find missing exponents ​

Answer: general public

Answer:

The answer is the general public.

Step-by-step explanation:

I got it right on the lesson!

Hope this helps! :D

Answer:

100-22=78

78+6=84

1500 x 84/100= 1260Dhs

Plss mark as brainliest

let's check each one of the options:

(A) is not equivalent

(B) is not equivalent

(C) is not equivalent

(D) it is equivalent to the given equation because

we divide in both sides of the equality by 3 and obtain:

then:

(1 - x_n^2 - y_n^2) is bounded away from zero for all n > N, and we can conclude that f(x_n, y_n) approaches a limit as n approaches infinity.

To determine the set of points at which the function is continuous, we need to check for continuity along all paths in the domain.

First, note that the expression in the numerator of the function, 1/(x^2 y^2), is continuous everywhere except at (0,0).

Next, we need to consider the denominator, 1-x^2-y^2. This expression is defined and continuous for all points (x,y) such that x^2 + y^2 < 1. However, it is undefined at the boundary of the disk, i.e., on the circle x^2 + y^2 = 1.

Therefore, the domain of the function is the open unit disk centered at the origin, {(x,y) | x^2 + y^2 < 1}.

To determine if the function is continuous at any point on this domain, we can use the limit definition of continuity. Specifically, we need to show that for any point (a,b) in the domain and any sequence {(x_n, y_n)} that converges to (a,b), the sequence {f(x_n, y_n)} converges to f(a,b).

Let's consider a few cases:

Case 1: (a,b) is not equal to (0,0)

In this case, we know that the numerator of the function is continuous and non-zero at (a,b). We also know that the denominator is continuous and non-zero in a neighborhood of (a,b) (since (a,b) is an interior point of the domain). Therefore, the quotient f(x,y) is continuous at (a,b).

Case 2: (a,b) = (0,0)

In this case, we need to be more careful. Let {(x_n, y_n)} be any sequence that converges to (0,0). We want to show that f(x_n, y_n) converges to a limit as n approaches infinity.

If we choose a sequence that stays away from the boundary of the unit disk, i.e., x_n^2 + y_n^2 < r^2 for some r < 1, then the same argument as in Case 1 applies and we can conclude that f(x_n, y_n) converges to f(0,0).

However, if we choose a sequence that approaches the boundary, then we need to be more careful. Without loss of generality, assume that x_n^2 + y_n^2 = 1 for all n (since any sequence that crosses the boundary can be split into two sub-sequences, one on each side of the boundary). Then, we have:

f(x_n, y_n) = (1/(x_n^2 y_n^2))/(1 - x_n^2 - y_n^2)

= 1/[(x_n y_n)^2 (1 - x_n^2 - y_n^2)]

As n approaches infinity, we know that x_n y_n approaches 0 (since |x_n y_n| <= 1/2 for all n). Therefore, we need to focus on the expression (1 - x_n^2 - y_n^2). If this is bounded away from zero, then the denominator of f(x_n, y_n) will approach zero and the function will not be continuous at (0,0).

To show that (1 - x_n^2 - y_n^2) is indeed bounded away from zero, note that we can write:

1 - x_n^2 - y_n^2 = (1 - x_n^2) - (1 - y_n^2)

Since x_n^2 + y_n^2 = 1, we know that 1 - x_n^2 and 1 - y_n^2 are both positive. Therefore, we can bound:

1 - x_n^2 - y_n^2 >= (1 - x_n^2) - (1 - y_n^2)

= y_n^2 - x_n^2

Since x_n and y_n both approach zero as n approaches infinity, we can choose an N such that |x_n| < 1/2 and |y_n| < 1/2 for all n > N. Then, for all n > N, we have:

|y_n^2 - x_n^2| <= |y_n^2| + |x_n^2| <= 1/4 + 1/4 = 1/2

Therefore, (1 - x_n^2 - y_n^2) is bounded away from zero for all n > N, and we can conclude that f(x_n, y_n) approaches a limit as n approaches infinity.

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The probability that the sample mean of 46 racing cars would be greater than 116.9 gallons is 0.0043, or 0.43%.

To solve this problem, we can use the central limit theorem, which states that the sampling distribution of the sample means approaches a normal distribution as the sample size increases.
First, we need to calculate the standard error of the mean, which is the standard deviation of the population divided by the square root of the sample size:
standard error = 7 / sqrt(46) = 1.032
Next, we can standardize the sample mean using the formula:
z = (sample mean - population mean) / standard error
In this case, the population mean is 114 and the sample mean we're interested in is 116.9. So:
z = (116.9 - 114) / 1.032 = 2.662
Finally, we can use a standard normal distribution table or calculator to find the probability that a z-score is greater than 2.662. This probability is approximately 0.0043, rounded to four decimal places.
Therefore, the probability that the sample mean of 46 racing cars would be greater than 116.9 gallons is 0.0043, or 0.43%.

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Answer:

0

Step-by-step explanation:

The graph of \(f(x) =2^{(x)\) shifted towards positive x-axes to produce the graph \(g(x) = 2^{(x-7)\).

An exponential function is a function which can be defined as f(x) = aˣ, Where 'x' is a variable and 'a' is a constant.

The given functions are,

\(f(x) =2^{(x)\)

And \(g(x) = 2^{(x-7)\)

The graph of f(x) and g(x) has shown below,

The graph of g(x) shifted towards positive x-axes.

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Answer:

Y=0/0

Step-by-step explanation: since their is no fall or rise in the slope it is 0

Answer:

3 times

Step-by-step explanation:

if the surface area of the job is 3 times what you had before, you will need 3 times the paint.

Option B is correct. The statement that is true about the values we have been given here is that They are the only values that make both equations true.

We have been given two equations

The first equation is

4y-3x=16

We have to plug in the values (-5, 0.25) in this equation

4(0.25) -3(-5) =16

Then we have

1 + 15 = 16

Hence 16 = 16 This satisfies the equation.

In the second equation we have to follow the same step

-x-8y=3

Such that

-(-5)-8*0.25 = 3

5-2 = 3

3 = 3 This also satisfies the equation.

Therefore the correct answer option is B

a. They satisfy equation 8 but not equation A.

B. They are the only values that make both equations true.

C. They show that the equations represent the same line.

D. They satisfy equation A but not equation B.

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This probability is greater than 1, which is not possible.

This suggests that there may be an error in the problem statement, or that some additional information is required to correctly solve the problem.

The probability of a store being a Kwik Save is:

P(Kwik Save) = 1000 / (1000 + 424) = 1000 / 1424 = 0.7037

The probability of a store closing, given that it is a closing store, is:

P(closing| Kwik Save) = 107 / (107 + 324) = 107 / 431 = 0.2488

The probability of a store being a closing store is:

P(closing) = (107 + 324) / (1000 + 424) = 431 / 1424 = 0.3029

The probability of a randomly chosen store being either closing or Kwik Save is:

P(closing or Kwik Save) = P(closing) + P(Kwik Save) - P(closing and Kwik Save)

= 0.7037 + 0.3029 - (107 / 1424)

= 0.8997

The probability that a randomly selected store is not closing given that it is a Somerfield is:

P(not closing| Somerfield) = 1 - P(closingl Somerfield)

To find P(closing| Somerfield), we can use Bayes' theorem:

P(closing| Somerfield) = P(Somer field| closing) * P(closing) / P(Somerfield)

P(Somerfield| closing) is the probability that a store is a Somerfield given that it is closing.

Since all the Somerfield stores are closing, this probability is 1.

P(Somerfield) is the probability that a store is a Somerfield.

It is:

P(Somerfield) = 424 / (1000 + 424) = 424 / 1424 = 0.2980

Therefore,

P(closing| Somerfield) = 1 * 0.3029 / 0.2980 = 1.0151.

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Answer:

96 is the number. x = 9

Step-by-step explanation:

x = 9, so y = 15–9 = 6, and so the answer is 96

brainliest?

Answer:

(7/8) - (6 3/4) = -5.875

The number of cubes used in the third, fourth, and fifth figures of the pattern are 17 cubes, 21 cubes, and 25 cubes, respectively.

The recursive function f(n + 1) = f(n) + 4 states that each subsequent figure in the pattern will have 4 more cubes than the previous figure.

To determine the number of cubes in each figure, we start with the given values and add 4 to each subsequent figure.

In Figure 1, there are 9 cubes. Adding 4 cubes, we get Figure 2 with 13 cubes. Continuing this pattern, we add 4 cubes to each subsequent figure.

Figure 3: Figure 2 + 4 = 13 + 4 = 17 cubes
Figure 4: Figure 3 + 4 = 17 + 4 = 21 cubes
Figure 5: Figure 4 + 4 = 21 + 4 = 25 cubes

Therefore, the number of cubes used in the third, fourth, and fifth figures of the pattern are 17 cubes, 21 cubes, and 25 cubes, respectively.

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QUESTION - Luis uses cubes to represent each term of a pattern based on a recursive function. The recursive function defined is f(n + 1) = f(n) + 4, where n is an integer and n ≥ 2. The number of cubes used in each of the first two figures is shown below. How many cubes does Luis use in the third, fourth, and fifth figures of the pattern? Fill in the blanks.
Figure 1: 9 cubes
Figure 2: 13 cubes
Figure 3:
Figure 4:
Figure 5:

Answer:

D. 5x-15y+20

Step-by-step explanation:

If you distribute 5 to all the numbers in the parenthesis, you get 5x-15y+20. This is because 5 multiplied by x is 5x, 5 multiplied by 3y is 15y, and 5 multiplied by 4 is 20. I hope this helps!

Answer:

D

Step-by-step explanation:

you have to distribute the number 5 between all of the numbers. Is because

5 * x is 5x  3y is 15y, and 5 *4 is 20.

To transform the regression equation, you would divide both sides of the equation by the square root of Var(ut), which is σ√(zt). This transformation helps in obtaining the transformed regression coefficients and standard errors that account for the heteroscedasticity in the error term.

When the variance of the error term (ut) is related to a known variable (zt), it implies the presence of heteroscedasticity in the regression model. Heteroscedasticity means that the variability of the error term is not constant across different levels of the independent variables.

To address this issue, we can transform the regression equation by dividing both sides by the square root of the variance of the error term, which is σ√(zt). This transformation is known as the weighted least squares (WLS) estimation.

By dividing both sides of the equation, we can obtain the transformed regression equation with the error term divided by its standard deviation. This transformation accounts for the heteroscedasticity by giving different weights to the observations based on the variability of the error term. It allows for a more appropriate estimation of the regression coefficients and standard errors, as it gives more weight to observations with smaller error variances and less weight to observations with larger error variances.

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Answer:

b. 3

Step-by-step explanation:

since a+b+c=180

so you do 100+50+c=180

150+c=180

180- 150=30

:)

Answer:

huh how can i compete mine cant but that one did

Step-by-step explanation:

Answer:

From my thinking prospection the answer should be 18....

Answer:

A

Step-by-step explanation:

given the x- intercepts x = a and x = b then the corresponding factors are

(x - a) and (x - b)

the equation of the quadratic is then the product of the factors , that is

y = a(x - a)(x - b) ← a is a multiplier

here the x- intercepts are x = 4 and x = 8 , then factors are

(x - 4) and (x - 8 ) , so

y = a(x - 4)(x - 8)

to find a substitute any other point on the graph into the equation

given vertex = (6, - 4 ) , then

- 4 = a(6 - 4)(6 - 8)

- 4 = a(2)(- 2) = - 4a ( divide both sides by - 4 )

1 = a , then

y = (x - 4)(x - 8) ← expand using FOIL

y = x² - 12x + 32

The probability that a randomly chosen student is a junior or has voted in the last presidential election is 0.8 or 80%.

To find the probability that a randomly chosen student is a junior or has voted in the last presidential election, we can use the formula

P(A or B) = P(A) + P(B) - P(A and B)

where A and B are two events.

Let's assume that there are 1000 students in the population, and 400 of them are juniors and 600 of them have voted in the last presidential election. Furthermore, let's assume that 200 students are both juniors and have voted in the last presidential election.

Then, the probability that a randomly chosen student is a junior or has voted in the last presidential election is

P(junior or voted) = P(junior) + P(voted) - P(junior and voted)

= 400/1000 + 600/1000 - 200/1000

= 0.8

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-- The given question is incomplete, the complete question is

"If total number of students are 1000, and 400 of them are juniors and 600 of them have voted in the last presidential election. Furthermore, 200 students are both juniors and have voted in the last presidential election. Then find the probability that a randomly chosen student is a junior or has voted in the last presidential election?" --

Answer:

12.25

Step-by-step explanation:

The commitment scheme for committing a single bit 0 or 1 is a bad commitment scheme.

Consider the following commitment scheme for committing a single bit 0 or 1: for 0, pick a random even element a (mod p) and commit a² (mod p). For 1, pick a random odd element and similarly commit its square. Let p be a big prime. To show that this is a bad commitment scheme, let's look at an example.

Suppose that p = 7 and the sender wants to send the value 1. Therefore, he chooses an odd number, say a = 3, and computes a² = 9 mod 7 = 2. Now he sends 2 to the receiver. The receiver has two possible options for guessing the number sent: 1 or 0. Let's assume that he guesses the number 0 and then he can choose any even number, say a = 2.

Now he computes a² = 4 mod 7. As 4 is the residue of an even number, it's impossible to distinguish between the values sent by the sender. Therefore, this is a bad commitment scheme.

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This is not a good commitment scheme.

A good commitment scheme should satisfy two properties: hiding and binding. Hiding means that the committed value should be computationally infeasible to determine without the commitment opening. Binding means that once the commitment is opened, it should be computationally infeasible to change the committed value.

In the given commitment scheme, the committed value is the square of a randomly chosen even or odd element modulo p. However, this scheme is not secure because it does not satisfy the hiding property.

To see why the hiding property is not satisfied, consider the case when the committed value is 0. Since we commit the square of a randomly chosen even element, any square root of the committed value modulo p will reveal the committed value. In this case, finding the square root of a modulo p, where a is even, is straightforward and does not require excessive computation. Therefore, an attacker can easily determine the committed value without knowing the opening.

This lack of hiding makes the commitment scheme insecure because an adversary can guess the committed value by calculating the square root. Thus, an attacker can break the hiding property of the commitment scheme, rendering it ineffective for secure communications.

In summary, the given commitment scheme is not a good one as it fails to satisfy the hiding property. It is important to use a commitment scheme that provides both hiding and binding properties to ensure the security and integrity of the committed values.

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OLAP is used for the analysis of accumulated data, and it requires a data warehouse.

while OLTP is used for transaction processing and requires a database optimized for real-time transaction processing. The analysis of accumulated data is known as Online Analytical Processing (OLAP), and it involves a data warehouse. OLAP is a business intelligence tool used for multi-dimensional analysis of large data sets, while a data warehouse is a central repository that stores historical and current data from multiple sources in a structured manner. OLAP allows users to perform complex queries on large data sets, analyze trends, and make informed business decisions. It is often used in data mining and decision support systems. OLAP tools can be used to slice and dice data, drill down to more detailed data, and roll up to higher levels of summary data. OLAP is primarily used for analytical purposes and is not designed for transaction processing. On the other hand, a data warehouse is designed to support OLAP and is used for storing large amounts of historical data that can be queried and analyzed. Data is extracted from various sources, transformed, and loaded into the data warehouse in a structured format, enabling efficient querying and analysis. OLTP (Online Transaction Processing), on the other hand, is a type of transaction processing that is designed for processing real-time transactions in databases. It is used for day-to-day operations such as order processing, inventory management, and customer management. OLTP is optimized for processing large volumes of transactions in real-time, and is not suitable for analytical purposes.

In summary, OLAP is used for the analysis of accumulated data, and it requires a data warehouse, while OLTP is used for transaction processing and requires a database optimized for real-time transaction processing.

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.

Given:

The equation is

\(\sin\left(x+\dfrac{7\pi}{2}\right)=-\dfrac{\sqrt{3}}{2}\)

To find:

The solutions of given equation over the interval \([0,2\pi]\).

Solution:

We have,

The equation is

\(\sin\left(x+\dfrac{7\pi}{2}\right)=-\dfrac{\sqrt{3}}{2}\)

\(\sin\left(x+\dfrac{7\pi}{2}\right)=-\sin \dfrac{\pi }{3}\)

\(\sin\left(x+\dfrac{7\pi}{2}\right)=\sin (-\dfrac{\pi }{3})\)

If \(\sin x=\sin y\), then \(x=n\pi +(-1)^ny\).

Over the interval \([0,2\pi]\).

\(x+\dfrac{7\pi}{2}=4\pi-\dfrac{\pi }{3}\) and \(x+\dfrac{7\pi}{2}=5\pi+\dfrac{\pi }{3}\)

\(x=\dfrac{11\pi }{3}-\dfrac{7\pi}{2}\) and \(x=\dfrac{16\pi}{3}-\dfrac{7\pi}{2}\)

\(x=\dfrac{22\pi-21\pi }{6}\) and \(x=\dfrac{32\pi-21\pi }{6}\)

\(x=\dfrac{\pi}{6}\) and \(x=\dfrac{11\pi }{6}\)

Therefore, the two solutions are tex]x=\dfrac{\pi}{6}\) and \(x=\dfrac{11\pi }{6}\).

Answer:

C. π/6 & 11π/6

Step-by-step explanation:

If you graph the equation ( Sin (x+7π/2)=-√3/2) and look between 0 & 2π, you'll see that the lines intersect the x-axis at π/6 & 11π/6.

Answer:

38 I'm 90% sure here so sorry if I am wrong

Step-by-step explanation:

Answer:

38

Step-by-step explanation:

I got it right

Answer:

Parallel

Step-by-step explanation:

In order to find out if two lines or parallel, perpendicular, or neither, we must first find the slope between the two lines and then compare them.

Lets take our line j first.

We can use the formula \(\frac{y_2-y_1}{x_2-x_1}\) and then plug in our coordinates to evaluate.

\(\frac{2-7}{3-10}\)

Evaluate. A negative divided by a negative is a positive.

\(\frac{-5}{-7}=\frac{5}{7}\)

Now lets do line k.

\(\frac{5-10}{3-10}\)

Evaluate

\(\frac{-5}{-7} =\frac{5}{7}\)

Now, to find out if a line is parallel or perpendicular we must first define what they mean in terms of slope.

Perpendicular: They form a right angle and they are the opposite reciprocal

Parallel: They are the same.

If we look at the slopes, we can conclude:

\(\frac{5}{7}=\frac{5}{7}\)

Therefore they are parallel.