the initial value problem is:

[tex]y(t) = -4e^t - e^{(2t)} + e^{(-2t)}[/tex]

Thus, the solution to the given initial value problem is[tex]y(t) = -4e^t - e^{(2t)} + e^{(-2t)}[/tex], with initial conditions y(0) = -4, y'(0) = -1, and y"(0) = -19.

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Answer:

D) 72 cm squared

Step-by-step explanation:

Separate into to shapes, a right triangle and a rectangle

Rectangular area = 6 x 8

6 x 8 = 48

Triangle area, you have to find the length

Equation A + B = C

B = 8 and C = 10

A + 8^2 = 10^2

Need to find A

10^2 - 8^2=

10 x 10 = 100

8 x 8 = 64
100 - 64= 36

Now we have to find the square root of 36
Finding the square root of means finding 2 numbers that are the same and multiplying them to get 36

6 x 6 = 36

So the 6 is A

Now to solve the triangle area

6 x 8 x 1/2 =

48 x 1/2 = 24

Now add the two areas

48 + 24 = 72 cm squared

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The system of inequalities that best represent the shaded feasible region shown on the graph is

x - 4y > -2

x + 2y > 4

Here, we have,

to determine the equation of the guess game

information gotten from the question include

inequality graph showing shaded regions

some interpretation on the information from the question include

shading above a line is greater than

dotted lines mean the inequality do not have equal to

This interpretation removes the first, the second and the last options

making the third option the correct choice

other consideration is the intercept

the first equation at x= 0, x - 4y > -2,  y > 2

the second equation at x= 0, x + 2y > 4,  y > 1/2

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The approximate solution for the system of equations is x = -48/29

From the question, we have the following parameters that can be used in our computation:

f(x) = 5/8x + 2

g(x) = -3x - 4

To calculate the solution for the system of equations, we have the following

f(x) = g(x)

Substitute the known values in the above equation, so, we have the following representation

5/8x + 2 = -3x - 4

Multiply through by 8

So, we have

5x + 16 = -24x - 32

Evaluate the like terms

29x = -48

Evaluate

x = -48/29

Hence, the solution is x = -48/29

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The convergence set of the power series ∑n=1∞ [tex](x-2)^n/n^2[/tex] is [1, 3). The series converges for x values in the interval [1, 3), and diverges for x values outside of this interval.

At the endpoints x = 1 and x = 3, the series converges for x = 1 and diverges for x = 3.

To find the convergence set of a power series, we can use the ratio test:

lim[n→∞] |[tex](x - 2)(n+1)^2 / n^2[/tex]| = lim[n→∞] |[tex](x - 2)(1 + 2/n)^2[/tex]| = |x - 2| lim[n→∞] [tex](1 + 2/n)^2[/tex]

Since lim[n→∞] [tex](1 + 2/n)^2 = 1[/tex], the series converges if |x - 2| < 1, and diverges if |x - 2| > 1.

If |x - 2| = 1, then the ratio test is inconclusive, so we need to check the endpoints x = 1 and x = 3 separately.

For x = 1, the series becomes:

∑n=1infinitynn2 = ∑n=1infinitynn2

which is the alternating harmonic series, which converges by the alternating series test.

For x = 3, the series becomes:

∑n=1infinitynn2 = ∑n=1[infinity]nn2

which diverges by the p-series test with p = 2.

Therefore, the convergence set of the series is:

1 ≤ x < 3

In interval notation, this can be written as:

[1, 3)

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The weekly rate of change in the locust population is that: Every week, the number of locusts grows by a factor of 3.58

The formula for exponential growth function is:

f(x) = a(1 + r)^{x}

where:

f(x) = exponential growth function

a = initial amount

r = growth rate

{x} = number of time intervals

Now, we are given the exponential equation as:

N(t) = 300 * (1.2)^(t)

where:

t is the elapsed time in days

N(t) is the total number of locust

Now, 7 days make a week and so, we have:

weekly rate of change =  (1.2)^(7) = 3.58

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Answer:

-157

Step-by-step explanation:

8-5[(4+3)^2-(2^3+8)], Your Answer Should Be "-157"

Hope this helps!

Answer:

The triangle is obtuse.

Step-by-step explanation:

Using the sine rule to determine the other angle:

[tex]\frac{sinA}{a} =\frac{sinB}{b} \\\frac{sin90}{10} =\frac{sinB}{8} \\10sinB=8sin90\\sinB=\frac{8}{10} \\B=sin^{-1} (\frac{8}{10} )\\B=53.1301[/tex]

180 - 53.1301 - 90 = 36.8699

Using sine rule again to determine the unknown length:

[tex]\frac{sinA}{a} =\frac{sinB}{b} \\\frac{sin90}{10} =\frac{sin36.8699}{b} \\10sin36.8699=bsin90\\b=6[/tex]

The value of the determinant det(B⁻¹ A) is -3.

We want to find the determinant of the product of matrices B⁻¹ and A, which can be written as det(B⁻¹ A).

Given that det(A) = 6 and det(B) = -2, you can use the following properties of determinants:

1. det(AB) = det(A) * det(B) for any square matrices A and B.
2. det(A⁻¹) = 1/det(A) for any invertible matrix A.

Now, let's find the determinant of the given product:
det(B⁻¹A) = det(B⁻¹) * det(A) by property 1.

Since det(B) = -2, we can find det(B⁻¹) using property 2:
det(B⁻¹) = 1/det(B) = 1/(-2) = -1/2.

Now, substitute the known values of det(A) and det(B⁻¹) into the equation:
det(B⁻¹ A) = det(B⁻¹) * det(A) = (-1/2) * 6 = -3.

So, the determinant of the product B⁻¹A is -3.

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