Complete the proof


Pleassseeee

Answer:

The length of the needle is 35cm

Step-by-step explanation:

Given

\(End\ 1 = 5cm\)

\(End\ 2 = 40cm\)

Required

The length of the needle (L)

This is calculated as thus:

\(L = End\ 2 - End\ 1\)

\(L = 40cm - 5cm\)

\(L = 35cm\)

The length of the needle is 35cm

The sector areas of the grey and white areas are 159.09 in² and 41.87 in²

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

The area of shaded sector is calculated as

Area = Central angle/360 * π * Radius^2

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

Hence, the area is 159.09

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Answer:

negative................

Answer:

Step-by-step explanation:

I really dont know but if I did I would say I'm a 6 grader

Answer:

False

Step-by-step explanation:

This is false although most prime numbers are not divisible two, but there is one exception to this... the number 2! The number 2 is prime and is divisible by 2. All other prime numbers cannot be divisible by 2 otherwise they would not be prime as prime numbers are only divisible by 1 and the number it self (this excludes 1).

Answer:

f(2) = 1

x = 0

Step-by-step explanation:

f(2) = 1 since y=1 when x=2.

f(0) = -3 since x=0 when y=-3

Answer:

3 4/5

Step-by-step explanation:

Answer: The pickles are $3.99

Step-by-step explanation:For percents,this is always done by simply dividing the percent (in this case 15%) by 100%. So, the conversational term "15%" becomes 15% / 100% = 0.15 in terms of a real mathematical number. 0.15 x $152.00=$3.99

9514 1404 393

Answer:

  18.98 pounds

Step-by-step explanation:

Let 'e' represent the number of pounds of expensive candy Ellen should use. Then 27-e is the number of pounds of less expensive candy. Her total cost is ...

  1.45(27 -e) +3.74e = 27(3.06)

  (3.74 -1.45)e = 27(3.06 -1.45) . . . . subtract 1.45(27)

  e = 27(3.06 -1.45)/(3.74 -1.45) = 27(1.61/2.29) ≈ 18.98

Ellen should use about 18.98 pounds of the expensive candy.

_____

Additional comment

We showed the solution this way because it is the generic solution to any mix problem. The fraction of the mix that is the highest-value contributor is the ratio of the difference between the mix value and the lowest contributor to the difference between contributors. Here, that fraction is ...

(3.06 -1.45)/(3.74 -1.45) = (1.61/2.29)

Answer:

to find the average rate of change for this function, you have to examine the formula of the average rate of change: (y2-y1)/(x2-x1).  

having that said, you can now plug in the values by solving the equation for x= -4 and x= 2:  

g(-4) = 50

g(2) = 14

so, average rate of change becomes: (14-50)/(2-(-4)) = 36/6 = 6.  

hope that helps!

Step-by-step explanation:

brainliest?

The solution to the inequality is x >= -2 for inequality 3x-5≥-11

The given inequality is 3x-5≥-11

Three times of x greater than or equal to minus eleven

x is the variable

3x - 5≥ -11:

Adding 5 to both sides, we get:

3x ≥ -6

Dividing both sides by 3, we get:

solution is  x≥-2

Therefore, the solution to the inequality is x >= -2.

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Answer:

Use the distributive property to multiply the factors on the right side of the equation.

Simplify the product by combining like terms.

Show that the right side of the equation can be written exactly the same as the left side.

Show that the right side of the equation simplifies to a cubed minus b cubed.

Step-by-step explanation:

The difference of two cubes identity a³ - b³ = (a - b)(a² + ab + b²) has been proved using; distributive property of algebra.

a³ - b³ = (a - b)(a² + ab + b²)

(a - b)(a² + ab + b²) = a(a² + ab + b²) - b(a² + ab + b²)

Multiplying out the brackets gives us;

a³ + a²b + ab² - a²b - ab² - b³

Like terms cancel out to give us;

a³ - b³.

This is same as the left hand side of our initial equation and thus it has been proved.

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The required simplification of the given expression \(=\lim_{x \to \infty} \frac{x^2-4x +4}{x^2-4}\) is 1.

Given that,
To determine the simplified value of the expression \(=\lim_{x \to \infty} \frac{x^2-4x +4}{x^2-4}\).

The process in mathematics to operate and interpret the function to make the function or expression simple or more understandable is called simplifying and the process is called simplification.

here,
Given expression,
\(=\lim_{x \to \infty} \frac{x^2-4x +4}{x^2-4}\)
\(=\lim_{x \to \infty} \frac{x^2(1-4/x +4/x^2)}{x^2(1-4/x^2)}\)
Now put limits, so 1 / x², 1 / x becomes zero so,
= 1

Thus, the required simplification of the given expression is 1.

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Answer:

I think A is the right answer

dy/dx = (y-3)(y+3) is the autonomous differential equation that satisfies all of the given properties.

The autonomous differential equation that satisfies all of the given properties is dy/dx = (y-3)(y+3). This equation has two equilibrium solutions at y = 0 and y = 3, and is positive for -inf < y < 0, and negative for 0 < y < 3, and positive for 3 < y < inf.

To demonstrate this, let's consider the equation at y=-3. Since y=-3 is less than 0, the equation can be simplified to dy/dx = 6. Since 6 is positive, y' is also positive, meaning that y is increasing. Similarly, if y=3, dy/dx = 0 which is neither positive nor negative, so y remains constant. Finally, for y>3, dy/dx = -6, which is negative, so y is decreasing.

Therefore, dy/dx = (y-3)(y+3) is the autonomous differential equation that satisfies all of the given properties.

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The required leaves values for the given stems are calculated below.

A plot where every information esteem is parted into a "leaf" (typically the last digit) and a "stem" (different digits). For instance "32" is parted into "3" (stem) and "2" (leaf). The "stem" values are recorded down, and the "leaf" values are recorded close to them.

For example , 42, 34, 48, and so on.

Here 42 and 48 qualities will be written in stem 4 as their most memorable digit is 4, while 2 and 8 will be placed in the leaf side.

Likewise, 3 will be the stem, and 4 will be the leaf for esteem 34.

4| 2, 8

3| 4

65​, 40​, 47​, 47​, 33​, 55​, 44​, 49​, 46​, 22​, 58​, 65​, 56​, 41​, 63​, 65​, 54​, 42​, 58​, 29​, 39​, 66 ​, 49​, 42​, 34​, 36​, 42​, 35​, 21​, 44​, 61​, 35​, 33​, 42​, 27​, 30​, 46​, 33​, 35​, 58.

As given in the question that the stem and Leaves are to be separated by a hyphen.

Stem-and-Leaf plot 6 -  5, 5, 3, 5, 6, 1

5 -  5, 8, 6, 4, 8, 8

4 -  0, 7, 7, 4, 9, 6, 1, 2, 9, 2, 2, 4, 2, 6

3 -  3, 9, 4, 6, 5, 5, 3, 0, 3, 5

2 -  2, 9, 1, 7

As we can see above there are 5 stems, where stem 6 has 6 leaves, stem 5 has 6 leaves, stem 4 has 14 leaves, stem 3 has 10 leaves, and stem 2 has 4 leaves.

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Answer:

The midsegment of a trapezoid is parallel to each base, and its length is one-half the sum of the lengths of the bases.

Step-by-step explanation:

The difference between their yearly incomes is $31,304.

To calculate each person's yearly income, we need to multiply their weekly income by the number of weeks in a year. Assuming there are 52 weeks in a year, the yearly income can be calculated as follows:

For the student who drops out of high school:

Yearly Income = Weekly Income x Number of Weeks

= 451 x 52

= 23,452

For someone with a bachelor's degree:

Yearly Income = Weekly Income x Number of Weeks

= 1053 x 52

= 54,756

The difference between their yearly incomes can be found by subtracting the student's yearly income from the bachelor's degree holder's yearly income:

Difference = Bachelor's Yearly Income - Student's Yearly Income

= 54,756 - 23,452

= 31,304

Therefore, the difference between their yearly incomes is $31,304.

It is important to note that these calculations are based on the given information and assumptions. The actual yearly incomes may vary depending on factors such as work hours, additional income sources, deductions, and other financial considerations.

Additionally, it is worth considering that educational attainment is just one factor that can influence income, and there are other variables such as experience, job type, and market conditions that may also impact individuals' earnings.

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Answer:

\(4y^{2}\)

Step-by-step explanation:

A = Area

w = width

l = length

\(A=lw\)

in this equation the area is \(32xy^{3}\) and the length is \(8xy\).

To find the equation we simply have to divide the area (\(32xy^{3}\)) by the length (\(8xy\)).

\(32xy^{3} =(8xy)w\)

When dividing, it's important to remember two things:

Using these two rules, we divide the common bases (number by number, x by x, y by y):

\(\frac{32}{8}=4\)

\(\frac{x}{x}=1\)

\(\frac{y^{3} }{y}=y^{2}\)

Multiplying all of them together, we find that the width is:

\(4y^{2}\)

Answer: second step.

Step-by-step explanation:

you then have to distribute the answer not subtract 9 1/4 from 1.5

you have to do the parenthesis next

4.5/0.5=45/5=9

then

1.5*9= 13.5

then

9.25-13.5 which I think you are capable of solving

Answer:

Step-by-step explanation:

Given work on simplification

Step 1 9 and one-fourth minus 1.5 times (4.5 divided by 0.5)

Step 2 7.75 times (4.5 divided by 0.5)

Continuous random variable is a random variable that can take on a continuum of worth. In alternative words, a random variable is said to be continuous if it supposes a value that falls between a particular interval.

A continuous random variable can be described as a random variable that can take on an infinite number of possible values. Due to this, the probability that a continuous random variable will take on a fixed value is 0.

The probability density function of a continuous random variable can be explained as a function that gives the probability that the value of the random variable will drop between a range of values. Let X be the continuous random variable, then the formula for the pdf, f(x).

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Refering to sales price in the store, the constant of proportionality will be the quotient between each sales price and its corresponding regular price

The constant of proportionality is 0.75

Answer:

Mean: 89

Median: 89

Mode: 89

Range: 22

Step-by-step explanation:

Answer:

89

Step-by-step explanation:

Mean-89

Median-89

Mode-89

Answer:

Step-by-step explanation:

In the Intermediate Phase of mathematics education, one topic that demonstrates a hierarchical and logical progression in patterns, functions, and algebra is the concept of "Linear Equations."

The topic of Linear Equations in the Intermediate Phase builds upon the foundation laid in earlier grades and serves as a stepping stone towards more advanced algebraic concepts. Here is an overview of the topic sequence and content area spread for Linear Equations:

Introduction to Variables and Expressions:

Students are introduced to the concept of variables and expressions, learning to represent unknown quantities using letters or symbols. They understand the difference between constants and variables and learn to evaluate expressions.

Solving One-Step Equations:

Students learn how to solve simple one-step equations involving addition, subtraction, multiplication, and division. They develop the skills to isolate the variable and find its value.

Solving Two-Step Equations:

Building upon the previous knowledge, students progress to solving two-step equations. They learn to perform multiple operations to isolate the variable and find its value.

Writing and Graphing Linear Equations:

Students explore the relationship between variables and learn to write linear equations in slope-intercept form (y = mx + b). They understand the meaning of slope and y-intercept and how they relate to the graph of a line.

Systems of Linear Equations:

Students are introduced to the concept of systems of linear equations, where multiple equations are solved simultaneously. They learn various methods such as substitution, elimination, and graphing to find the solution to the system.

Word Problems and Applications:

Students apply their understanding of linear equations to solve real-life word problems and situations. They learn to translate verbal descriptions into algebraic equations and solve them to find the unknown quantities.

The content area spread for Linear Equations includes concepts such as variables, expressions, equations, operations, graphing, slope, y-intercept, systems, and real-world applications. The progression from simple one-step equations to more complex systems of equations reflects a logical sequence that builds upon prior knowledge and skills.

By following this hierarchical progression, students develop a solid foundation in algebraic thinking and problem-solving skills. They learn to apply mathematical concepts and procedures in a systematic and logical manner, paving the way for further exploration of patterns, functions, and advanced algebraic topics in later phases of mathematics education.