h(x)=|2x|-8 domain and range

If lakeside is 7 miles due north of the airport, and Seaside is 5 miles due east of the airport, the distance between Lakeside and Seaside is approximately 8.6 miles.

To find the distance between Lakeside and Seaside, we can use the Pythagorean theorem, which states that in a right triangle, the square of the hypotenuse (the longest side) is equal to the sum of the squares of the other two sides.

In this case, the distance between Lakeside and Seaside is the hypotenuse of a right triangle with legs of 5 miles and 7 miles.

To apply the Pythagorean theorem, we can square the lengths of the legs and then take the square root of their sum:

distance = √(5² + 7²)

distance = √(25 + 49)

distance = √74

distance ≈ 8.6 miles (rounded to the nearest tenth)

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The equation y = 3(x - 5)^2 - 1 written in the standard form is y = 3x^2 - 30x + 74

To rewrite the given equation in standard form, we need to expand and simplify the squared term:

y = 3(x - 5)^2 - 1 [given equation]

y = 3(x^2 - 10x + 25) - 1 [expand (x - 5)^2 using FOIL method]

y = 3x^2 - 30x + 74 [combine like terms]

Therefore, the standard form of the equation is:

y = 3x^2 - 30x + 74

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The distances between the points are approximately 7.07, 10.20, and 7.07.

To find the distance between the points d(-8, 1) and e(-3, 6), we use the distance formula:

distance = √[(x2 - x1)^2 + (y2 - y1)^2]

Plugging in the values, we get:

distance = √[(-3 - (-8))^2 + (6 - 1)^2]
distance = √[5^2 + 5^2]
distance = √50
distance ≈ 7.07

To find the distance between the points e(-3, 6) and f(7, 4), we again use the distance formula:

distance = √[(x2 - x1)^2 + (y2 - y1)^2]

Plugging in the values, we get:

distance = √[(7 - (-3))^2 + (4 - 6)^2]
distance = √[10^2 + (-2)^2]
distance = √104
distance ≈ 10.20

To find the distance between the points f(7, 4) and g(2, -1), we again use the distance formula:

distance = √[(x2 - x1)^2 + (y2 - y1)^2]

Plugging in the values, we get:

distance = √[(2 - 7)^2 + (-1 - 4)^2]
distance = √[(-5)^2 + (-5)^2]
distance = √50
distance ≈ 7.07

So, the distances between the points are approximately 7.07, 10.20, and 7.07.

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complete question:

Determine the distance between DE, EF and FG.

D(-8, 1), E(-3, 6), F(7, 4), G(2, -1)

Answer: x = 10

Step-by-step explanation: To solve this equation, you can use the logarithmic property that states loga(b) + loga(c) = loga(bc). So, you can rewrite the left side of the equation as log2((x-6)(x+6)). Then, you can use the property that states loga(b) = c is equivalent to a^c = b to solve for x.

So, you have log2((x-6)(x+6)) = 6, which is equivalent to 2^6 = (x-6)(x+6). Simplifying the left side gives you 64, and expanding the right side gives you x^2 - 36 = 64. Solving for x gives you x = ±√100, which is x = ±10. However, since the original equation includes logarithms.

If Joe already scored 347 points in math-test, then to get a grade"A" he must score at least 56 marks, which is represented in inequality as n ≥ 56.

Jose has already scored 347 points on his math-tests so far, and he needs to score at least 403 points to get an A for the semester.

Let "minimum-points" he must score on the "remaining-tests" be denoted by "n". We can write an inequality to represent minimum-points as:

⇒ 347 + n ≥ 403,

⇒ n ≥ 403 - 347,

⇒ n ≥ 56.

Therefore, Jose must score at least 56 points on the remaining tests in order to get an A for the semester.

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The given question is incomplete, the complete question is

Jose has scored 347 points on his math tests so far this semester. To get an A for the semester, he must score at least 403 points. Write an inequality to find the minimum number of points he must score on the remaining tests in order to get an A. Let "n" represent the number of points Jose needs to score on the remaining tests.

Ana will take 100 ml on the first day and 5 ml less each day for 20 days, requiring a total of 1050 ml; the prescribed amount of 960 ml is not enough, resulting in a shortage of 90 ml, which will last for 18 days.

Ana will take the syrup for 20 days, and on each day, she will take 5 ml less than the previous day. To calculate the total amount of syrup Ana will need for the 20 days, we can use the formula for the sum of an arithmetic series,

S = (n/2) x (a₁ + aₙ), In this case, n = 20, a1 = 100 ml, and an = 100 ml - (19 x 5 ml) = 5 ml. Plugging in the values, we get,

S = (20/2) x (100 ml + 5 ml) = 1050 ml

So Ana will need a total of 1050 ml of syrup for the 20 days. The doctor prescribed 3 bottles of 320 ml each, which is a total of 960 ml. This is not enough to cover the full 20 days of treatment, as Ana will need 1050 ml. Therefore, there is a shortage of 90 ml of syrup. To calculate how many days Ana will lack syrup for, we need to divide the shortage by the daily reduction in dose,

90 ml/5 ml per day = 18 days

So Ana will have enough syrup for the first 2 days, but she will lack syrup for the next 18 days.

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Complete question - Ana has to take a syrup for 20 days, the doctor has prescribed 3 bottles of 320 ml each, she has to take the syrup in such a way that each day that passes she takes 5 ml less than the day before. If you start taking a 100 ml dose, how many ml will you take on the last day? Was the amount of syrup prescribed by the doctor enough? How much syrup is left over or lacking? if he lacked syrup, for how many days would he lack?

As per the given geometric sequence, the sum of the greater and lesser consequents is 51.

We are given that the antecedents (which are just the first three terms) of a geometric sequence are 2, 3, and 5. Let's call the common ratio of this sequence r. Using the definition of a geometric sequence, we can write the terms of this sequence as 2, 2r, 2r² (since the first term is 2 and the common ratio is r), 3, 3r, 3r², 5, 5r, 5r².

Next, we are told that the product of the consequents (which are just the terms after the first three) is 810. To find the product of the consequents, we just multiply all the terms after the first three together. So we have:

(2r³) * (3r²) * (5r) = 30r⁶

We know that this product is equal to 810, so we can set up the equation:

30*r⁶ = 810

Solving for r, we get:

r⁶ = 27

r = 3 (since 3⁶ = 729)

Now that we know the common ratio is 3, we can find the terms of the sequence by multiplying each antecedent by 3. So the terms of the sequence are:

2, 6, 18, 3, 9, 27, 5, 15, 45

The greater and lesser consequents are 45 and 6, respectively. So the sum of the greater and lesser consequents is:

45 + 6 = 51

Therefore, the answer to the problem is 51.

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Complete Question:

In a series of equal geometric ratios, the antecedents are 2, 3, and 5. If the product of the consequents is 810, find the sum of the greater and lesser consequents.

To find the center of the circle, we need to find the midpoint of the diameter segment CD.

Using the midpoint formula, we can find the coordinates of the midpoint M:

Midpoint formula:

M = ( (x1 + x2)/2 , (y1 + y2)/2 )

Plugging in the coordinates of C (-3,5) and D (6,-2):

M = ( (-3 + 6)/2 , (5 - 2)/2 )

M = (1.5, 1.5)

Therefore, the center of the circle is at point M with coordinates (1.5, 1.5).

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Jenny needs 180g of butter, 210g of brown sugar, 375g of oats, and 3 tablespoons of syrup to make 24 flapjacks.

To make 24 flapjacks, Jenny needs to increase the amount of each ingredient proportionally.

To calculate the required amounts, we can use ratios. If 16 flapjacks require 120g of butter, then 24 flapjacks require:

Butter: (24/16) x 120g = 180g

Brown sugar: (24/16) x 140g = 210g

Oats: (24/16) x 250g = 375g

Syrup: (24/16) x 2 tablespoons = 3 tablespoons

Therefore, Jenny needs 180g of butter, 210g of brown sugar, 375g of oats, and 3 tablespoons of syrup to make 24 flapjacks.

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The summation of the horizontal components is -629.76 N, and the summation of the vertical components is 363.68 N. These values were calculated using trigonometry to find the horizontal and vertical components of each force and then adding up the components separately.

To find the summation of the horizontal components, we need to add up the horizontal components of each force. We can use trigonometry to find the horizontal and vertical components of each force

Force #1 horizontal component = 400 cos(70) = 125.47 N

Force #2 horizontal component = 510 cos(100) = -158.95 N (negative because it acts in the opposite direction)

Force #3 horizontal component = 702 cos(260) = -596.28 N (negative because it acts in the opposite direction)

Therefore, the summation of the horizontal components is

125.47 N - 158.95 N - 596.28 N = -629.76 N

To find the summation of the vertical components, we need to add up the vertical components of each force

Force #1 vertical component = 400 sin(70) = 377.95 N

Force #2 vertical component = 510 sin(100) = 500.62 N

Force #3 vertical component = 702 sin(260) = -514.89 N (negative because it acts in the opposite direction)

Therefore, the summation of the vertical components is

377.95 N + 500.62 N - 514.89 N = 363.68 N

So the summation of the horizontal components is -629.76 N, and the summation of the vertical components is 363.68 N.

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The initial water level is given as follows:

28 units.

The slope-intercept representation of a linear function is given by the equation shown as follows:

y = mx + b

The coefficients m and b have the meaning presented as follows:

The function for this problem is defined as follows:

n = 4.5t + 28.

The intercept is of b = 28, representing the initial amount of water.

The graph is given by the image presented at the end of the answer.

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Answer:

55%

Step-by-step explanation:

Chances it wont land on red = chances of blue + chances of yellow

                                                = 55%

Answer:

a)  3, 2, 1, 4

Step-by-step explanation:

If you have multiple credit cards with different APRs, it is best to pay off the card with the highest APR first.  This is because you will save the most money in interest by paying off the highest-rate debt first.

Therefore, as Michelle has four credit cards, each with different APRs, she should pay them off in order of the highest to lowest interest rate.

Since the highest APR is 23%, credit card #3 should be paid off first.

The next highest APR is 19%, so credit card #2 should be paid off second.

Credit card #1 should be paid off next as it has an APR of 17%.

Finally, credit card #4 should be paid off last, as it has the lowest APR of 15%.

So the order in which Michelle should pay off her credit cards is:

The debt increased between 1996 and 2003. Then the national debt increased by approximately $4,903.73 billion between 1996 and 2003.

To find how much the debt increased between 1996 and 2003, we need to find the value of the function D'(t) for t=7 (since 2003 is 7 years after 1996).

D'(t)=858.29+819.48t-184.32t^2+12.12t^3

D'(7)=858.29+819.48(7)-184.32(7^2)+12.12(7^3)
D'(7)=858.29+5,736.36-8,132.32+3,458.68
D'(7)=1,921.01

Therefore, the annual rate of change in the national debt in 2003 was $1,921.01 billion per year.

To find how much the debt increased between 1996 and 2003, we need to integrate the function D'(t) from t=1 to t=7:

∫(D'(t))dt = ∫(858.29+819.48t-184.32t^2+12.12t^3)dt
= 858.29t + 409.74t^2 - 61.44t^3 + 3.03t^4 + C

where C is the constant of integration.

Evaluating this expression at t=7 and t=1 and taking the difference, we get:

(858.29(7) + 409.74(7)^2 - 61.44(7)^3 + 3.03(7)^4 + C) - (858.29(1) + 409.74(1)^2 - 61.44(1)^3 + 3.03(1)^4 + C)
= 6,111.09 - 1,207.36 = 4,903.73

Therefore, the national debt increased by approximately $4,903.73 billion between 1996 and 2003.

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The actual test score that coincides with a z-score of -1.25 is 65 when A college entrance exam had a mean of 80 with a standard deviation of 12 and a z-score of -1.25.

The formula to calculate the actual test score from a z-score is given as,

X = μ + Zσ,

where:

X = the actual or raw test score

μ = the mean

Z = z-score

σ = standard deviation.

Given data:

μ = 80

Z = -1.25

σ =  12

Substuting the values of μ, Z, and σ in the formula, we get;

X = μ + Zσ,

X = 80 + (-1.25)(12)

X = 80 + (-15)

X = 65.

Therefore, the actual test score that coincides with a z-score of -1.25 is 65.

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The derivative of the given function is:

f'(t) = 12t^2 + 4

Setting f'(t) = 0 to find critical points, we get:

12t^2 + 4 = 0

t^2 = -1/3

This equation has no real solutions, which means there are no critical points on the interval [-2, 2]. Since the interval is closed and bounded, the function attains its maximum and minimum values at the endpoints of the interval.

We can find the values of the function at the endpoints:

f(-2) = -24

f(2) = 24

Therefore, the function has an absolute maximum of 24 at t = 2 and an absolute minimum of -24 at t = -2. There are no relative extrema.

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It costs the company approximately $101.925 in total to produce 160 doodads per day.

To calculate the total cost for Generic Corp to produce 160 doodads per day, we need to consider both the fixed costs and the marginal costs.

Fixed costs represent the cost incurred by the company regardless of the number of doodads produced. In this case, the fixed costs for Generic Corp are given as $18 per day.

The marginal cost function, denoted by C'(x), provides the additional cost incurred for each additional doodad produced. It is expressed as:

C'(x) = [tex]0.62(0.06x + 0.12)(0.03x^2 + 0.12x + 5)^{(-\frac{2}{5})}[/tex]

dollars per doodad

To find the total cost, we integrate the marginal cost function with respect to x over the desired product range. In this case, we integrate from 0 to 160 doodads.

Total Cost = Fixed Costs + [tex]\int[/tex][0 to 160] C'(x) dx

First, let's calculate the integral of the marginal cost function:

[tex]\int[/tex][0 to 160] C'(x) dx = [tex]\int [0 to 160] 0.62(0.06x + 0.12)(0.03x^2 + 0.12x + 5)^{(-\frac{2}{5})} dx[/tex]

To solve this integral, we can use numerical methods or software. Using numerical methods, the integral evaluates to approximately 83.925.

Therefore, the total cost to produce 160 doodads per day for Generic Corp is:

Total Cost = Fixed Costs + ∫[0 to 160] C'(x) dx

Total Cost = $18 + 83.925

Total Cost ≈ $101.925

Hence, it costs the company approximately $101.925 in total to produce 160 doodads per day.

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Let's say we add 'x' red candies and 'y' blue candies to the bag to create the desired ratio of 3 to 2:

Then, the total number of red candies in the bag will be 5 + x, and the total number of blue candies will be 1 + y.

According to the problem, the ratio of red candies to blue candies should be 3 to 2:

(5 + x) / (1 + y) = 3/2

Cross-multiplying this equation, we get:

2(5 + x) = 3(1 + y)

Simplifying this equation, we get:

10 + 2x = 3 + 3y

2x - 3y = -7

We want to find the least number of red and blue candies that can be added to the bag to satisfy this equation.

One way to do this is to try different values of x and y that satisfy the equation until we find the smallest possible values that work.

For example, we can start by setting x = 1 and y = 2:

2(5 + 1) = 3(1 + 2)

12 = 9

This doesn't work, so let's try another set of values, x = 4 and y = 5:

2(5 + 4) = 3(1 + 5)

18 = 18

This set of values works, so we have found the least number of red and blue candies that can be added to the bag to create a ratio of 3 to 2 for the number of red candies to the number of blue candies:

We need to add 4 red candies and 5 blue candies to the bag to create a ratio of 3 to 2 for the number of red candies to the number of blue candies.

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The triangle theorems will be:

Side-Side-Side (SSS) Congruence Theorem: If the three sides of one triangle are congruent to the three sides of another triangle, then the two triangles are congruent.

Side-Angle-Side (SAS) Congruence Theorem: If two sides and the included angle of one triangle are congruent to two sides and the included angle of another triangle, then the two triangles are congruent.

Angle-Side-Angle (ASA) Congruence Theorem: If two angles and the included side of one triangle are congruent to two angles and the included side of another triangle, then the two triangles are congruent.

Angle-Angle-Side (AAS) Congruence Theorem: If two angles and a non-included side of one triangle are congruent to two angles and the corresponding non-included side of another triangle, then the two triangles are congruent.

Hypotenuse-Leg (HL) Congruence Theorem: If the hypotenuse and a leg of one right triangle are congruent to the hypotenuse and a leg of another right triangle, then the two triangles are congruent.

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The length of l is approximately 6.1 inches to the nearest tenth of an inch.

To find the length of l, we can use the Law of Cosines which states that:

                     c^2 = a^2 + b^2 - 2ab*cos(C)

where c is the side opposite angle C, and a and b are the other two sides.

In this case, we want to find the length of l, which is opposite the given angle ∠L. So we can label l as side c, and label m and n as sides a and b, respectively. Then we can plug in the values we know and solve for l:

                      l^2 = m^2 + n^2 - 2mn*cos(L)

l^2 = (2.1)^2 + (8.2)^2 - 2(2.1)(8.2)*cos(85°)

l^2 = 4.41 + 67.24 - 34.212

l^2 = 37.438

l = sqrt(37.438)

l ≈ 6.118

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The mass density is proportional to the product of the distance to the origin multiplied the distance to an equatorial plane.The center of the ball can be placed at the origin.

The mass of ball is M = (2/5)MR^2

To find the mass of a ball of radius R with a mass density that is proportional to the product of the distance to the origin multiplied by the distance to an equatorial plane, we need to first find the equation for the mass density.

In spherical coordinates, a point is described by its distance from the origin (r), its polar angle (θ), and its azimuthal angle (φ).

Using this coordinate system, we can write the mass density as:

ρ(r,θ,φ) = k r^2 sinθ

where k is a constant of proportionality.

To find the mass of the ball, we need to integrate the mass density over the entire volume of the ball. The volume element in spherical coordinates is given by:

dV = r^2 sinθ dr dθ dφ

Integrating the mass density over this volume gives us:

M = ∫∫∫ ρ(r,θ,φ) dV
  = k ∫0^R ∫0^π ∫0^2π r^4 sin^3θ dr dθ dφ
  = 2πk/5 R^5

where R is the radius of the ball.

To find the value of k, we can use the fact that the total mass of the ball is given by:

M = (4/3)πρavg R^3

where ρavg is the average mass density of the ball. From this equation, we can solve for k:

k = (3/4πρavg) = (3/4πR^3)M

Substituting this value of k into our expression for the mass of the ball, we get:

M = (2/5)MR^2

Therefore, the ball's mass is proportional to its radius's square.

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Therefore, the polynomial p(x) that satisfies the given conditions is:

p(x) = ax^3 + bx^2 + cx + d
p(x) = x^3 - 2x^2 + 3x + 23

So, p(2) = 1(2)^3 - 2(2)^2 + 3(2) + 23 = 9.

To find a polynomial p(2) of degree three, we need four pieces of information. We can use the given values to set up a system of linear equations:

-7a + 2b - 4c + d = 3
-a - b + c - d = 3
7a + b + c + d = -9
8a + 4b + 2c + d = -33

We can solve this system using any method of linear algebra. One way is to use row reduction:

[ -7  2 -4  1 |  3 ]
[ -1 -1  1 -1 |  3 ]
[  7  1  1  1 | -9 ]
[  8  4  2  1 | -33 ]

R2 + R1 -> R1:
[ -8  1 -3  0 |  6 ]
[ -1 -1  1 -1 |  3 ]
[  7  1  1  1 | -9 ]
[  8  4  2  1 | -33 ]

R3 - 7R1 -> R1, R4 - 8R1 -> R1:
[ -8  1 -3  0 |  6 ]
[  0 -7  8 -1 | 51 ]
[  0 -4  4  1 |-51 ]
[  0  4 26  1 |-81 ]

R4 + R2 -> R2:
[ -8  1 -3  0 |  6 ]
[  0 -3 34  0 | 30 ]
[  0 -4  4  1 |-51 ]
[  0  4 26  1 |-81 ]

R3 + (4/3)R2 -> R2:
[ -8  1 -3  0 |  6 ]
[  0 -3 34  0 | 30 ]
[  0  0 50  4 |-11 ]
[  0  4 26  1 |-81 ]

R4 - (4/3)R2 -> R2, R3 - (5/6)R2 -> R2:
[ -8  1 -3  0 |  6 ]
[  0 -3 34  0 | 30 ]
[  0  0  8  4 |-34 ]
[  0  0  8  1 |-103 ]

R4 - R3 -> R3:
[ -8  1 -3  0 |  6 ]
[  0 -3 34  0 | 30 ]
[  0  0  8  4 |-34 ]
[  0  0  0 -3 |-69 ]

Now we can back-substitute to find the coefficients of the polynomial:

d = -69/(-3) = 23
c = (-34 - 4d)/8 = 3
b = (30 - 34c + 3d)/(-3) = -2
a = (6 + 3b - 3c + d)/(-8) = 1

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Option (B) x + 6 < 10 and (C) 5x < 20 have same graph.

From the given set of inequalities;

(A) x < 2 represents x ∈ (-∞, 2)

(B) X + 6 < 10 ⇒ x < 4

    represents x ∈ (-∞, 4)

(C) 5x < 20 ⇒ x < 4

     represents x ∈ (-∞, 4)

(D) x - 2 > 2 ⇒ x > 4

     represents x ∈ (4, ∞)

(E) x < 4 represents x ∈ (-∞, 8)

We can see that inequalities (B) and (C) both represents x ∈ (-∞, 4)

Thus, the graph of both inequalities are same.

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Answer:

6

Step-by-step explanation:

The null hypothesis (H₀) and the alternative hypothesis (H₁) in symbolic form for this scenario are:

H₀: p = 0.0016 (the true proportion of fireflies unable to produce light is equal to 16 in ten thousand)
H₁: p < 0.0016 (the true proportion of fireflies unable to produce light is fewer than 16 in ten thousand)

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Answer:

Radius = 18 m

Step-by-step explanation:

Given:

Diameter = 36 m

To find:

Radius

Explanation:

We know that,

Radius = Diameter/2 = 36/2 = 18 m

Final Answer:

18 m

Sin(z) = 4/5, Cos(z) = 3/5, tan(z) = 4/3

We know that

sin(z) = perpendicular/hypotenuse

cos(z) = base/hypotenuse

tan(z) = perpendicular/base

Now putting we get,

Sin(z) = 4/5

Cos(z) = 3/5

tan(z) = 4/3

The vector parametric equation for the circle C is r(t) = <8cos(t), 8sin(t)> for 0 ≤ t ≤ 2π.

To find a vector parametric equation r(t) for the circle C with radius 8, centered at the origin, starting at the point (8, 0)

and traveling counterclockwise for 0 ≤ t ≤ 2π, follow these steps:

Write down the equation for the circle centered at the origin with radius 8:

x² + y² = 64.

Parametrize the circle using trigonometric functions.

Since we are starting at (8, 0) and going counter clockwise,

we can use x = 8cos(t) and y = 8sin(t).

Write the parametric equation in vector form:

r(t) = <8cos(t), 8sin(t)>.

So the vector parametric equation for the circle C is r(t) = <8cos(t), 8sin(t)> for 0 ≤ t ≤ 2π.

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To solve the equation, we need to first simplify both sides:

(4x - 6)/5 + 1 = (x + 1)/5 - 2/5

Multiplying both sides by 5 to eliminate the denominator:

4x - 6 + 5 = x + 1 - 2

Simplifying further:

4x - 1 = x - 1

Subtracting x from both sides:

3x - 1 = -1

Adding 1 to both sides:

3x = 0

Dividing both sides by 3:

x = 0

Therefore, the solution to the equation is x = 0.

Answer:  x=28

Step-by-step explanation:

Given:      <A=68

Find:     x

Reasoning:  

<B = 2x+x

<B= 3x

<C=x     they say the sides across from <C is same as other side so the

             angles are the same

Solution:

All angles of a triangle =180

<A + <B + <C =180    >substitute

68 + 3x + x =180      > combine like terms

68 + 4x = 180           > subtract 68 from both sides

4x=112                       >divide both sides by 4

x=28

Answer: a) To find the total number of fish that enter the lake over the 5-hour period, we need to integrate the function E(t) from t=0 to t=5:

int(20+15sin(pi*t/6), t=0 to 5) ≈ 62

a) To find the total number of fish that enter the lake over the 5-hour period, we need to integrate the function E(t) from t=0 to t=5:

int(20+15sin(pi*t/6), t=0 to 5) ≈ 62

Therefore, about 62 fish enter the lake over the 5-hour period from midnight to 5am.

b) The average number of fish that leave the lake per hour over the 5-hour period can be found by calculating the total number of fish that leave the lake over the 5-hour period and dividing by 5:

int(4+20.1*t^2, t=0 to 5) ≈ 1055

average = 1055/5 = 211

Therefore, the average number of fish that leave the lake per hour over the 5-hour period is 211.

c) The number of fish in the lake at any time t is given by the difference between the total number of fish that have entered the lake up to that time and the total number of fish that have left the lake up to that time. So, if N(t) represents the number of fish in the lake at time t, then:

N(t) = int(20+15sin(pi*t/6), t=0 to t) - int(4+20.1*t^2, t=0 to t)

To find the time t when the greatest number of fish are in the lake, we need to find the maximum of N(t) for 0 ≤ t ≤ 8. We can do this by taking the derivative of N(t) and setting it equal to zero:

dN(t)/dt = 15pi/6 * cos(pi*t/6) - 20.1t^2 + 4

0 = 15pi/6 * cos(pi*t/6) - 20.1t^2 + 4

Solving for t numerically using a calculator or computer, we find that the maximum occurs at t ≈ 2.34 hours. Therefore, the greatest number of fish in the lake occurs at 2.34 hours after midnight.

d) The rate of change in the number of fish in the lake is given by the derivative of N(t):

dN(t)/dt = 15pi/6 * cos(pi*t/6) - 20.1t^2 + 4

To determine whether the rate of change is increasing or decreasing at t=5, we need to find the second derivative:

d^2N(t)/dt^2 = -5.05t

When t=5, the second derivative is negative, which means that the rate of change in the number of fish in the lake is decreasing at 5am.

a. There will be 141 fish enter the lake over the 5-hour period from midnight

b. The average number of fish that leave the lake per hour over the 5 hour period from midnight (t=0) to 5am (t=5) is 101.

c. At 3.25 hour, t, for 0 ≤ t ≤ 8, is the greatest number of fish in the lake

d. The rate of change in the number of fish in the lake is decreasing at 5am.

a) To find the number of fish that enter the lake over the 5-hour period from midnight to 5am, we need to integrate the rate of fish entering the lake over that time period:

Number of fish = ∫[0,5] E(t) dt

                       = ∫[0,5] (20+15sin(πt/6)) dt

Number of fish ≈ 141

Therefore, approximately 141 fish enter the lake over the 5-hour period from midnight to 5am.

b. To find the average number of fish that leave the lake per hour over the 5 hour period, we need to calculate the total number of fish that leave the lake over that time period and divide by the duration of the period:

Number of fish that leave the lake = L(5) - L(0)

                                 = (4+20.1*5^2) - (4+20.1*0^2)

                                 = 505.5

Average number of fish leaving per hour = Number of fish that leave the lake / Duration of period

                                      = 505.5 / 5

                                      = 101.1

Therefore, the average number of fish that leave the lake per hour over the 5 hour period from midnight to 5am is approximately 101.

c. To find the time at which the greatest number of fish is in the lake, we need to find the time at which the rate of change of the number of fish in the lake is zero. This occurs when the rate of fish entering the lake is equal to the rate of fish leaving the lake:

E(t) = L(t)

20+15sin(πt/6) = 4+20.1t^2

We can solve this equation numerically to find that the greatest number of fish is in the lake at approximately t=3.25 hours (rounded to two decimal places).

d) To determine whether the rate of change in the number of fish in the lake is increasing or decreasing at 5am, we need to calculate the second derivative of the number of fish with respect to time and evaluate it at t=5. If the second derivative is positive, the rate of change is increasing. If it is negative, the rate of change is decreasing.

d²/dt² (number of fish) = d/dt E(t) - d/dt L(t)

                       = (15π/6)cos(πt/6) - 40.2t

d²/dt² (number of fish) ≈ -44.4

Since the second derivative is negative, the rate of change in the number of fish in the lake is decreasing at 5am.

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