Show how to multiply 6 x 298 using friendly numbers and then use properties and mental math.

Therefore, the largest prime whose multiples need to be considered to extend the Sieve of Eratosthenes to 200 is 13.

The Sieve of Eratosthenes is a method used to find all prime numbers up to a given limit by iteratively marking the multiples of each prime number. To extend the Sieve of Eratosthenes to 200, determine the largest prime whose multiples need to be considered.

Begin by creating a list of numbers from 2 to 200. Mark 2 as the first prime number. Proceed to eliminate all multiples of 2 from the list (4, 6, 8, ...).

Move to the next unmarked number, which is 3. Mark it as a prime and eliminate all its multiples from the list (6, 9, 12, ...).

Continue this process for each unmarked number: marking it as a prime and eliminating its multiples. The next unmarked number after 3 is 5.

Continue this process until you reach the square root of the upper limit, which is √200 ≈ 14.14. Therefore, the largest prime whose multiples need to be considered to extend the Sieve of Eratosthenes to 200 is 13. Multiples of primes greater than 13 that fall below 200 have already been eliminated during the sieving process.

(x) is continuous and twice differentiable for ±≥ 0. Suppose f”( ±) ≤ 5 for all ±≥ 0 and f (0) = f’(0) = 0. The linear function with a derivation of 0 is y = 0. Thus we have g(t) = 0. The quadratic function with concavity 5 and initial slope 0 is y = 5/2 x^2 + 0x. Therefore, h(t) = (5/2)x^2

Given the conditions, the function g(t)=3t is always greater than or equal to f'(t), as it represents the maximum possible velocity considering the maximum constant acceleration of 3 over time.

Given that f(t) is a continuous and twice-differentiable function for t≥0 and that f''(t)≤3 for all t≥0, the function f(t) represents the position of an object over time, with f'(t) and f''(t) being the velocity and acceleration, respectively. Further, given that f(0)=f'(0)=0, it is stated that at time t=0, the object is at the origin and is at rest.

Using the Racetrack Principle, we know that velocity must be less than or equal to the maximum acceleration times time. In plain terms, the actual speed of a body at a time 't' (f'(t)) could not be more than the distance it would have covered if it was accelerating at the maximum possible rate from rest. Since f''(t)≤3, the maximum possible acceleration is 3.

Thus, for all t≥0, if we consider the linear function g(t)=3t, we can say g(t) is greater than or equal to f'(t). Because even if the acceleration was at the maximum value of 3 for the entire duration, the velocity wouldn't exceed 3t.

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[tex]x+2y=8[/tex] can be written in a slope-intercept form as [tex]y = 4-\dfrac{x}{2}[/tex].

A slope is also known as the gradient of a line is a number that helps to know both the direction and the steepness of the line.

[tex]x+2y=8[/tex]

To find the function, simply isolate y.

[tex]x+2y=8\\2y = 8-x\\y = \dfrac{8-x}{2}\\\\y = \dfrac{8}{2}-\dfrac{x}{2}\\\\y = 4-\dfrac{x}{2}[/tex]

Hence, [tex]x+2y=8[/tex] can be written in a slope-intercept form as [tex]y = 4-\dfrac{x}{2}[/tex].

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To determine which fraction is greater, we convert both fractions to decimal form. Six over ten is greater than one over two.

To determine which fraction is greater, six over ten or one over two, we can convert both fractions to decimal form and compare them. Six over ten can be simplified to three over five, which is equal to 0.6. One over two is equal to 0.5. Since 0.6 is greater than 0.5, we can conclude that six over ten is greater than one over two.

Answer:

the answer is 12, 15. and 18.

Step-by-step explanation:

The value of the piece of equipment after 2 years is $6,200.00.

A linear equation is an algebraic equation of the form y=mx+b, where m is the slope and b is the y-intercept, and only a constant and a first-order (linear) term are included. Sometimes, the aforementioned is referred to as a "linear equation of two variables," with y and x serving as the variables.

The equipment is depreciated linearly, which means that its value decreases at a constant rate each year. We can find the rate of depreciation by subtracting the scrap value from the purchase price, and then dividing by the number of years of useful life:

Rate of depreciation = (Purchase price - Scrap value) / Useful life

Rate of depreciation = (9000 - 2000) / 5

Rate of depreciation = 1400

This means that the equipment loses $1,400 of value each year. After 2 years, the value of the equipment will be:

Value after 2 years = Purchase price - (Rate of depreciation x Number of years)

Value after 2 years = 9000 - (1400 x 2)

Value after 2 years = 6200

Therefore, the value of the piece of equipment after 2 years is $6,200.00.

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To find the maximum of the function f(x, y) = y² + xy – x² on the square 0 ≤ x ≤ 6, 0 ≤ y ≤ 6, we need to find the critical points by taking the partial derivatives with respect to x and y.

To find the maximum of the function f(x, y), we need to find the critical points by taking the partial derivatives with respect to x and y:

[tex]f_x[/tex] = y - 2x

[tex]f_y[/tex] = 2y + x

Setting these derivatives equal to zero and solving for x and y, we get the critical point (2,2). To determine if this point is a maximum, minimum, or point of inflection, we can use the second derivative test. Taking the second partial derivatives:

[tex]f_{xx[/tex] = -2, [tex]f_{yy[/tex] = 2, [tex]f_{xy[/tex] = 1

Using the discriminant D = [tex]f_{xx} \times f_{yy} - {(f_{xy})}^2[/tex], we have D = [tex](-2) \times 2 - {1}^2[/tex] = -4. Since D is negative, the critical point (2,2) is a saddle point, which means there is no maximum or minimum on the given square.

Answer: The answer is B, just got this one on a test.

Step-by-step explanation:

Final answer:

Anne Treisman is associated with the Attenuation Model, which posits that unattended information is not lost but simply weakened. This model contrasts with filter theories, emphasizing that meaningful information can penetrate the attention barrier for further processing.

Explanation:

The theorist associated with the idea that information not actively attended to is not lost but rather decreases in signal strength and is therefore not given any significance or future attention is Anne Treisman. Treisman proposed the Attenuation Model, which challenges the filter theory, suggesting that selection of information begins at the physical or perceptual level. However, unlike the filter theory that assumes unattended information is completely blocked, Treisman believes that unattended information is just weakened or attenuated. This means that any highly meaningful or pertinent information, such as one's own name, can still get through the filter for further processing at the level of meaning.

Supporting this model, late selection models like that of Deutsch and Deutsch also propose that all information is processed on the basis of meaning, but only task-relevant information reaches conscious awareness. This concept is in line with the idea of subliminal perception where unconscious information can be fully processed for meaning.

Therefore, the attenuation model and late selection models suggest that our attention selects information but does not completely eliminate the processing of unattended stimuli, particularly if the unattended information holds some form of intrinsic importance or meaning to the individual.