Considere as funções com domínio dos números reais dadas por f(x) = -x + 5 e g(x) = 2x + 2 Calcule o valor de : F(0) + g(1) ________ F(2)

A 3.
B4.
C9.
D5.
E 7

Binding constraints directly influence the optimal solution in a linear Programming problem, whereas constraints with positive slack are non-binding and do not directly impact the solution.

binding constraints and positive slack in the context of linear programming. In a linear programming problem, we aim to find the optimal solution for an objective function, given a set of constraints. The terms "binding constraints" and "positive slack" are related to these constraints.

1. Binding constraints: These are constraints that directly impact the optimal solution of the problem. In other words, they "bind" the feasible region (the area where all the constraints are satisfied) and affect the maximum or minimum value of the objective function. Binding constraints are active constraints, as they influence the final solution.

2. Positive slack: Slack is the difference between the left-hand side and right-hand side of a constraint when the constraint is satisfied. If this difference is positive, it means that there is some "extra" or "unused" resource in that constraint. Positive slack indicates that the constraint is non-binding, meaning it does not directly impact the optimal solution. It shows that there is some room for the constraint to be further tightened without affecting the final outcome.

In summary, binding constraints directly influence the optimal solution in a linear programming problem, whereas constraints with positive slack are non-binding and do not directly impact the solution. Knowing the difference between these terms can help you better understand and analyze linear programming problems.

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Among the given monomial functions, the function with the maximum value is y=5x6.

To determine which monomial function has a maximum value, we need to find the leading coefficient and exponent of each function. The leading coefficient of a monomial function is the coefficient of the term with the highest degree. The degree of a monomial function is the sum of the exponents of its variables.

Looking at the given functions, we can see that y=5x6 has the highest degree and a positive leading coefficient. This means that the function has an upward-facing parabola and its maximum value is at x=0.

Therefore, y=5x6 is the monomial function that has a maximum value.

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Answer: True

Use the vertical line test to determine whether or not a graph represents a function. If a vertical line is moved across the graph and, at any time, touches the graph at only one point, then the graph is a function. If the vertical line touches the graph at more than one point, then the graph is not a function.

Answer:

True

Step-by-step explanation:

Using the vertical line test since it only passes through the graph once it is a function

Answer:

$87.50

Step-by-step explanation:

30% = 0.3

We take

125 - (125 x 0.3) = $87.50

So, the retail price $87.50

Answer:

27/2 km/hr

Step-by-step explanation:

Answer:

B, D and E.

Step-by-step explanation:

These numbers are irrational because their exact values can only be represented by those square roots. If you try to "do the maths", it will be something like:

\sqrt{15} = 3.872983...\\

\sqrt{57} = 7.549834...\\

\sqrt{8} = 2.828427...

Answer:

A,D and E and this is the correct answer

well, from 1990 to 1998 is 8 years, and we know the amount went from $4800 to $6300, let's check for the rate of growth.

\(\qquad \textit{Amount for Exponential Growth} \\\\ A=P(1 + r)^t\qquad \begin{cases} A=\textit{accumulated amount}\dotfill & \$6300\\ P=\textit{initial amount}\dotfill &\$4800\\ r=rate\to r\%\to \frac{r}{100}\\ t=\textit{years}\dotfill &8\\ \end{cases} \\\\\\ 6300=4800(1 + \frac{r}{100})^{8} \implies \cfrac{6300}{4800}=(1 + \frac{r}{100})^8\implies \cfrac{21}{16}=(1 + \frac{r}{100})^8\)

\(\sqrt[8]{\cfrac{21}{16}}=1 + \cfrac{r}{100}\implies \sqrt[8]{\cfrac{21}{16}}=\cfrac{100+r}{100} \\\\\\ 100\sqrt[8]{\cfrac{21}{16}}=100+r\implies 100\sqrt[8]{\cfrac{21}{16}}-100=r\implies \stackrel{\%}{3.46}\approx r\)

now, with an initial amount of $4800, up to 2007, namely 17 years later, how much will that be with a 3.46% rate?

\(\qquad \textit{Amount for Exponential Growth} \\\\ A=P(1 + r)^t\qquad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{initial amount}\dotfill &4800\\ r=rate\to 3.46\%\to \frac{3.46}{100}\dotfill &0.0346\\ t=years\dotfill &17\\ \end{cases} \\\\\\ A=4800(1 + 0.0346)^{17} \implies A=4800(1.0346)^{17}\implies A \approx 8558.02\)

A)  (-infinity,1) U (1,infinity).

b)  There is a horizontal asymptote at y=1.

c)  The derivative is always negative (except at x=1 where it is undefined), so the function is decreasing on the interval (-infinity,1) and increasing on the interval (1,infinity).

d)   The function has no local maximum or minimum values

e)   The function is concave up on the interval (-infinity,1) U (1,infinity).

f) This has no real solutions, so there are no inflection points.

(a) The domain of f(x) is all real numbers except x = 1. Thus, the domain can be expressed as (-infinity,1) U (1,infinity).

(b) There is a vertical asymptote at x=1 since the denominator becomes zero at x=1. To find the horizontal asymptote, we can use the fact that as x approaches infinity or negative infinity, the function approaches the value of its leading term. In this case, the leading term is x/x, which simplifies to 1. Therefore, there is a horizontal asymptote at y=1.

(c) To find the intervals of increase and decrease, we can take the derivative of the function:

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

The derivative is always negative (except at x=1 where it is undefined), so the function is decreasing on the interval (-infinity,1) and increasing on the interval (1,infinity).

(d) The function has no local maximum or minimum values since it does not change direction.

(e) To find the intervals of concavity, we can take the second derivative of the function:

f''(x) = 4/(x-1)^3

The second derivative is positive for all x except x=1 where it is undefined, so the function is concave up on the interval (-infinity,1) U (1,infinity).

(f) To find the inflection points, we can set the second derivative equal to zero:

4/(x-1)^3 = 0

This has no real solutions, so there are no inflection points.

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Answer:

(-2/5, 11/5)

Step-by-step explanation:

We can use the formula for finding a point that divides a line segment into a given ratio. Let P be the point on the line segment QR that partitions it in the ratio 2:3. Then we have:

P = ( (3x2 + 2x1)/(3+2), (3y2 + 2y1)/(3+2) )

where (x1, y1) = (-8, -8) is the coordinates of Q and (x2, y2) = (2, 7) is the coordinates of R.

Substituting the values, we get:

P = ( (32 + 2(-8))/(3+2), (37 + 2(-8))/(3+2) )

P = ( (-2/5), (11/5) )

Therefore, the coordinates of the point P on the directed line segment QR that partitions it in the ratio 2:3 are (-2/5, 11/5).

By the above proof, we know that the dimension of this subspace is less than or equal to the rank of A. Therefore, rank(AB) ≤ rank(A).

To prove that dim(AV) ≤ rank(A), where A: X → Y and V is a subspace of X, we need to show that the dimension of the subspace AV is less than or equal to the rank of the transformation A.

Proof:

Let {v1, v2, ..., vk} be a basis for V, where k is the dimension of V.

We want to show that the set {Av1, Av2, ..., Avk} is linearly independent in Y.

Suppose there exist coefficients c1, c2, ..., ck such that c1Av1 + c2Av2 + ... + ckAvk = 0. We need to show that c1 = c2 = ... = ck = 0.

Applying the transformation A to both sides, we get A(c1v1 + c2v2 + ... + ckvk) = A(0).

Since A is a linear transformation, we have A(c1v1 + c2v2 + ... + ckvk) = c1Av1 + c2Av2 + ... + ckAvk = 0.

But we know that {Av1, Av2, ..., Avk} is linearly independent, so c1 = c2 = ... = ck = 0.

Therefore, the set {Av1, Av2, ..., Avk} is linearly independent in Y, and its dimension is at most k.

Hence, dim(AV) ≤ k = dim(V).

From the above proof, we can deduce that rank(AB) ≤ rank(A) for any linear transformations A and B. This is because if we consider the transformation A: X → Y and the transformation B: Y → Z, then rank(AB) represents the maximum number of linearly independent vectors in the image of AB, which is a subspace of Z.

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Step-by-step explanation:

He would have 211 left. 750- 539= 211

The slope-intercept form of the equation is y = - mx + b.

The equation of a line can be expressed in a variety of ways. Although they may differ in appearance, all of them describe the same line because multiple equations can describe the same line. However, all linear equations describing the same line are equivalent.

The formula y = mx + b can be used to represent any straight line. Any straight line equation, known as a linear equation, can be expressed as y = mx + b, where m denotes the line's slope and b its y-intercept. The value of y at the line's intersection with the y-axis is its y-intercept.

Given a graph of the equation, choose two points on the line and use them to calculate the slope in order to put the equation in slope-intercept form. This is the answer to the equation's question. Next, determine the y-coordinates, intercept's which should be of the type (0,b). The value in the equation is the y-coordinate.

Thus, the equation derived from the graph is as follows:

Since we are aware that the slope here is negative,

Therefore Our formula is y = -mx + b.

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By congruent complement theorem ∠4 is congruent to ∠1.

According to the given question.

Angle 4 and angle 5 are complements and angle1 and angle 5 are complements.

Now, according to congruent complements theorem

"If two angles are complementary to the same angle, then these angles are congruent to each other."

Since, it is given that ∠4 and ∠5 are the complements and ∠1 and ∠5 are compliments.

⇒ ∠1 and ∠4 are complimentary to the same angle ∠5.

Therefore, by congruent complement theorem ∠4 is congruent to ∠1.

Thus option 1 is correct.

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The solution to the Laplace equation V²u = 0 with the given function f(θ) as \(u(r, \theta) = \sum[0 * r^\lambda+ (100 * \sqrt{2}) * r^{(-\lambda)}][Ccos(\lambda\theta) + Dsin(\lambda\theta)]\)

To solve the Laplace equation V²u = 0 with the given boundary condition u(1, θ) = f(θ), where f(θ) is defined as:

f(θ) = \(\left \{ {100 {if} 0 \leq \theta < \pi/4} \atop {0 if \pi/4 < \theta < 0}} \right}} \right.\)

We will use separation of variables to find the solution. Let's assume the solution can be written as u(r, θ) = R(r)h(θ), where R(r) represents the radial component and h(θ) represents the angular component.

Using separation of variables, we can write the Laplace equation as:

\((1/r)(d/dr)(r(dR/dr)) + (1/r^2)(d^2h/d\theta^2) = 0\)

To separate the variables, we set each term equal to a constant, denoted by -λ²:

\((1/r)(d/dr)(r(dR/dr)) = \lambda^2 (1)\)

\((1/r^2)(d^2h/d\theta^2) = -\lambda^2 (2)\)

Solving equation (1), we obtain the radial equation:

\(r(d^2R/dr^2) + (dR/dr) - \lambda^2R = 0\)

This is a standard differential equation with solutions of the form \(R(r) = Ar^{\lambda} + Br^{-\lambda}\), where A and B are constants.

Solving equation (2), we obtain the angular equation:

d²Θ/dθ² + λ²Θ = 0

This is a standard differential equation with solutions of the form

Θ(θ) = Ccos(λθ) + Dsin(λθ), where C and D are constants.

Now, we can combine the radial and angular components to form the general solution:

\(u(r, \theta) = \sum[Ar^\lambda + Br^{-\lambda}][Ccos(\lambda\theta) + Dsin(\lambda\theta)]\)

Next, we apply the boundary condition u(1, θ) = f(θ):\(u(1, \theta) = \sum[Ar^\lambda + Br^{-\lambda}][Ccos(\lambda\theta) + Dsin(\lambda\theta)] = f(\theta)\)

Comparing the terms on both sides, we can determine the coefficients A, B, C, and D using the given function f(θ).

To solve these equations, we'll use trigonometric identities and properties. Let's begin with the first equation:

\(Acos(\lambda\theta) + Bsin(\lambda\theta) = 100\)

We can rewrite this equation using the identity sin(π/4) = cos(π/4) = \(\sqrt2\):

\(Acos(\lambda\theta) + Bsin(\lambda\theta) = 100\)

(A/\(\sqrt2\) ) * \(\sqrt2\)  * cos(λθ) + (B/\(\sqrt2\) ) * \(\sqrt2\) * sin(λθ) = 100

Now, we can equate the coefficients of cos(λθ) and sin(λθ) separately to determine A and B. Since cos(λθ) and sin(λθ) are orthogonal functions, their coefficients must be zero:

(A/\(\sqrt2\)) * \(\sqrt2\) = 0 (coefficient of cos(λθ))

(B/\(\sqrt2\)) * \(\sqrt2\)= 100 (coefficient of sin(λθ))

From the first equation, we can conclude that A = 0. Substituting this into the second equation:

(B/\(\sqrt2\)) * \(\sqrt2\) = 100

B = \(\sqrt2\) * 100

B = 100 * \(\sqrt2\)

Therefore, we have A = 0 and B = 100 * \(\sqrt2\).

For the interval π/4 < θ < 2π, we have:

\(Acos(\lambda\theta) + Bsin(\lambda\theta) = 0\)

Again, equating the coefficients of cos(λθ) and sin(λθ) separately:

\(Acos(\lambda\theta) + Bsin(\lambda\theta) = 0\)

A * 0 + B * 1 = 0

B = 0

From this equation, we conclude that B = 0.

To summarize, we found A = 0, B = 100 * \(\sqrt2\) for the interval 0 ≤ θ < π/4, and B = 0 for the interval π/4 < θ < 2π.

Using these coefficients, we can now write the solution to the Laplace equation V²u = 0 with the given function f(θ) as:

\(u(r, \theta) = \sum[0 * r^\lambda+ (100 * \sqrt{2}) * r^{(-\lambda)}][Ccos(\lambda\theta) + Dsin(\lambda\theta)]\)

Therefore, the solution to the Laplace equation V²u = 0 with the given function f(θ) as \(u(r, \theta) = \sum[0 * r^\lambda+ (100 * \sqrt{2}) * r^{(-\lambda)}][Ccos(\lambda\theta) + Dsin(\lambda\theta)]\)

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Answer:

13.20$ :)

Step-by-step explanation:

40% off of 20$ is 12$ and 10% of 12$ is 1.20$ add that and you get 13.20$

Answer:

the answer should be 360° n

Answer: y=4x+11

y=mx+b

for the line that passes through the points

Answer: 45 minutes for BOTH workouts

Step-by-step explanation:

Let time for plan a be x

Let time for plan b be y

On Wednesday

# of clients for plan a are 5

# clients for plan b are 3

On Thursday

# of clients for plan a are 7

# of clients for plan b are 9

Wednesday eqn --> 5x+3y=6

Thursday eqn --> 7x+9y=12

Multiply Wednesday equation by -3 and use elimination to solve

-15x-9y=-18

7x+9y=12

-8x=-6

x=6/8

Reduce: x=3/4 hour

Substitute x=3/4 into Thursday equation

7(3/4)+9y=12

9y=27/4

y=3/4

Therefore, the length of BOTH the plan a and plan b workouts were 45 minutes

Step-by-step explanation:

y = 30

120y = 120×30 = 3600

√3600 = 60

The value of y is 60 which makes 120y a perfect square.

A perfect square is a number that represents the square of a natural number.

For example 9,16,36,64,121 etc

Here given that we have to calculate the value of y for which 120y will be a perfect square.

so 120y = n² where n is the natural number

For calculating the value of y first factorize 120

120= 2*2*2*3*5

= 2² * 2*3*5

=4 *30

from the above, it is clear that 120 can be written as 4*30 where 4 is a perfect square but 30 is not a perfect square.

So if we multiply 30 by 120 then the product has 2²*30² which will be a perfect square.

Therefore the value of y is 30 for which 120*30=n² where n=60.

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Besides vertical asymptotes, the zeros of the denominator of a rational function give rise to vertical holes or removable discontinuities in the graph of the function.

When the denominator of a rational function becomes zero at a specific value of x, it results in an undefined value for that particular \(x\) value. This means that the function is not defined at that point, creating a potential discontinuity.

However, in some cases, the numerator of the rational function may also have the same zero value as the denominator, which cancels out the undefined behavior and makes the function well-defined at that point. These canceled zeros give rise to vertical holes in the graph.

A vertical hole occurs when there is a "hole" or an open circle in the graph of the rational function at the x value where the denominator is zero. This indicates that the function has a removable discontinuity at that point.

To clarify, if the zero of the denominator is not canceled out by the numerator, it results in a vertical asymptote. On the other hand, if the zero of the denominator is canceled out by the numerator, it leads to a vertical hole or a removable discontinuity in the graph of the function.

Therefore, the zeros in the denominator of a rational function result in vertical holes or detachable discontinuities on the graph of the function in addition to vertical asymptotes.

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Answer:

Im guessing

Step-by-step explanation:

The same

Equal

the mid point

The end

your welcome and not for sure

Try it out

Answer:

No

(x + 4) ^2 = (x^2 + 8x + 16)

Step-by-step explanation:

You have to factor the equation (X + 4) ^2.

Expanded: (x + 4) * (x + 4)

Using the foil technique:

First: (x) * (x) = x^2

Outer: (x) * (4) = 4x

Inner: (4) * (x) = 4x

Last: (4) * (4) = 16

Factor equations should look like this

a^2 + bx + c, where the Outer and Inner values are added together. Resulting in:

x^2 + 8x + 16

Answer:

Irrational

Step-by-step explanation:

Because it is on going

Hi! 14.19274128 is IRRATIONAL. hope this helps you! Good luck and have a great day/night. ❤️✨

Answer:

56, There are 56 daisies.

Step-by-step explanation:

Answer:

The word problem is "How many $25 are there in $125?"

Step-by-step explanation:

Given

\(n(\$25) = \$125\)

Required

Write a word problem for the expression

We start by solving the given equation

\(n(\$25) = \$125\)

Divide both sides by $25

\(\frac{n(\$25)}{\$25} = \frac{\$125}{\$25}\)

\(n = \frac{\$125}{\$25}\)

\(n = 5\)

This implies that there are 5, $25 in $125

Hence; The word problem is "How many $25 are there in $125?"

Answer:

y=6+2x

Step-by-step explanation:

ANSWER :-

There are total 8 types of pets represented by the dot plot . They are :

ANSWER :-

Dog is the most popular pet

Step-by-step explanation :

Liza's classmates have owned two rabbits , six dogs , three cats , two horses , five birds , one mouse , three fishes and one other pet.

Therefore,

Now, we can easily count or see the pet which is in larger quantity as compared to others.

The probability of the sample mean being less than 79 is 77.64%

In order to solve the given problem we have to take the help of Standard error mean

SEM = ∑/√(n)

here,

∑ = population standard deviation

n = sample size

hence, the z-score can be calculated as

z = ( x' - μ)/σ/√(n)

here,

x' = sample mean

μ = population mean

σ = population standard deviation

n = sample size

adding the values into the formula

SEM = σ / √(n)

= 21/√64

= 2.625

z = (x' - μ)/SEM

= (79-77)/2.625

= 0.76

now, using standard distribution table we find that probability of a z-score is less than 0.77 then converting it into percentage

0.77 x 100

= 77%

The probability of the sample mean being less than 79 is 77.64%

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