This is a picture of a cube and the net for the cube.

What is the surface area of the cube?

Responses

Answer: I did not do the math, I just tell people how to do it

Hope I helped.

Step-by-step explanation:

Multiply the number of triangles you created by 180.

the approximate difference in the mean circumferences of the Earth and the Moon is about 29153.88 kilometers.

A circle is a particular type of ellipse in mathematics or geometry where the eccentricity is zero and the two foci are congruent.

A circle is also known as the location of points that are evenly spaced apart from the centre. The radius of a circle is measured from the centre to the edge. T

he line that splits a circle into two identical halves is its diameter, which is also twice as wide as its radius.

A circle is a fundamental 2D form whose radius is quantified. The interior and exterior parts of the aircraft are separated by the circles.

The formula for the circumference of a circle is:

C = 2πr

where C is the circumference, π (pi) is a constant approximately equal to 3.14, and r is the radius.

The difference in the mean circumferences of the Earth and the Moon can be found by subtracting the circumference of the Moon from the circumference of the Earth.

The circumference of the Earth is:

C earth = 2π * 6371.0 km = 40075.04 km (rounded to two decimal places)

The circumference of the Moon is:

C moon = 2π * 1737.5 km = 10921.16 km (rounded to two decimal places)

The difference in the mean circumferences of the Earth and the Moon is:

C earth - C moon = 40075.04 km - 10921.16 km ≈ 29153.88 km

Therefore, the approximate difference in the mean circumferences of the Earth and the Moon is about 29153.88 kilometers.

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The equation of the circle is determined as (x + 3)²  + (y - 9)² = 25.

The equation of the circle is calculated as follows;

For the equation;

y = 3/4x+5

Slope = 3/4,  the slope of the line perpendicular to the line = -4/3.

it is pass through the (0, 5).

The equation of the line: y - 5 = -4x/3

For the equation;

3x + 4y = 38

4y = -3x + 38

y = -3x/4 + 38

slope = -3/4, the slope of the line perpendicular to the line = 4/3

it is pass through the (6, 5)

The equation of the line: y - 5 = 4/3(x - 6)

Solve the two equations together to find the point of intersection;

(-4/3)x + 5 = (4/3)(x - 6) + 5

-4x/3 = 4x/3 - 8

8x/3 = -8

8x = -24

x = -3

y - 5 = -4x/3

y - 5 = -4(-3)/3

y - 5 = 4

y = 9

The center of the circle = (-3, 9)

The radius of the circle is calculated as;

r = √ (-3 -0)² + (9 - 5)²

r = √(9 + 16)

r = 5

The equation of the circle becomes;

(x - a )² + (y - b)² = r²

(x + 3)²  + (y - 9)² = 5²

(x + 3)²  + (y - 9)² = 25

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Answer:

Step-by-step explanation:

Answer:

Experiencing anxiety on the day of the examination is a common problem for many students, especially those in grade 12 who may feel a lot of pressure to perform well. However, there are several strategies that they can use to minimize the impact of anxiety and enhance their performance in the examination. Some of these strategies include:

1. Get enough sleep: Getting enough sleep is essential for students to perform well on the day of the examination. Students should aim to get at least 7-8 hours of sleep before the exam day. A well-rested brain will be able to function better, and students will be less likely to feel anxious or stressed.

2. Eat a balanced breakfast: Eating a balanced breakfast on the day of the examination can help students feel more alert and focused. A healthy breakfast can also provide the necessary energy needed to get through the examination.

3. Plan your day: Plan your day well ahead of the examination. This will help you feel more in control and reduce feelings of anxiety. Set aside time to review notes, relax and mentally prepare for the exam.

4. Practice relaxation techniques: Practice relaxation techniques such as deep breathing or progressive muscle relaxation to help calm your nerves. These techniques can help students reduce stress levels and anxiety, which in turn can enhance their performance.

5. Avoid last-minute cramming: Cramming at the last minute can increase feelings of anxiety and negatively impact performance. Instead, spend time reviewing notes and practicing past papers in the days leading up to the examination.

6. Stay positive: It is essential to stay positive and avoid negative thoughts or self-doubt. Remind yourself of your hard work and preparation for the exam. Believe in your abilities, and focus on what you can do rather than what you cannot.

By implementing these strategies, grade 12 students can minimize the impact of anxiety on the day of the examination, and enhance their overall performance. Remember that exam performance is not just about what you know, but also about how you manage your anxiety and stress levels.

On performing the given row-operation on the given augmented-matrix:

(a) multiplying row1 by -2 and adding to row2, we have [tex]\left[\begin{array}{ccc}5&-2&6\\-10&10&-15\end{array}\right][/tex],

(b) interchanging row1 and row2, we have [tex]\left[\begin{array}{ccc}0&6&-3\\5&-2&6\end{array}\right][/tex],

(c) Multiplying row1 by 2, the matrix becomes [tex]\left[\begin{array}{ccc}10&-4&12\\0&6&-3\end{array}\right][/tex].

The augmented matrix on which we have to perform the row operation is

⇒ [tex]\left[\begin{array}{ccc}5&-2&6\\0&6&-3\end{array}\right][/tex],

Part(a) : We have to multiply row1 by -2 and and add to row2,

The row-operation is performed as : [tex]\left[\begin{array}{ccc}5&-2&6\\(5\times-2)+0&(-2\times-2)+6&(6\times-2)-3\end{array}\right][/tex];

Simplifying further ,

We get,

[tex]\left[\begin{array}{ccc}5&-2&6\\-10&10&-15\end{array}\right][/tex].

Part(b) : We have to interchange the "row1" and "row2",

On interchanging,

We get,

[tex]\left[\begin{array}{ccc}0&6&-3\\5&-2&6\end{array}\right][/tex].

Part(c) : We have to multiply the elements of "row1" by 2,

So, On multiplying the "row1" by 2,

We get,

⇒ [tex]\left[\begin{array}{ccc}5\times 2&-2\times 2&6\times 2\\0&6&-3\end{array}\right][/tex]

⇒ [tex]\left[\begin{array}{ccc}10&-4&12\\0&6&-3\end{array}\right][/tex].

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Answer:

3

Step-by-step explanation:

The inverse function returns the original value for which a function gave the output.

This means if there is a function f(x) and at x = a, f(a) = k then f⁻¹(k) = a

This may be difficult to grasp with variables but essentially returning to the specific question
if f(3) = 17

then

f⁻¹(17) = 3

And that is the answer

The equation in standard form of the eclipse is expressed as:

D. x²/9 + y²/8 = 1.

The standard form equation for an ellipse with center at the origin, a vertex at (a, 0), and a focus at (c, 0) is:

x²/a² + y²/b² = 1, where c² = a² - b².

In this case, the center is at the origin, a vertex is at (3, 0), and a focus is at (1, 0).

Since the center is at the origin, we know that a vertex is located a distance of a units to the right and left of the center, so a = 3.

Since the focus is located a distance of c units to the right and left of the center, we know that c = 1.

Using c² = a² - b², we can solve for b²:

1² = 3² - b²

b² = 8

So the equation in standard form is:

x²/9 + y²/8 = 1.

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Answer:

<2  = 30°

Step-by-step explanation:

Let angle 1 = x

Let angle 2 = y

x + y = 95

x = 35 + y  Substitute 35 + y for x in x + y = 95

x + y = 95

35 + y + y = 95  Subtract 35 from both sides

y + y = 60    Combine like terms

2y = 60  Divide both sides by 2

y = 30

Helping in the name of Jesus.

Answer:

Step-by-step explanation:

The value of P(x=1) from the given discrete data is 0.5.

From the given data

Let P(x=2)=x

p+p+x=1

x=1-2p

E(x²)=∑PiXi²=P(0)²+p(1)²+(1-2p)(2)²

= 0+p+(1-2p)4

= 4-7p

E(x)=∑PiXi=P(0)+P(1)(1-2p)(2)

= 0+p+2-4p

= 2-3p

Given that, E(x²)=E(x)

4-7p=2-3p

4p=2

p=0.5

Therefore, the value of P(x=1) from the given discrete data is 0.5.

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Answer:(a) The rate of decay is 14.4% per day, which can be written as a decimal by dividing by 100: r = 0.144.

(b) The formula for continuous decay is given by:

A = A₀ * e^(-rt)

where A is the remaining amount of the substance after time t, A₀ is the initial amount, r is the rate of decay, and e is the mathematical constant approximately equal to 2.71828.

Plugging in the given values, we get:

A = 140 * e^(-0.144*15)

A = 140 * e^(-2.16)

A ≈ 47.23

Therefore, after 15 days, approximately 47.23 mg of the radioactive substance will be left, rounded to two decimal places.

Step-by-step explanation:

The equation of p(x) in vertex form is p(x) = 3(x + 1)² - 3.

Option C is the correct answer.

Since, Polynomial is an equation written as the sum of terms of the form kxⁿ.

where k and n are positive integers.

We have,

From the table,

From p(x) = 24 to p(x) = -3 the value of p(x) is decreasing and from p(x) = -3 to p(x) = 24 it is increasing.

So,

The vertex is (- 1, -3).

The vertex form of p(x).

p(x) = a ( x - h)² + k

Where (h, k) is the vertex.

Now,

(-1, -3) = (h, k)

Make (- 2, 0) = (x, p(x))

0 = a (- 2 + 1)² + (-3)

0 = a x 1 - 3

a = 3

Thus, The vertex equation of p(x).

p(x) = 3(x + 1)² - 3

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Answer:

Step-by-step explanation:

When x_1 =7 then you can figure out by given condition say X_n+1= 2X_n-3

then X_2= 11 by substituting the value of X_1 as 7 similarly you can find out the value of X_3 by substituting X_2 as 11 so X_3 =19.

Similarly you can find out b)

The polygons that can be formed by the intersection of a plane and a cylinder either parallel or perpendicular to the base include :

If a plane intersects three distinct vertices of the cube while bypassing any other vertices, a triangle is produced. To visualize this, suppose a corner of the cube were removed with a knife; picture that cut surface as a triangle.

On the other hand, if a plane coincides with four points on the cube and outlines a square shape, it creates what's called a "square cross-section." This transpires when said plane parallels one side of the cube while intersecting only the midpoints of its opposite face's edges.

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Full question is:

Select all the polygons that can be formed by the intersection of a plane and a cylinder either parallel or perpendicular to the base

A triangle

B square

C. rectangle

D pentagon

E circle

The dimensions that minimizes the perimeter are undefined and DNE

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

Base = w

Heights = h and T

Where

T = 1.6w

The area of the window is calculated as

A = 1/2Tw + hw

So, we have

A = 1/2 * 1.6w * w + hw

A = 0.8w² + hw

Differentiate with respect to w

A' = 1.6w + h

Set to 0

1.6w + h = 0

Solve for h

h = -1.6w

The above means that the perimeter cannot be minimized

This is because h and w cannot be negative

Hence, h = undefined and w = undefined

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Answer:

[tex]\dfrac{\boxed{2}}{\boxed{3}} \times \dfrac{\boxed{1}}{\boxed{2}} = \dfrac{\boxed{1}}{\boxed{3}}[/tex]

Step-by-step explanation:

The square is 1 unit long and 1 unit wide
There are 3 smaller rectangles in the horizontal direction

Length of each rectangle is therefore 1/3 unit

There are 6 rectangles along the vertical direction

The height of each rectangle is therefore 1/6 unit

The shaded area is 2 rectangles wide and 3 rectangles in height

Width of shaded area = 2 x 1/3 = 2/3

Height of shaded region = 3 x 1/6 = 1/2

Area of shaded rectangle is

[tex]\dfrac{\boxed{2}}{\boxed{3}} \times \dfrac{\boxed{1}}{\boxed{2}} = \dfrac{\boxed{1}}{\boxed{3}}[/tex]

Answer:

Let's use "K" to represent Kevin's current age, and "M" to represent his mother's current age.

According to the problem, if you triple Kevin's age, the result is 1 less than his mother's age:

3K = M - 1

We can simplify this equation by adding 1 to both sides:

3K + 1 = M

This equation tells us that if we add 1 to three times Kevin's age, we get his mother's age. We can use this equation to solve for Kevin's age if we know his mother's age, or we can use it to solve for his mother's age if we know Kevin's age.

An equivalent fraction to 1 whole and 7/8 is 15/8.

A fraction that has the numerator higher than or equal to the denominator is said to be inappropriate. It is, in other words, a "top-heavy" fraction since the fraction represents a value greater than one full unit.

1 whole and 7/8 can be written as a mixed number, which is a combination of a whole number and a fraction:

1 whole and 7/8 = 1 + 7/8

To find an equivalent fraction, we can convert the mixed number to an improper fraction by multiplying the whole number by the denominator of the fraction, and then adding the numerator. The result is the numerator of the improper fraction, and the denominator stays the same:

1 + 7/8 = (1 × 8 + 7)/8 = 15/8

Therefore, an equivalent fraction to 1 whole and 7/8 is 15/8.

For instance, the improper fraction 7/4 results from the numerator's (7), which is greater than the denominator (4).

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The probability of landing on red is given by the number of red sections divided by the total number of sections. In this case, there are 2 red sections out of a total of 6 sections. Therefore, the probability of landing on red is:

P(red) = 2/6 = 1/3

So the probability of landing on red is 1/3.

Answer:

$74.66

Step-by-step explanation:

Plot the points on the graphing calculator (x = length of HDMI cord, y = cost). Then generate a linear regression model:

y = 3.68x + 1.06 (approximately)

x = 20, so y = 74.66

$74.66 is the correct answer.

Step-by-step explanation:

How do you have two equal to signs omg?

Anyway,

x+6+2x=2x+6+4x+6+2x=2x+6+4

3x+6=8x+12=2x+10

0=5x+6=2x+10

0=3x-4=0

3x-4=0

3x=4

x = 4/3 = 1 ⅓

Note that the five number summary and range for the data are given as follows:

The five-number summary and range were calculated using the given stem and leaf plot. The minimum value is 102, the first quartile (Q1) is 226, the median (Q2) is 340, the third quartile (Q3) is 492, and the maximum value is 670. The range is the difference between the maximum and minimum values, which is 568.

The range is calculated by subtracting the minimum value from the maximum value:

Range: 607 - 102 = 505

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The option that is not a run-on sentence is this: According to data collected throughout film history, the average length for a feature film is between 80 and 100 minutes, although some far exceed the standard running time.

A run-on sentence is a sentence that continues without using appropriate punctuation to join independent clauses. In the sentence above, the comma is used in the right place to join the independent clause, so we may not refer to this as a run-on sentence.

The other three options continue on end without appropriate punctuation to break off the independent clauses.

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The matrix row operation on the matrix A gives us: 0x + 10y - 25z = 4

There are 3 primary basic operations that are utilized on the rows of a matrix when we utilize the matrix to solve a particular system of linear equations . The aim is usually to get the left region of the matrix to resemble the identity matrix .

The three primary operations are:

Switching Rows

Multiplying a Row by a Number

Adding Rows

Now, from the given matrix, the equations are:

x - y + 5z = -1

x - y - 4z = 5

5x + y + 0z = -1

We are told to add -5 times row 1 to row 3 and this gives:

-5x + 5x + 5y + 5y - 25z + 0z = 5 - 1

0x + 10y - 25z = 4

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Answer:

f(x) = x²-8x+15

g(x) = x -3

h(x) = f(x) ÷ g(x)

= x²-8x+15 ÷ x-3

x-3√x²- 8x +15 | x-5

x² - 3x

-5x + 15

- 5x + 15

-----------

h(x) = x-5

Mr. Moketsi could help students in their learning of properties of 2D shapes through hands-on activities and discovery, as suggested by the constructivist perspective.

Mr. Moketsi initiates the class by inquiring about their comprehension of 2D shapes and their properties, provoking conversation amongst students as they share their thoughts with their peers and their teacher.

He subsequently introduces an assortment of manipulatives, such as pattern blocks, geoboards, and tangrams to encourage experimentation and inquiry into these geometric figures, while posing thought-provoking questions throughout the lesson to help learners make connections between different shapes with a focus on discovering their unique attributes.

Mr. Moketsi encourages reflection, questioning how this newfound knowledge may be applied in reality, stimulating critical thinking abilities during classroom interaction with surfaces representing planes containing lengths and widths.

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