Put the quadratic


y=2x^2-4x+2


into the quadratic formula


enter the number that belongs in the green box.

Answer: A

Step-by-step explanation:

The equation \(y=\frac{2}{3}x-1\) is in slope-intercept form(\(y=mx+b\)).  For a line to be parallel to another, it must have the same slope.  The slope of the line given is 2/3.  Thus, simply plug in 2/3 for the slope to get: \(y=\frac{2}{3} x+b\).  Then, because the line passes through -6, 6, plug those values in for x and y, respectively.  \(6=\frac{2}{3} *-6+b\)

\(6=-4+b\)

\(10=b\)

Thus, the equation of the line is \(y = \frac{2}{3} x+10\), or A

Hope it helps :D

If you turn off the spaceship's engine and it continues to glide in a straight line in the direction it was when the engine was stopped, we first have to find the speed of the spacecraft, which can be given by the formula:

At t=1, the coordinates of the spaceship are (1,0). If we switch off the engine at this moment, the spaceship will continue to move in a straight line in the direction of its velocity vector, which is (1,0) at t=1.

At t=2, the coordinates of the spaceship are (2,1). If we switch off the engine at this moment, the spaceship will continue to move in a straight line in the direction of its velocity vector, which is (1,2) at t=2.

Therefore, at t=1 the spacecraft will move horizontally in a straight line. At t=2, the ship will move on a straight line with slope 2, passing through the point (2,1).

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Answer: 6.4

Step-by-step explanation:

square root of 41 is 6.403124 and so on so round down to 6.4

Answer:

Step-by-step explanation:

A: dividend;

B: divisor;

C: quotient

Two ways of writting the linear function are:

A general linear function can be written as:

y = a*x + b

Where a is the slope and b is the y-intercept.

Here we know that the slope is 2/5, replacing that we will get:

y = (2/5)*x + b

We also know that this line passes through (12, -5), then we can replace these values to get:

-5 = (2/5)*12 + b

-5 = 24/5 + b

-5 - 24/5 = b

-25/5 - 24/5 = b

-49/5 = b

Then the linear equation is:

y = (2/5)*x - 49/5

Other form of writting this is the standard form, multiply both sides by 5.

5y = 2x - 49

5y - 2x = -49

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The correct answer for the first question is D. The data are discrete because the data can take on any value in an interval.

Quiz scores from a college level statistics course are analyzed to determine student progress. This is not a voluntary response sample because the students taking the quiz are required to participate, and their scores are collected for the purpose of analyzing their progress.

For the second question, the data described, which is the number of donations a charity receives each month, is discrete. Discrete data can only take on specific values and cannot be divided into smaller, meaningful intervals. In this case, the number of donations can only be whole numbers, such as 0, 1, 2, and so on. It cannot take on any value in an interval or be represented by fractions or decimals. Therefore, the data is discrete.

It is important to distinguish between discrete and continuous data when analyzing and interpreting data, as different statistical methods and techniques are used for each type. Discrete data is usually counted or measured in whole numbers, while continuous data can take on any value within a given range or interval.

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Answer:

The probability is really close.

Step-by-step explanation:

10/2 = 5. So both die must land on a 5.

Answer:

there is a 18.75% chance of rolling a sum of 10

Step-by-step explanation:

Answer:

x = 2

Step-by-step explanation:

2x-3 = 1

use addition property of equality (add 3 to each side to keep balance)

2x = 4

use division property of equality (divide both sides by 2)

4/2 = 2

so,

x = 2

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The probability that the diameter of a selected bearing is greater than 123 millimeters is 0.4641 (rounded to four decimal places).Hence, the correct option is A) 0.4641.

The given question is asking to find the probability that the diameter of a selected bearing is greater than 123 millimeters, when the diameters of ball bearings are distributed normally with the mean diameter of 126 millimeters and the standard deviation of 33 millimeters.

Given, Mean diameter, µ = 126 millimeters

Standard deviation, σ = 33 millimeters

We need to find the probability that the diameter of a selected bearing is greater than 123 millimeters.

Now, the given distribution is a normal distribution. Therefore, to solve the problem, we will use the z-distribution formula, which is:

z = (x - µ) / σ,

where x is the value of the variable, µ is the mean, and σ is the standard deviation.

We will substitute the given values in the above formula,

z = (x - µ) / σ=> z = (123 - 126) / 33=> z = -0.09

Using a standard normal distribution table, the probability that z is greater than -0.09 is 0.4641, which

(z > -0.09) = 0.4641

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The total profit will be maximized by using the objective function Z = 22x₁ + 53x₂ + 124x₃ + 25x₄ + 156x₅ + 37x₆ + 108x₇ + x₈

To formulate this problem mathematically, we need to define the decision variables, objective function, and constraints.

We will define a binary decision variable, \(x_i\), which indicates whether or not to invest in project i. If \(x_i\) = 1, the project i will be funded, and if \(x_i\) = 0, the project i will not be funded.

The objective is to maximize the total profit earned by investing in the selected projects. Thus, the objective function can be written as:

Maximize Z = 22x₁ + 53x₂ + 124x₃ + 25x₄ + 156x₅ + 37x₆ + 108x₇ + x₈

This objective function adds the profit of all the projects that will be funded based on the decision variable \(x_i\).

The total investment cannot exceed the available investment budget of $1 million, which can be written as:

0.50x₁ + 1.40x₂ + 1.70x₃ + 1.10x₄ + 4.00x₅ + 0.60x₆ + 2.50x₇ + 1.10x₈ <= 1

Additionally, the decision variables are binary, which means that x_i can only be either 0 or 1. Therefore, we can add binary constraints for each decision variable as follows:

x_i = 0 or 1 for i = 1,2,3,4,5,6,7,8

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The volume of the 'scooped' pyramid is approximately 26.6667 cubic units.

To find the volume of the 'scooped' pyramid, we first need to determine the area of its base. Since the cross-sectional area of the pyramid is x 4 at a distance x from the tip, we can assume that the area at the tip is zero. This means that the area of the base is 4 times the area at a distance of 5 from the tip (since the distance from tip to base is 5).

Therefore, the area of the base is 4x4 = 16 square units. To find the volume, we can use the formula for the volume of a pyramid, which is:

Volume = (1/3) x Base Area x Height

In this case, the height of the pyramid is 5 units. So, we can substitute the values we have:

Volume = (1/3) x 16 x 5

Volume = 26.6667 cubic units

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To prove this identity, we start with Fermat's Little Theorem, which states that if p is a prime number and a is any integer coprime to p, then a^(p-1) ≡ 1 (mod p).

Using this theorem, we can rewrite the given identity as a^(p-1) * a(p-2) ≡ 1 (mod p^2).

Next, we can multiply both sides by a to get a^(p-1) * a(p-1) ≡ a (mod p^2).

Since a and p are coprime, we can use Euler's Totient Theorem, which states that a^φ(p) ≡ 1 (mod p) where φ(p) is the Euler totient function. Since p is prime, φ(p) = p-1, so a^(p-1) ≡ 1 (mod p).

Using this result, we can rewrite our identity as a^(p-1) * a(p-1) * a^-1 ≡ a^(p-1) ≡ 1 (mod p), which implies that a^(p-1) ≡ 1 (mod p^2).

Therefore, we have proven the identity a p(p−1) ≡ 1 (mod p 2 ), where a is coprime to p, and p is prime.

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Answer:

-1 -3

Step-by-step explanation:

The expressions are simplified to give;

7. 4n³(3n² + 4)

8. -3x(3x² - 4)

9. 5(k² - 8k + 2)

10. -10(6 + 6n² + 5n³)

11. 3(6n³ - 4n - 7)

12. 9(7n³ + 9n + 2)

Algebraic expressions are defined as mathematical expressions that are composed of variables, coefficients, terms, factors and constants.

These algebraic expressions are also made up of arithmetic operations, such as;

To factorize the expressions, we have;

12n⁵ + 16n³

Find the common term

4n³(3n² + 4)

-9x³ - 12x

find the common term

-3x(3x² - 4)

5k² - 40k + 10

find the common terms

5(k² - 8k + 2)

-60 + 60n² + 50n³

find the common term

-10(6 + 6n² + 5n³)

18n³ -12n - 21

find the common term

3(6n³ - 4n - 7)

63n³ + 81n + 18

Find the common term

9(7n³ + 9n + 2)

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Answer:

248 tickets were students and 124 were adults! 372 ÷ 3 = 124 and 124 × 2 = 248 and 248 + 124 = 372 so

124 tickets for adults

and 284 tickets for students

Answer:

y = 4x - 25

Step-by-step explanation:

y = 4x + b

-1 = 4(6) + b

-1 = 24 + b

-25 = b

The function arithmetic, if \(f(x)=3x and g(x)\)\(=x^2\), then \((f+g)(x)\) is\(x^2 + 3x\)

Function arithmetic is a mathematical operation of combining one or more functions to produce a new, equivalent function. Function arithmetic can also involve combining the same function multiple times, such as adding the same function twice, or taking the square root of a function. These operations allow a mathematician to manipulate functions in order to simplify or solve equations. Function arithmetic is also important in calculus, where operations such as differentiation and integration are used to calculate the area under or above a curve.

if\(f(x)=3x\) and \(g(x)=x^2\)then find\((f+g)(x)\)

\((f+g)(x) = f(x) + g(x)\)

\(= 3x + x^2\)

\(= x^2 + 3x\)

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Answer:

i think option d is the correct answer of this question.

Let's verify

As LHS not equal to RHS

Answer:

The real zeros of the quadratic function f(x) = -4x^2 - 6x + 1 are approximately -0.15 and -1.35.

Step-by-step explanation:

To find the real zeros of the quadratic function f(x) = -4x^2 - 6x + 1, we need to find the values of x that make f(x) equal to zero. We can do this by using the quadratic formula:

x = [-b ± sqrt(b^2 - 4ac)] / 2a

where a, b, and c are the coefficients of the quadratic equation ax^2 + bx + c.

In this case, a = -4, b = -6, and c = 1. Substituting these values into the quadratic formula, we get:

x = [-(-6) ± sqrt((-6)^2 - 4(-4)(1))] / 2(-4)

x = [6 ± sqrt(52)] / (-8)

x = [6 ± 2sqrt(13)] / (-8)

These are the two solutions for the quadratic equation, which we can simplify as follows:

x = (3 ± sqrt(13)) / (-4)

Therefore, the real zeros of the quadratic function f(x) = -4x^2 - 6x + 1 are approximately -0.15 and -1.35.

We are 99% confident that the true average stance duration among elderly individuals lies within the range of 744.56 ms to 857.44 ms.

To test whether the true average stance duration is larger among elderly individuals than among younger individuals, we can perform a one-tailed independent samples t-test. The null hypothesis (H0)

Using the t-test, we compare the means and standard deviations of the two samples and calculate the test statistic

a) To calculate a 99% confidence interval for the true average stance duration among elderly individuals, we can use the sample mean, sample standard deviation, and the t-distribution.

Given:

Older adults: n = 28, sample mean = 801, sample standard deviation = 117

Using the formula for a confidence interval for the mean, we have:

Margin of error = t * (sample standard deviation / √n)

Since the sample size is relatively large (n > 30), we can use the z-score instead of the t-score for a 99% confidence interval. The critical z-value for a 99% confidence level is approximately 2.576.

Calculating the margin of error:

Margin of error = 2.576 * (117 / √28) ≈ 56.44

The confidence interval is then calculated as:

Confidence interval = (sample mean - margin of error, sample mean + margin of error)

Confidence interval = (801 - 56.44, 801 + 56.44) ≈ (744.56, 857.44)

b) To test whether the true average stance duration is larger among elderly individuals than among younger individuals, we can perform a one-tailed independent samples t-test.

The null hypothesis (H0): The true average stance duration among elderly individuals is equal to or less than the true average stance duration among younger individuals.

The alternative hypothesis (Ha): The true average stance duration among elderly individuals is larger than the true average stance duration among younger individuals.

. With the given data, perform the t-test and obtain the p-value.

c) To construct a 95% confidence interval for the difference in means between older and younger adults, we can use the formula for the confidence interval of the difference in means.

Given:

Older adults: n1 = 28, sample mean1 = 801, sample standard deviation1 = 117

Younger adults: n2 = 16, sample mean2 = 780, sample standard deviation2 = 72

Calculating the standard error of the difference in means:

Standard error = √((s1^2 / n1) + (s2^2 / n2))

Standard error = √((117^2 / 28) + (72^2 / 16)) ≈ 33.89

Using the t-distribution and a 95% confidence level, the critical t-value (with degrees of freedom = n1 + n2 - 2) is approximately 2.048.

Calculating the margin of error:

Margin of error = t * standard error

Margin of error = 2.048 * 33.89 ≈ 69.29

The confidence interval is then calculated as:

Confidence interval = (mean1 - mean2 - margin of error, mean1 - mean2 + margin of error)

Confidence interval = (801 - 780 - 69.29, 801 - 780 + 69.29) ≈ (-48.29, 38.29)

Comparison with part (b): In part (b), we performed a one-tailed test to determine if the true average stance duration among elderly individuals is larger than among younger individuals. In part (c), the 95% confidence interval for the difference in means (-48.29, 38.29) includes zero. This suggests that we do not have sufficient evidence to conclude that the true average stance duration is significantly larger among elderly individuals compared to younger individuals at the 95% confidence level.

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Answer:

y=68

Step-by-step explanation:

54=y−14

54+14=y

68=y

Answer:

18

Step-by-step explanation:

6*6/2 = 18

We have to find the area of the region between y=x^(1/2) and y=x^(1/3) for 0≤x≤1.

To find the area of the region between y=x^(1/2) and y=x^(1/3) for 0≤x≤1, we have to integrate x^(1/2) and x^(1/3) with respect to x. That is, Area = ∫0¹ [x^(1/2) - x^(1/3)] dx= [2/3 x^(3/2) - 3/4 x^(4/3)] from 0 to 1= [2/3 (1)^(3/2) - 3/4 (1)^(4/3)] - [2/3 (0)^(3/2) - 3/4 (0)^(4/3)]= 0.2857 square units

Therefore, the area of the region between y=x^(1/2) and y=x^(1/3) for 0≤x≤1 is 0.2857 square units. Note: The question  but the answer has been provided in the format requested.

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To determine an expression in terms of m and l for the moment of inertia of the masses about axis a, we need some additional information about the configuration of the masses and the axis.

The moment of inertia depends on the distribution of masses relative to the axis of rotation. It is a measure of an object's resistance to rotational motion. The formula for the moment of inertia varies depending on the specific shape and distribution of masses.

If you can provide more details about the arrangement of masses and the axis of rotation, I can help you derive the expression for the moment of inertia in terms of m and l.

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Answer:

y = -28/5

Step-by-step explanation:

R = 7x +5y

35 = 7(9) + 5y

35 = 63 +5y

-5y= 63- 35

-5y/-5 = 28/-5

Y = -28/5

Answer:

Y = - \(\frac{28}{5}\)

Step-by-step explanation:

Given

R = 7X + 5Y ← substitute X = 9, R = 35

35 = 7(9) + 5Y

35 = 63 + 5Y ( subtract 63 from both sides )

- 28 = 5Y ( divide both sides by 5 )

- \(\frac{28}{5}\) = Y

Answer:

C

Step-by-step explanation:

Multiplying 1.6 gets you how far he walks, and then dividing the answer gives you how many times he would be stops, roughly 6.4 times

Answer: 7/74

Step-by-step explanation:

P(G)=15/37*14/36

      = 1/37*7/2

      = 7/74

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"After hearing his puppy yelp, for a moment after, Mr. Wright has a vivid auditory memory of the puppy's high pitch bark. Mr. Wright's experience clearly demonstrates echoic memory."

Echoic memory is a form of sensory memory characterized by the temporary retention of auditory stimuli or sounds. It involves the capacity to retain and retrieve auditory information that has recently been perceived. As an example, Mr. Wright's strong recollection of the high-pitched bark of the puppy following its auditory encounter serves as evidence of his echoic memory. This suggests that the auditory details of the puppy's bark were briefly stored in his sensory memory, enabling him to mentally replay or recollect the sound within his mind for a short duration.

To put it differently, echoic memory pertains to the transient preservation of auditory input. It allows individuals to temporarily hold and recall auditory information that they have just heard. To illustrate, when Mr. Wright heard the puppy's shrill bark, he demonstrated his echoic memory by vividly recalling the sound afterward. This indicates that the sensory memory briefly encoded and stored the auditory aspects of the puppy's bark, granting Mr. Wright the ability to mentally replay or retrieve the sound within his consciousness for a brief period.

In summary, echoic memory represents a specific type of sensory memory that focuses on the brief retention of auditory stimuli. It enables individuals to remember and reproduce auditory information that has been recently perceived. In the case of Mr. Wright, his ability to vividly recall the puppy's high-pitched bark highlights his possession of echoic memory. This suggests that the auditory details of the bark were momentarily stored in his sensory memory, allowing him to mentally replay or remember the sound within his mind for a short period of time.

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Answer:

Step-by-step explanation:

place a point at (0,0)- the starting point

next point at (2,1) - in 2 minutes she is 1 block away

Next point (4,2)- in 4 minutes she will be 2 blocks away

Connect those 3 points

then place a point at (9, 2)- 5 minutes later she was still at the store

Connect points

Point at (11,1) and (12,0) and connect