what is -4 7/8 × 2/3 =​

The population that is most dense is the one for  5000 individuals per 200 square meters

Given the following information:

100 individuals per 100m^2

5000 individuals per200m^2

100,000 individuals per 20,000m^2

15,000 individuals per 10,000m^2

We find the population that is most dense using the formula:

Population density=(No. of people/Land area)

For 100 individuals per 100m^2,

Population density=100/100=1

For 5000 individuals per200m^2

Population density=5000/200=25

For 100,000 individuals per 20,000m^2

Population density=100,000/20,000=5

For 15,000 individuals per 10,000m^2

Population density=15,000/10,000=1.5

Hence, highest population density is 25 for 5000 individuals per200m^2  

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Answer:

3x²

im lazy to solve right now

hope it helps

Answer:

Step-by-step explanation:

half of the circle is 180 degrees

so x + 3x = 180

4x = 180

x = 45

Answer:

The width is 7 ft

The length is 17 ft

Step-by-step explanation:

width = w

length = w+10

A = l*w

119 = w*(w+10)

Distribute

119 = w^2 + 10 w

Subtract 119 from each side

0 = w^2 +10w-119

Factor

0 = ( w-7) (w+17)

Using the zero product property

0 = w-7     0 = w+17

w = 7       w = -17

Since the width cannot be negative

w =7

l = w+10

l = 17

The width is 7 ft

The length is 17 ft

Answer:

Quadratic equation:   x² + 10x - 119 = 0

  Length:   17 feet

    Width:     7 feet

Step-by-step explanation:

length:  x + 10

width:  x

Area = Length * Width

119 = (x + 10) * x

119 = x² + 10x

x² + 10x - 119 = 0

x² + 10x - 119 = (x + 17)(x - 7) = 0

x + 17 = 0 and x - 7 = 0

      x = -17  and  x = 7

Positive length and width makes more sense...

  length:  -17 + 10 = -7

   width:  -17

  length:  7 + 10 = 17

    width:  7

Answer:

O(-4,-2) is the answer

Step-by-step explanation:

because it lies between a horizontal line

Answer:

270 cm²

Step-by-step explanation:

The question is not complete, but a similar question is attached.

Solution:

A prism is a three dimensional polygon with two identical and opposite faces. These bases are called the base of the polygon.

A triangular prism is a prism in which its bases are triangular in shape and it also has three rectangular sides.

The total surface area (TSA) of a triangular prism is given by:

TSA = area of triangle 1 + area of triangle 2 + area of rectangle 1 + area of rectangle 2 + area of rectangle 3

Area of triangle 1 = (1/2) * base * height = 1/2 * 6 * 5 = 15 cm²

Area of triangle 2 = (1/2) * base * height = 1/2 * 6 * 5 = 15 cm²

Area of rectangle 1 = length * breadth =15 * 6 = 90 cm²

Area of rectangle 2 = length * breadth =15 * 5 = 75 cm²

Area of rectangle 3 = length * breadth =15 * 5 = 75 cm²

TSA = 15 cm² + 15 cm² + 90 cm² + 75 cm² + 75 cm² = 270 cm²

Answer:

I don't understand the question but I think the answer is either 10 or 30

Answer:

5

Step-by-step explanation:

Answer:

When you plot the graph, you may see a line or a curve. This presents the trend of the relationship of the two quantities when the independent variable changes.

Graphs visually depict patterns and trends, while equations provide precise mathematical descriptions, both revealing valuable information about the relationship between two quantities.

Graphs visually represent data and patterns, allowing us to identify the nature of the relationship between two quantities. The slope and intercepts of a graph provide information about the rate of change and the values when one quantity is zero.

Equations, on the other hand, offer precise mathematical descriptions of the relationship. The form of the equation reveals the nature of the relationship, while coefficients and constants provide specific information. Solving equations yields solutions that satisfy the relationship.

Both graphs and equations work together to provide a comprehensive understanding of the relationship between two quantities, with graphs offering visual insights and equations providing precise mathematical information.

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The length of a truck is an example of a continuous data set, and 20 feet is just one possible value within that range.

The given value, "a truck with a length of 20 feet," is an example of a continuous data set.

Continuous data refers to values that can take on any numerical value within a range, with no gaps or interruptions. In this case, the length of a truck can take on any value between 0 and some maximum length, with no gaps or interruptions in between.

In contrast, discrete data refers to values that can only take on certain specific values, usually integers. For example, the number of tires on a truck is a discrete data set because it can only take on integer values (e.g. 4, 6, 8).

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There are 96 ways for four people to sit in six chairs so that the leftmost chair is occupied and the rightmost is empty.

There are 4 guests and the leftmost seat is to be occupied, so there are 4  such ways to choose the occupant of that seat.

Then, for each of those 4 ways, there are 4 ways to decide the occupant for the second seat -- the remaining three people plus the possibility of that seat being empty, that makes a total of 16 ways to decide the configuration of the first two seats.

Then, for each of those 16 ways, there are 3 ways to choose the occupant for the third seat (either 3 remaining guests if the second seat was left empty, or the two remaining guests plus the possibility of leaving the 3rd seat empty). Hence, 16 times 3 is 48. For each of those ways, there are two choices for the next seat, i.e. 48 times 2 = 96, and finally one choice for the fifth seat. So in all, there are 96 ways.

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Answer:

(A) -2

Step-by-step explanation:

Answer:

3. < ACB=<DCE

Step-by-step explanation:

3. < ACB=<DCE, because both are vertical angles which are equal

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To prove that A* search always finds the optimal goal, we need to show that it satisfies two properties 1. Completeness: A* search is guaranteed to find a solution if one exists. 2. Optimality: If a solution is found by A* search, it is guaranteed to be the optimal solution.

1. Completeness:

  To prove completeness, we need to show that A* search is guaranteed to find a solution if one exists.

  A* search explores the search space by expanding nodes based on the estimated cost of reaching the goal, which is determined by the heuristic function. The heuristic function used in A* search is admissible, meaning it never overestimates the actual cost to reach the goal.

  A* search maintains a priority queue of nodes to be expanded, and it always selects the node with the lowest estimated cost (f-value) to expand next. Since the heuristic is admissible, the f-value of the goal node will never decrease as we explore the search space.

  If a solution exists, A* search will eventually reach the goal node because it explores nodes in order of increasing estimated cost. Once the goal node is reached, A* search will terminate and return the solution. Therefore, A* search is complete.

2. Optimality:

  To prove optimality, we need to show that if a solution is found by A* search, it is guaranteed to be the optimal solution.

  Suppose there exists an optimal solution that is different from the one found by A* search. Let's assume this alternative solution has a lower cost than the one found by A* search.

  Since the heuristic function used in A* search is admissible, it never overestimates the actual cost to reach the goal. This implies that the estimated cost (h-value) of any node in the search space is less than or equal to the actual cost (g-value) of reaching the goal from that node.

  Now, consider the node in the alternative solution where it deviates from the path found by A* search. This node must have a lower estimated cost (h-value) than the corresponding node in the A* search path because the alternative solution has a lower overall cost.

  However, since A* search always selects the node with the lowest estimated cost (f-value) to expand next, it would have chosen the node in the alternative solution before the corresponding node in the A* search path. This contradicts our assumption that the alternative solution has a lower cost.

  Therefore, we can conclude that if a solution is found by A* search, it is guaranteed to be the optimal solution.

By establishing both the completeness and optimality properties of A* search, we have shown that A* search always finds the optimal goal.

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Answer:

JNM=42

Step-by-step explanation:

JNM is equal to LNK so you can set the equations equal to each other.

7x-21=4x+6, subtract 4x from both sides

3x-21=6, add 21 to both sides

3x=27, divide by 3

x=9, then you plug it into an equation

4 times 9 + 6

36+6

42

<JNM = \(42\) degrees.

The angle made by two ground lines is measured horizontally , and is called a horizontal angle.

According to the question

Line J L and line M K slightly inclined with horizontal intersects at N.

<JNM = <LNK

\(4x + 6 = 7x - 21\)

Subtract 4x from both sides

\(6 = 3x -21\)

\(3x=27\)

Divide by 3

\(x=\frac{27}{3}\)

\(x =9\)

<JNM = \(4x+6\)

<JNM = \(4(9)+6\)

<JNM = \(42\) degrees

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The matrix is now in row echelon form. The inverse is the right side of the augmented matrix: \($$\begin{bmatrix}1 & 1 & 0 \\2 & 1 & 3 \\0 & 1 & 2\end{bmatrix}$$\)

We will use the algorithm for finding the inverse of a matrix by row reducing.

Step 1: Augment the matrix with the identity matrix.

\($$\begin{bmatrix}1 & 1 & 0 & | & 1 & 0 & 0\\2 & 1 & 3 & | & 0 & 1 & 0\\0 & 1 & 2 & | & 0 & 0 & 1\end{bmatrix}$$\)

Step 2: Use row operations to transform the matrix into the identity matrix.

1. Multiply the first row by \($\frac{1}{2}$\) to get:

\($$\begin{bmatrix}\frac{1}{2} & \frac{1}{2} & 0 & | & \frac{1}{2} & 0 & 0\\2 & 1 & 3 & | & 0 & 1 & 0\\0 & 1 & 2 & | & 0 & 0 & 1\end{bmatrix}$$\)

2. Subtract two times the first row from the second row to get:

\($$\begin{bmatrix}\frac{1}{2} & \frac{1}{2} & 0 & | & \frac{1}{2} & 0 & 0\\0 & 0 & 3 & | & -1 & 1 & 0\\0 & 1 & 2 & | & 0 & 0 & 1\end{bmatrix}$$\)

3. Subtract the first row from the third row to get:

\($$\begin{bmatrix}\frac{1}{2} & \frac{1}{2} & 0 & | & \frac{1}{2} & 0 & 0\\0 & 0 & 3 & | & -1 & 1 & 0\\0 & 0 & 2 & | & -\frac{1}{2} & 0 & 1\end{bmatrix}$$\)

4. Multiply the third row by \($\frac{1}{2}$\) to get:

\($$\begin{bmatrix}\frac{1}{2} & \frac{1}{2} & 0 & | & \frac{1}{2} & 0 & 0\\0 & 0 & 3 & | & -1 & 1 & 0\\0 & 0 & 1 & | & -\frac{1}{4} & 0 & \frac{1}{2}\end{bmatrix}$$\)

5. Subtract \($\frac{3}{2}$\)times the second row from the third row to get:

\($$\begin{bmatrix}\frac{1}{2} & \frac{1}{2} & 0 & | & \frac{1}{2} & 0 & 0\\0 & 0 & 3 & | & -1 & 1 & 0\\0 & 0 & 0 & | & \frac{7}{4} & -1 & \frac{1}{2}\end{bmatrix}$$\)

Step 3: The matrix is now in row echelon form. The inverse is the right side of the augmented matrix:

\($$\begin{bmatrix}1 & 0 & 0\\-1 & 1 & 0\\\frac{7}{4} & -1 & \frac{1}{2}\end{bmatrix}$$\)

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