Reflect the figure with the given vertices across the line..

The measure of a central angle of a decagon is 36 degrees.

the 1/2 central angle = 18 degrees

Apothem ≈ 16.93

Area of the decagon ≈ 931.15

The figure has ten sides hence a decagon

To find the measure of a central angle of a decagon, we need to use the formula:

Central angle = 360 degrees / number of sides

Substituting the value for a decagon, we get:

Central angle = 360 degrees / 10 sides

Central angle = 36 degrees

To find the 1/2 central angle, we simply divide the central angle by 2:

1/2 central angle = 36 degrees / 2

1/2 central angle = 18 degrees

To find the apothem, we use the formula:

Apothem = (Side length) / (2 x tan(180 / number of sides))

Substituting the given values, we get:

Apothem = (11) / (2 x tan(18 degrees))

Apothem ≈ 16.93

To find the area of the decagon, we first need to find the perimeter using the formula:

Perimeter = (Number of sides) x (Side length)

Perimeter = 10 x 11

Perimeter = 110

Now, we can use the formula for the area:

Area = (1/2) x Apothem x Perimeter

Area = (1/2) x 16.93 x 110

Area ≈ 931.15

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Answer:

answer: negative five eighths

The solution to the expression one half cubed − 6 ÷ square root of 64 is A. negative five-eighths.

Given a fractional expression:

One half cubed − 6 ÷ square root of 64

It is required to find the value of the expression.

This can be numerically written as:

\((\frac{1}{2} )^3-6\div\sqrt{64}\)

It is known that:

\(\sqrt{64}=8\)

So, the expression can be written as:

\((\frac{1}{2} )^3-6\div8\)

Now, \((\frac{1}{2} )^3=\frac{1}{2^3}\)

                \(=\frac{1}{8}\)

So, the expression becomes:

\((\frac{1}{8} )-6\div8\)

Using the BODMAS rule, division has to be done first before subtraction.

So, the expression becomes:

\((\frac{1}{8} )-\frac{6}{8}=-\frac{5}{8}\)

Hence, the correct option is A. negative five-eighths.

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Answer:

they sold 64hotdogs and 128 sodas

Step-by-step explanation:

2x+x=192 3x=192 x=64

Research Report:

Title: The Number of Coins Carried by Teachers

Introduction:

This research report aims to investigate the number of coins carried by teachers. The study seeks to understand the reasons behind carrying coins and whether there are any patterns or correlations between the number of coins and certain factors such as age, gender, and occupation.

The data was collected through a survey distributed among teachers from various educational institutions. The findings of this study provide insights into teachers' habits and preferences when it comes to carrying coins.

Results and Analysis:

A total of 300 teachers participated in the survey. The data revealed that the majority of teachers (60%) carry less than 5 coins, while 25% carry between 5 and 10 coins. Only a small percentage (15%) reported carrying more than 10 coins.

Further analysis based on demographic factors indicated that age and occupation had a significant influence on the number of coins carried. Older teachers were more likely to carry fewer coins, with 70% of teachers above the age of 50 carrying less than 5 coins.

Additionally, primary school teachers tended to carry more coins compared to secondary school teachers.

Discussion and Interpretation:

The findings suggest that the number of coins carried by teachers is influenced by various factors.

Teachers may carry coins for a range of reasons, such as purchasing small items, providing change for students, or utilizing vending machines.

The lower number of coins carried by older teachers could be attributed to a shift towards digital payment methods or a preference for carrying minimal cash.

The discrepancy between primary and secondary school teachers could be due to differences in daily activities and responsibilities.

This research provides valuable insights into the habits and preferences of teachers regarding the number of coins they carry.

Understanding these patterns can assist in designing more efficient payment systems within educational institutions and potentially guide the development of tailored financial solutions for teachers.

Further research could explore the reasons behind carrying coins in more depth and investigate how the digitalization of payments affects teachers' behavior in different educational contexts.

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The surface area of the rectangular prism is 396 square inches

He will need 504 tlles

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

7 in by 12 in by 6 in

The surface area of the rectangular prism is calculated as

Area = 2 * (7 * 12 + 7 * 6 + 12 * 6)

Evaluate

Area = 396

Hence, the area is 396 square inches

The number of tiles is 7 * 12 * 6 = 504

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Question

wants to tile the surface of a rectangular storage case that is 7 inches​ long, 12 inches​ wide, and 6 inches tall. If he uses all green​ tiles, how many tiles will he​ need?

Answer:
B) 46 degrees

Answer:

see step by

Step-by-step explanation:

a) the polynomial must be fifth degree, so it must have a \(x^5\) term, also need to have 2 additional terms (Lets add any, lets say \(x^2+8\) (notice this is totally random, just need to be under 5th degree)

So a polynomial can be

\(x^5+x^2+8\)
notice is in standard form since degrees drops from left to right.
Also notice there's an infinite amount of possible answers

b) p-q is the same as -q+p
For example, lets say

\(p=x+1\)

\(q=3x^2-5\)
\(p-q=x+1-(3x^2-5)=x+1-3x^2+5=-3x^2+x+6\)

also

\(-q+p=-(3x^2-5)+x+1=-3x^2+5+x+1=-3x^2+x+6\)

notice is the same expression.

Answer:

24°

Step-by-step explanation:

For an n-sided regular polygon, the sum of internal angles = (n-2)×180°

Since all internal angles are equal, the internal angle can is given as:

Here, it is given that the polygon has 15 sides.

So n = 15

So each internal angle is 156°.

Internal and external angles are always supplementary. So the sum of the internal and external angles is 180°.

So external angle = 180° - 156° = 24°.

The measure of each external angle is 24°.

Answer:

I can't find the answer.

Step-by-step explanation:

You should put a photo of the graph and I would be able to tell :)

Sorry

Using given function, we conclude that it will take 3 seconds the ball to reach a height of 45 meters.

In this question, we have been given a function h = 30t - 5t²

This function is used to find the height (h) of the ball in meters after 6 seconds.

We need to find the amount of time required to reach ball to a height of 45 meters.

here, height h = 45 meters

to find : time t

h = 30t - 5t²                          ........................(Given function)

45 = 30t - 5t²

30t - 5t² - 45 = 0

5t² - 30t + 45 = 0

t² - 6t + 9 = 0

(t - 3)² = 0

t - 3 = 0

t = 3 seconds

Therefore, it will take 3 seconds  the ball to reach a height of 45 meters.

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Answer:

x<3

Step-by-step explanation:

\( - 2(5 - 4x) < 6x - 4 \\ - 10 + 8x < 6x - 4 \\ 8x - 6x < - 4 + 10 \\ 2x < 6 \\ x < 3\)

ANSWER:

1. b = 5

2. h = 6

3. b = 30

STEP-BY-STEP EXPLANATION:

We know the ratio between height and base.

By means of the following proportions we can determine the missing dimensions in each case:

Answer: 3.5 drinks

Step-by-step explanation:

If you follow the graph and go over to 140 pounds

and go up to where the line goes through the 140 line

it lands on 3.5

Answer:

6 in ^2

Step-by-step explanation:

divide side lengths by two and add those numbers

Answer:

D

Step-by-step explanation:

108.5

Answer:

Option (A)

Step-by-step explanation:

AE and RV are the two line segments intersecting at a point N.

Therefore, ∠ANR ≅ ∠ENV [Vertical angles are equal in measure]

(8x + 12)° = (5x + 57)°

8x - 5x = 57 - 12

3x = 45

x = 15

Since, m∠ENV = 5(15) + 57

                        = 75 + 57

                        = 132°

Therefore, Option (A) will be the answer.

The slant height of the cone is 12.6 units

What are solids ?

A three-dimensional object that is closed (which may, according to some terminology conventions, be self-intersecting). A solid is any constrained area of space that is bounded by surfaces, according to Kern and Bland (1948, p. 18). The sphere, cube, cone, and cylinder, as well as the polyhedra more broadly, are some of the most basic solids.

h= 12

diameter = d = 8

so, radius, r = d/2 = 8/2 = 4

Formula for slant height , l is:

l = √(h² + r²) = √(12² + 4² )

 = √(144 + 16) = √160 = 12.649 ≈ 12.6 units

The slant height of the cone is 12.6 units

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Answer:

Step-by-step explanation:

sin (B) = \(\frac{2}{5}\)

Answer:

12x^3 is equivalent to

12x*12x*12x which if we multiply is

1728x

we divide by 4x

1728x divided by 4x=432x

Hope This Helps!!!

Answer:

3x^2

Step-by-step explanation:

when u divide 12x^3/4x....u divide 12/4=3 along with the x also..tat is x^3/x=x^2

Answer: Choice A)  y - 10 = 2(x - 3)

============================================================

Explanation:

We can rule out choices C and D because this diagonal line has a positive slope (as it moves uphill when moving to the right).

So m = 2 must be the slope.

---------

Recall that

y - y1 = m(x - x1)

represents the point slope form of a linear equation.

The point shown on this graph is (3,10) meaning that x1 = 3 and y1 = 10 pair up together.

So,

y - y1 = m(x - x1)

y - 10 = 2(x - 3)

which points to choice A as the final answer

The height of the can of cylinder shape is 9.14 cm.

What is a cylinder?

In mathematics, a cylinder is a three-dimensional solid that maintains two parallel bases separated by a curved surface at a specific distance. These bases frequently have a circular shape (like a circle), and an axis connects their respective centres.

We are given the diameter as 4 cm.

So, the radius is 2cm.

Also, it is given that the surface area of the can is 140 square centimeters.

So, using the surface area of cylinder, we get

⇒Area = 2πr (h + r)

⇒140 = 2π * 2 (h + 2)

⇒140 = 4π (h + 2)

⇒140 = 4 * 3.14 * (h + 2)

⇒140 = 12.56 * (h + 2)

⇒11.14 = h + 2

⇒h = 9.14

Hence, the height of the can of cylinder shape is 9.14 cm.

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The maximum height of the golf ball on Mars is about 48.8 meters.

To find how much longer it would take for the golf ball hit on Mars to reach the ground, we can use the given equation h = -1.9t² + 26t, where h is the height of the object and t is the time in seconds. First, we need to find the time it takes for the golf ball to reach the ground on Mars. Setting h = 0, we get:

0 = -1.9t² + 26t

Solving for t using the quadratic formula, we get t = 13.684 or t = 0.856. Since the golf ball is launched upward, we need the positive value of t, so it takes about 13.7 seconds for the golf ball to reach the ground on Mars. Therefore, it takes 13.7 − 5.4 = 8.3 seconds longer for the golf ball hit on Mars to reach the ground than the golf ball hit on Earth.

To find the maximum height of the golf ball on Mars, we can use the formula h = -1.9t² + 26t, where t is the time when the height is maximum. Since the formula is a quadratic equation, the maximum height occurs at the vertex of the parabola, which is at t = -b/2a.

In this case, a = -1.9 and b = 26, so the time at maximum height is t = -26 / 2(-1.9) = 6.842. Plugging this value into the equation for h, we get:

= -1.9(6.842)² + 26(6.842) ≈ 48.8 meters

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Answer:

2

Step-by-step explanation:

BC = 2

BD = 7

AC = 5.

BD - AC = 2

( so BC is equals to 2 )

Answer:

85

Step-by-step explanation:

the mean is equal to=99+66+90=255

255÷3=85

Answer:

Anna's test average would be 85%.

Step-by-step explanation:

To find the average, you have to do 66 + 90 + 99 to get an answer of 255. The next thing you do is divide 255 by 3 to get a final answer of 85%. :) I hope this helps! ^^

Veremos que la propocición es falsa. La magnitud de un ángulo no tiene dependencia con la longitud de sus lados.

La magnitud de un ángulo depende unicamente de la apertura entre sus lados. Un ejemplo de esto puede ser un par de triangulos equivalentes, donde las figuras son las mismas, pero las dimensiones de los lados no lo son.

En triangulos equivalentes podemos ver que todos los angulos interiores son exactamente igulaes, pese a que la longitud de los lados es diferente, asi vemos que la magnitud de un ángulo no tiene dependencia con el largo de sus lados.

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It is false that if the rank of the augmented matrix of a system of n linear equations in n unknowns is greater than the rank of the matrix of coefficients, then the matrix of coefficients is nonsingular

The augmented matrix is an extension of a matrix in which we add a column ( non homogeneous part of the system)

Say , Ax = b

Here , b is column part

So , Augmented matrix will be [ A : b ]

The Rank of augmented matrix can be found by performing elementary row operations on a augmented matrix and counting the number of the rows without zero comparing the rank of an augmented matrix

The rank of coefficient matrix can determined if there is a  solution to the system of solution

In the question this statement,

If the rank of the augmented matrix of a system of n linear equations in n unknowns is greater than the rank of the matrix of coefficients, then the matrix of coefficients is nonsingular is false

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The quotient of  10⁻⁴/10² is 10⁻⁶

Quotient is the result obtained by dividing one quantity by another. It represents the value that is obtained after division.

For example, if you divide 20 by 2, the quotient is 10 because 20 divided by 2 equals 10.

Division law states that:

10ᵃ / 10ᵇ = 10ᵃ⁻ᵇ

In this case have:

10⁻⁴/10² = 10⁻⁴⁻²   (Division law)

10⁻⁴/10² = 10⁻⁶

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