I need help pleaseee

Answer: 16-4√x/16-x is the answer.

Step-by-step explanation:

=> 4/4+√x*4-√x/4-√x

=> 4(4-√x)/(4)²-(√x)²  {using identity - (a+b) (a-b)=a² - b²}

=> 16-4√x/16-x

The translation of ABC by (Rx + Ry) is UVW and the rigid composition of XYZ to X'Y'Z' cannot be determined

The rule (Rx + Ry) represents a translation image of a shape, where R is a vector representing the direction and distance of the translation, and (x, y) are the coordinates of the original shape.

To apply this rule to a shape, we simply add the vector R to each point of the shape, which moves the shape in the direction and distance specified by R.

Using the above as a guide, we have the following the translation of ABC by (Rx + Ry) to be UVW

The triangle X'Y'Z' is not given in the figure

So, the rigid composition of XYZ to X'Y'Z' cannot be determined

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Answer:

Step-by-step explanation:

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:

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 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: D

Step-by-step explanation:

The answer is D. The cost of gasoline is higher in the U.S. than anywhere else in the world.

According to the U.S. Energy Information Administration, the average price of a gallon of regular unleaded gasoline in the United States was $3.53 on March 8, 2023. This is higher than the average price of gasoline in many other countries, including Canada ($1.56), Mexico ($1.97), and the United Kingdom ($1.73). However, the cost of gasoline in the United States is lower than the average price in some countries, such as Norway ($6.62), Sweden ($5.62), and Denmark ($5.51).

Here are the facts about the other statements:

A. Most Americans commute to work in a personally owned vehicle. According to the U.S. Census Bureau, in 2019, 77.4% of workers commuted to work by driving alone.

B. Most Americans who commute to work in car's drive alone. According to the U.S. Census Bureau, in 2019, 77.4% of workers who commuted to work by driving did so alone.

C. Americans spend more income on transportation than anything else except housing. According to the U.S. Bureau of Labor Statistics, in 2020, Americans spent an average of $10,564 on transportation, more than any other category except housing ($16,916).

The formula of the nth term of the sequence is f(n) = 8 *0.7r^(n - 1)

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

a=_1=8, r=0.7

Express properly

So, we have

First term, a = 8

Common ratio, r = 0.7

The above definition is of a geometric sequence that has a first term of 8 and a common ratio of 0.7

Using the above as a guide, we have the following:

f(n) = a * r^(n - 1)

substitute the known values in the above equation, so, we have the following representation

f(n) = 8 *0.7r^(n - 1)

Hence, the nth term of the sequence is f(n) = 8 *0.7r^(n - 1)

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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:

Both sides are equal, and are equvalent to 5 square root of 2

Step-by-step explanation:

This is a special right angled triangle known as the 45-45-90 triangle

The hypotenuse of a 45-45-90 triangle is x square root 2, with x being both sides of the triangle

10=x square root of 2 (Divide both sides by square root of 2)

10/square root of 2=x (Rationalize)

x=5 square root of 2

Answer:

Step-by-step explanation:

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)

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

Answer: Cut 3 inch per person

Step-by-step explanation:

there is a 4 foot sub you take the quadrilatiral and divide it by 10 and you get 3 inches per person remember to multiply the reciprocal by 12 as there are 12 inches in a foot and you have 4 foot so the product of that reciprocal will be equal to 4 divided by your sum from adding the recriptorcal into the quadrilateral remember a quadrilateral not a duolateral or a triolatiral as there is 4 feet in the sub. >:)

Answer:

7 Cuts

Step-by-step explanation:

This is because if you do 48 inches / 6 inches = 8 sections Since each cut will create two sections, you need to make 7 cuts (one fewer than the number of sections) to get 8 equal portions.

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.

The Slope of x = -3 is undefined; 6x-3y=18 is 2; (4,-2) and (1,7) is -3; and (7,5) and (8,5) is 0.

Slope of a line (m) = rise/run = change in y / change in x

Note that the slope of vertical lines are always undefined.

Therefore, we have the following:

x = -3 represents the equation of a vertical line, therefore, the slope is undefined.

6x - 3y = 18 is rewritten as y = 2x - 6 in slope-intercept form. The slope is 2.

For (4,-2) and (1,7):

Slope (m) = (7 -(-2))/(1 - 4) = 9/-3 = -3

For (7,5) and (8,5):

Slope (m) = 5 - 5/8 - 7 = 0/1 = 0

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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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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 outlier does not affect the mode

The outlier will increase the mean.

The outlier will not affect the median.

The outlier will increase the mean because  it will significantly increase the sum of the values.

The presence of an outlier will not affect the median, as it only changes the position of the middle value.

The mode is the value that occurs most frequently in a data set. In this case, the mode will not be affected because the outlier is not the most frequent in the data set and only appears once.

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The number of games that the team did play in the last season is 23.

The frequency is the number of times an event happens. Therefore, frequency is the number of repetitions of a digit or an event.

The given frequency table represents the frequency of different runs scored by the team in the entire season.

The frequency is written in tally form, which means that each of the vertical lines represents the count of 1, while a group of fours lines such that the four lines are vertical while the fifth line cuts the other four represents a count of 5 as shown in the second column of the second row.

Therefore, the frequency table can be made as:

Number of runs     Frequency

            0                         8

            1                          5

            2                         3

            3                         6

            4                         1

Total:                              23

Hence, the team played 23 games.

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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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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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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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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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Θ(x) = tan⁻¹(5x / (4(x - 13))), and Θ is maximized when the distance the cow must stand from the billboard x₀ = 26.

To find Θ, use similar triangles. Let the distance from the cow to the top of the billboard be h. Then, using similar triangles:

h / x = 5 / (x - 13)

Solving for h:

h = 5x / (x - 13)

The vertical angle subtended by the billboard at the cow's eye is equal to the angle whose tangent is given by:

tan(Θ) = h / 4

Substituting for h:

tan(Θ) = 5x / (4(x - 13))

To maximize Θ, find the value of x that maximizes tan(Θ). Taking the derivative of tan(Θ) with respect to x and setting it equal to zero:

d/dx (tan(Θ)) = 5(52 - 26x) / (4(x - 13))² = 0

Solving for x:

x₀ = 26

Substituting x₀ into the expression for Θ:

Θ(x₀) = tan⁻¹(5(26) / 4(13)) = tan⁻¹(5/2)

Therefore, Θ(x) = tan⁻¹(5x / (4(x - 13))), and Θ is maximized when x₀ = 26.

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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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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]