factor completely

5y²-2y-3​

Answer:

\(147.6\:\mathrm\)

Step-by-step explanation:

The perimeter of a polygon is simply sum of the lengths of all sides of the polygon.

Therefore, we can find this polygon's perimeter by adding up the length of each of its sides:

\(9.8+9.8+11.7+9.8+21.5+21+11.7+9.8+31.3+11.2=\boxed{147.6\:\mathrm{m}}\)

Since there is a lot to keep track of, I'd recommend picking any side and moving clockwise/counter-clockwise until you get all sides.

Answer:

We know that the equation for the speed is:

Speed = Distance/time.

First, we know that he walks 2 miles in 15 minutes.

distance = 2miles

time = 15 minutes

Then his speed in that interval is:

Speed = (2 mi)/(15 min) = (2/15) miles per minute.

Now, at this same speed, he wants to walk 3 more miles. And we want to find the equation that represents how much time she needs to walk 5 miles (the 2 first miles plus the other 3 miles)

We use again the equation:

Speed = Distance/Time

But we isolate Time, to get:

Time = Distance/Speed

Where:

Distance = 5 miles

Speed = (2/15) miles per min

Time = (5 miles)/((2/15) miles per min) = 37.5 minutes

She needs 37.5 minutes to walk the 5 miles.

Answer:

1/4

Step-by-step explanation:

Dividing a number by something is the same thing as multiplying it by the reciprocal.  To find the reciprocal, you flip the number.

In this case, you are dividing 83 by 4.  The reciprocal of 4 is 1/4.  So the missing number is 1/4.

-20.75

Step-by-step explanation:

Answer:

B. The population is declining at a rate of 11.5%

Step-by-step explanation:

if your graph the equation you will clearly see that it is declining and if you don't know how to graph the equation then go to the demos calculator, its an app & website, super easy to use. most people would assume the answer would be one of the 88.5% just because it's in the original equation but the .885 in the equation doesn't represent the rate of change there for the answer is 11.5%.

Answer:

d

Step-by-step explanation:

The rule that was used to translate the image is (x, y) → (x – 6, y + 11)

Translation is a way of moving an object from one location to another on the xy- plane

Given the coordinate of the right triangle LMN with vertices L(7, –3), M(7, –8), and N(10, –8)

If the triangle is translated on the coordinate plane so the coordinates of L’ are (–1, 8), the translation used will be:

L'(1, 8) = (x, y) - L(7, -3)

(X, Y) =  L'(1, 8) +  L(7, -3)

(X, Y) = (1+7, 8+3))

(X, Y) = (-6, 11)

Hence the rule that was used to translate the image is (x, y) → (x – 6, y + 11)

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The average rate of speed of the wind is 18 mph and the average rate of speed of the plane is 36 mph.

To find the average rate of speed of the wind and the plane, we can set up a system of equations.

Let x be the average airspeed of the plane and y be the average wind speed.

From the initial trip, we have the equation: (x + y) * (1/3) = 18.
This is because the total distance traveled is the sum of the plane's speed and the wind's speed, multiplied by the time taken.

From the return trip, we have the equation: (x - y) * (3/5) = 18.
This is because the total distance traveled is the difference between the plane's speed and the wind's speed, multiplied by the time taken.

Now, we can solve these two equations to find the values of x and y.

Simplifying the equations, we get:
1/3 * (x + y) = 18
3/5 * (x - y) = 18

Cross-multiplying and simplifying further, we get:
x + y = 54
3x - 3y = 90

Next, we can solve this system of equations using any method (substitution, elimination, etc.).

Adding the two equations, we get:
4x = 144
x = 36

Substituting the value of x into one of the equations, we get:
36 + y = 54
y = 18

Therefore, the average rate of speed of the wind is 18 mph and the average rate of speed of the plane is 36 mph.

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Dalia had an average airspeed of

42

miles per hour.

The average wind speed was

12

miles per hour.

Answer: A

Step-by-step explanation:

In the diagram, we see that we have two pairs of congruent angles, so it is going to be either ASA or AAS congruence.

To determine if it is ASA or AAS congruence, we can use the fact that for ASA, the side is included between the congruent angles, whereas in AAS, the side isn't.

Since the congruent side is included between the two angles, there is ASA congruence.

Answer:

A) 80°

Step-by-step explanation:

55° + 45° + (5x + 30)° = 180°

55° + 45° + 30° + 5x = 180°

130 + 5x = 180°

5x = 50°

x = 10°

=> (5x + 30)° = 5*10° + 30° = 50° + 30° = 80

Answer:

The answer is (A)

Step-by-step explanation:

:D

Answer:

Solve for in the equation 3 (x- 0.27)= 1.35

Step-by-step explanation:

First you look at a diagram and notice that you are moving forward so you are multiplying the 3 pack of tubes by the 0.27 ounce of glue  to get your answer of 1.35

The maximum value of the objective function Z = 40x₁ + 5x₂, subject to the constraints x₁ + x₂ ≤ 5 and 6x₁ + x₂ ≤ 20, where x₁, x₂ are integers between 0 and 25, can be found using the branch and bound method.

To solve this problem, we start with an initial solution by considering the constraints and finding the feasible region. Then, we compute the objective function for this initial solution. Next, we branch the problem into subproblems by considering different combinations of x₁ and x₂. We calculate the lower bounds for each subproblem to determine the potential for improvement. We continue branching and bounding until we find the optimal solution with the maximum value of Z.

The branch and bound method systematically explores the solution space, optimizing the objective function by considering different feasible regions. It provides a structured approach to find the optimal solution for linear programming problems.

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Answer:

b

Step-by-step explanation:

=0-(-x)

=0-×-x

=0+x

=x

The relation R is not reflexive, but it is symmetric and transitive.

A homogeneous relation R over the set A, which comprises the elements x, y, and z, is known as a transitive relation. If R relates x to y and y to z, then R likewise relates x to z.

For x, y ∈ Z, xRy if and only if (x+y)² ≡ ±1.

(a) Reflexivity: For x ∈ Z, we have (x + x)² = 4x² ≡ 0 (mod 1), which is not equal to ±1. Therefore, xRx does not hold for any x ∈ Z, and R is not reflexive.

(b) Symmetry: For x, y ∈ Z, if xRy, then (x + y)² ≡ ±1. This implies that (y + x)^2 ≡ (x + y)² ≡ ±1. Therefore, yRx also holds, and R is symmetric.

(c) Transitivity: For x, y, z ∈ Z, if xRy and yRz, then (x + y)² ≡ ±1 and (y + z)² ≡ ±1. Expanding these expressions, we get:

(x + y)² ≡ ±1  =>  x² + 2xy + y² ≡ ±1

(y + z)² ≡ ±1  =>  y² + 2yz + z² ≡ ±1

Adding these two equations, we get:

x² + 2xy + y² + y² + 2yz + z² ≡ ±2

Simplifying, we get:

x² + 2xy + 2yz + z² ≡ ±2 - 2y²

Now, we need to show that (x + z)² ≡ ±1. Expanding (x + z)², we get:

(x + z)² = x² + 2xz + z²

Substituting x² + 2xy + 2yz + z² ≡ ±2 - 2y², we get:

(x + z)² ≡ 2 - 2y² + 2xz

To complete the proof, we need to show that there exists some integer k such that 2 - 2y² + 2xz - k² ≡ ±1. We can rewrite this expression as:

2xz - k² ≡ 2y² - 3 (mod 4)

Since the left-hand side is even, the right-hand side must also be even. Therefore, y² ≡ 1 (mod 4), which implies that y is odd.

Now, we can substitute y = 2m + 1 for some integer m, and simplify:

2xz - k² ≡ 8m² + 8m - 1 (mod 4)

We can rewrite the right-hand side as 4(2m² + 2m) - 1, which is congruent to -1 (mod 4). Therefore, there exists some integer k such that 2xz - k² ≡ ±1, which implies that (x + z)² ≡ ±1. Hence, xRz holds, and R is transitive.

In summary, the relation R is not reflexive, but it is symmetric and transitive.

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Question: In a sale, the price of a book is reduced by 25%. The price of the book in the sale is £12. Work out the original price of the book

Answer: £16

Step-by-step explanation:

To determine the original price of the book, we can use the fact that the sale price is 75% (100% - 25%) of the original price. Let's denote the original price as x.

75% of x = £12

To solve for x, we can set up the equation:

0.75x = £12

To isolate x, we divide both sides of the equation by 0.75:

x = £12 / 0.75

x = £16

Therefore, the original price of the book was £16.

1) Tash is multiplying each number of the bracket while Taylor solves the bracket first and then applied multiplication.

2) The solution to the expression 7(x + 2) is 7x + 14.

According to the distributive property the same expression can be written in different ways by using the parenthesis or common multiplications.

Expression in maths is defined as the relation of numbers variables and functions by using mathematical signs like addition, subtraction, multiplication and division.

Part(1),

Taylor's solution is,

5(9 + 2)

5(11)

Tash's solution is,

5(9 + 2)

5(9) + 5(2)

Tash is multiplying each number of the bracket while Taylor solves the bracket first and then applied multiplication.

Part(2),

The solution to the expression 7(x + 2 is,

E = 7(x + 2

E = 7x + 14

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Answer:

Step-by-step explanation: 15

Mean means average, add all numbers and divide by amount

Mean= (15+12+14+19+x)/5

        =(60+x)/5

Median means to list numbers in order and pick the middle one

12 14 15 19       x is somewhere in the list in middle

Mode means the number that occurs the most, so 1 of these numbers is repeated

It must be 14 or 15 because the median must be the same as the mean

let's try both

(60+14)/5 =14.8                  (60+15)/15=15

The answer is 15.

The median of the data is M = 15 and the value of x = 15

The median is the value that's exactly in the middle of a data set when it is ordered. It's a measure of central tendency that separates the lowest 50% from the highest 50% of values. The steps for finding the median differ depending on whether you have an odd or an even number of data points

Arrange the data points from smallest to largest.

If the number of data points is odd, the median is the middle data point in the list.

If the number of data points is even, the median is the average of the two middle data points in the list.

Given data ,

The mean is calculated by summing all the numbers and dividing by the total count of numbers. In this case, we have 5 numbers, including x.

Mean = (15 + 12 + 14 + 19 + x) / 5

Since the mean is also equal to the median, we can rearrange the numbers in ascending order:

12, 14, 15, 19, x

The median is the middle number in a sorted list of numbers. In this case, the median is 15.

Since the mode is also equal to the mean and median, and we can see that 15 is the only number that appears more than once in the list, we can conclude that 15 is also the mode.

Setting the mean equal to 15 and solving for x:

15 = (15 + 12 + 14 + 19 + x) / 5

Multiplying both sides by 5 to eliminate the fraction:

75 = 15 + 12 + 14 + 19 + x

Subtracting the known numbers from both sides:

75 - (15 + 12 + 14 + 19) = x

15 = x

Hence , the value of x is 15 and median is 15

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No, it is not possible for both Ray and Kelsey to be correct. A third-degree polynomial function can have at most 3 zeros.

This is because the degree of a polynomial function represents the maximum number of times the function can change direction. Since a third-degree polynomial can change direction at most 3 times, it can have at most 3 zeros. As for the function g(x), there is no information provided on the list of possible functions for me to pick one. Please provide more information so I can assist you better.

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Answer: 18.92

Step-by-step explanation: Multiply  

22 by  0.86

Answer:

18.92

Step-by-step explanation:

Multiply 0.86 times 22

\(\sqrt[3]{2}+\sqrt[3]{4}\) is the root of the cubic polynomial  \(x^3-6 x-6\) .

A cubic function in mathematics is a function with the formula: f(x)=ax3+bx2+cx+d where the variable x has real values, the coefficients a, b, c, and d are complex integers, and a, a, and 0 are equal to zero. In other words, it is a real function as well as a polynomial function of degree three. The set of real numbers is specifically the domain and the codomain. All odd-degree polynomials have at least one real root, while cubic functions have one to three real roots (which may not be distinct).

A cubic function's graph always has one inflection point. It might have a local minimum and local maximum as its two essential points. A cubic function is monotonic if it is not.

Let  \(r=\sqrt[3]{2}+\sqrt[3]{4}\)

\(\text { Then, since }(a+b)^{\wedge} 3=a^{\wedge} 3+3 a^{\wedge} 2 b+3 a b^{\wedge} 2+b^{\wedge} 3=a^{\wedge} 3+3 a b *(a+b)+b^{\wedge} 3\)

\(\begin{aligned}\mathbf{r}^{\wedge} 3 &=\left(\frac{3}{2} \sqrt{2}+3 \cdot \sqrt[3]{2} \cdot \sqrt[3]{4} \cdot(\sqrt[3]{2}+\sqrt[3]{4})+(\sqrt[3]{4}) 3=\right.\\&=2+3 \cdot \sqrt[3]{8} \cdot(\sqrt{2}+\sqrt{4})+4=2+3 * 2 * r+4=6+6 r .\end{aligned}\)

\(\text { It means that } r=\sqrt[3]{2}+\sqrt[3]{4} \text { is the root to this equation }\)

\(r^3-6 r-6=0 .\)

Therefore, \(\sqrt[3]{2}+\sqrt[3]{4}\) is the root of the cubic polynomial  \(x^3-6 x-6\)

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Answer:

162.5

Step-by-step explanation:

250%=2.5

65×2.5=162.5

Answer:

162.5

Step-by-step explanation:

250% is the same as 2.5

2.5 * 65 = 162.5

I hope this helped and please mark me as brainliest!

The equation is formed and solved below.

What is an equation?

A weighing scale, balance, or seesaw is equivalent to an equation. Each equation relates to a single side of the balance. Different quantities can be placed on either side: if the weights on both sides are equal, the scale balances, and the equality that displays the balance is balanced as well (if not, then the lack of balance corresponds to an inequality represented by an inequation). If two equations or two systems of equations have the same set of solutions, they are equivalent. The operations listed below convert an equation or a system of equations into an equivalent one, assuming that the operations are meaningful for the expressions to which they are applied.

The equation is 6(x)² + 5x = 4, which is solved below as

6x² - 5x - 4 = 0

or, 6x² - 8x + 3x - 4 = 0

or, 2x(3x-4) + 1(3x-4) = 0

or, (2x+1)(3x-4) = 0

either 2x + 1 = 0 => x = -1/2

or 3x - 4 = 0 => x = 4/3

The values could be -1/2 or 4/3

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Answer:

82

Step-by-step explanation:

10z^2-2z-2

10x3^2-2x3-2

90-6-2

90-8

82

Answer:

82

Step-by-step explanation:

Evaluate 10z² - 2z - 2 where z = 3:

\( \longrightarrow \) 10z² - 2z - 2 = 10 × 3² - 2 × 3 - 2

3² = 9:

\( \longrightarrow \) 10 × 9 - 2 × 3 - 2

10 × 9 = 90:

\( \longrightarrow \) 90 - 2 × 3 - 2

-2 × 3 = -6:

\( \longrightarrow \) 90 - 6 - 2

90 - 6 - 2 = 90 - (6 + 2):

\( \longrightarrow \) 90 - (6 + 2)

6 + 2 = 8:

\( \longrightarrow \) 90 - 8

\( \longrightarrow \) 82

Answer: C

The radius of the inscribed circle of AHJK is equal to the inradius of the triangle, which is denoted as r.

Since P is the incenter of the triangle, the inradius is equal to the distance from P to each of the sides of the triangle. Let's call these distances a, b, and c.

You know that LP = 4x + 10 and MP = 8x - 2. The length of the inradius can be expressed as:

r = (a+b-c)/2

You can now use this equation and the given information to solve for r.

Since LP is the distance from P to side L, we can set a = 4x + 10.

Since MP is the distance from P to side M, we can set b = 8x - 2.

Substituting these values into the equation for r, you get:

r = ((4x + 10) + (8x - 2) - c) / 2

= (12x + 8 - c) / 2

You can simplify this to:

r = 6x + 4 - c/2

Thus, the radius of the inscribed circle of AHJK is equal to 6x + 4 - c/2. To find the exact value of the radius, you need to know the value of c.

With a running time of 4.5 hours and speeds of 24 mph and 30 mph, respectively, the bus covered a one-way distance of 72 miles.

With a running time of 4.5 hours and speeds of 24 mph outgoing and 30 mph inbound, the bus covered a one-way distance of 72 miles.

As the outgoing and inbound timings were equal, the computation was performed by multiplying the outbound trip's speed by the entire running time, and then dividing the result by two. Therefore, 108 miles are equal to 24 mph times 4.5 hours. The one-way distance is equal to 72 miles when you divide this result in half. When both directions' speeds are known, this calculation can be used to determine the overall distance of a round trip.

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Answer:

one is the value of x this is correct answer

This is a fundamental result in set theory and is useful in understanding the behavior of infinite sets.

To prove that a set S is infinite if and only if there exists a proper subset A of S with a one-to-one correspondence between A and S, we need to show both directions of the statement.

First, suppose that S is infinite. Then, we can construct a proper subset A of S by removing any element from S. Since S is infinite, there are infinitely many elements left in S after removing one. Therefore, A is a proper subset of S. Now, to show that there is a one-to-one correspondence between A and S, we can define a function f: A → S, where f(x) = x for all x in A. This function is clearly one-to-one and onto, since every element in S is in A or is the removed element. Hence, there exists a one-to-one correspondence between A and S.

Conversely, suppose that there exists a proper subset A of S with a one-to-one correspondence between A and S. This means that we can list the elements of S in a sequence, with A being a subset of that sequence. Since A is proper, there must be at least one element in S that is not in A. We can then append this element to the sequence to obtain a longer sequence of elements in S. We can repeat this process infinitely many times, since there are infinitely many elements in S. Therefore, S is infinite.

In conclusion, we have shown that a set S is infinite if and only if there exists a proper subset A of S with a one-to-one correspondence between A and S. This is a useful result in understanding the properties of infinite sets.

A set S is infinite if and only if there exists a proper subset A of S with a one-to-one correspondence between A and S. This means that we can construct a one-to-one function from A to S and vice versa. The proof for this statement involves showing that if S is infinite, we can construct a proper subset A of S with a one-to-one correspondence between A and S, and conversely, if such a subset exists, we can use it to show that S is infinite. This result is important in understanding the properties of infinite sets.

Therefore, we have proved that a set S is infinite if and only if there exists a proper subset A of S with a one-to-one correspondence between A and S. This is a fundamental result in set theory and is useful in understanding the behavior of infinite sets.

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Answer + Step-by-step explanation:

Two algebraic expressions are equivalent if they always lead to the same result when you evaluate them, no matter what values you substitute for the variables. To check whether a more complex expression is equivalent to a simpler expression, you can distribute any coefficients and combine any like terms on each side of the equation. Then arrange the terms in the same order, usually x-term before constants. If all of the terms in the two expressions are identical, then the two expressions are equivalent

In this case, we can see that when x-3, both expressions have a value of 30; when x-5, both expressions have a value of 42; when x-1, both expressions have a value of 18; when x-8, both expressions have a value of 60; and when x-2, both expressions have a value of 15; and when x-6, both expressions have a value of 39 https://the-equivalent.com/expressions-are-equivalent/. Therefore, we can conclude that these two expressions are equivalent

Both algorithms provide worst-case O(n) time complexity, making them efficient for finding Pareto optimal points in a set of n points.

To find a single Pareto optimal point in a set S of n points, we can use the following algorithm with a worst-case O(n) time complexity:

1. Initialize a variable (xi, yi) as the first point in S.

2. For each point (xj, yj) in S, starting from the second point:

  - If xj > xi and yj > yi, update (xi, yi) to be (xj, yj).

  - If xj <= xi or yj <= yi, continue to the next point.

3. Return the final (xi, yi) as the Pareto optimal point.

The algorithm works by iteratively comparing each point with the current Pareto optimal point. If a point has both a higher x-coordinate and a higher y-coordinate than the current Pareto optimal point, it becomes the new Pareto optimal point. Otherwise, it is skipped.

The time complexity of this algorithm is O(n) because we iterate through the set S once, comparing each point with the current Pareto optimal point. Since each comparison takes constant time, the overall time complexity is linear in the number of points.

If the points in S are already sorted by their x-coordinates and each point has a unique x-coordinate, we can modify the algorithm to find all Pareto optimal points in O(n) time as well. Here's the modified algorithm:

1. Initialize an empty result list.

2. Initialize (xi, yi) as the first point in S.

3. Add (xi, yi) to the result list.

4. For each point (xj, yj) in S, starting from the second point:

  - If yj > yi, update (xi, yi) to be (xj, yj) and add it to the result list.

  - If yj <= yi, continue to the next point.

5. Return the result list containing all Pareto optimal points.

In this modified algorithm, we only consider the y-coordinate comparison since the points are sorted by their x-coordinates.

Whenever we find a point with a higher y-coordinate, we update the current point and add it to the result list. The time complexity remains O(n) as we still iterate through the set S once.

Both algorithms provide worst-case O(n) time complexity, making them efficient for finding Pareto optimal points in a set of n points.

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