HELP ME ASAP PLEASE!!!!!!!!!

Answer:

To solve for r, we can start by simplifying the equation:

1475/2pi = (3/4r^2pi) + (1/4pi*(r-15)^2) + (1/4pi(r-25)^2)

Multiplying both sides by 2*pi:

1475 = 3/4r^2pi2 + 1/4pi*(r-15)^22 + 1/4pi*(r-25)^2*2

1475 = 3/2r^2pi + 1/2pi(r-15)^2 + 1/2pi(r-25)^2

Multiplying both sides by 2:

2950 = 3r^2pi + pi*(r-15)^2 + pi*(r-25)^2

Distributing pi:

2950 = 3r^2pi + pir^2 - 30pir + 225pi + pir^2 - 50pir + 625pi

Combining like terms:

2950 = 5r^2pi - 80pir + 850*pi

Rearranging:

5r^2pi - 80pir + 850*pi - 2950 = 0

Simplifying:

5r^2pi - 80pir + 675*pi = 0

Dividing both sides by 5*pi:

r^2 - 16*r + 135 = 0

This is a quadratic equation, which can be solved using the quadratic formula:

r = (-(-16) ± sqrt((-16)^2 - 4(1)(135))) / (2(1))

r = (16 ± sqrt(256 - 540)) / 2

r = (16 ± sqrt(284)) / 2

r ≈ 1.7321 * 16 or r ≈ 8.2679

Since r represents the distance from the center of the octagon to a vertex, only the larger value of r makes sense in this context.

Therefore, r ≈ 8.2679 feet.

To find the area of the region in which the cow can graze, we can divide the octagon into eight congruent isosceles triangles with base 25 feet and height equal to the distance from the center to a side (which is equal to r).

The area of each triangle is (1/2)bh = (1/2)(25)(8.2679) = 103.3494 square feet.

Multiplying by 8 to account for all eight triangles:

8 * 103.3494 = 826.7952 square feet.

Rounding to the nearest square foot:

The area in which the cow can graze is approximately 827 square feet

Answer:

True

Step-by-step explanation:

Vertical angles are right angle that is 90°

A supplementary angle is an angle that forms up by 2 angles with the sum of 180°.

It is true because 2 vertical angles form a supplementary angle.

Answer:

The Davenports' medical expenses contribute to their total itemized deductions are $375 (7.5% for 2019).

The costs you incurred for state and local income or sales taxes, real estate taxes, personal property taxes, mortgage interest, and disaster losses are all included in itemised deductions. You can also count charitable donations and a portion of your out-of-pocket medical and dental costs.

Currently for the 2019 (due 2020), you can deduct medical expenses that exceed 7.5% of your AGI, but back then in 2019, the threshold was 7.5%, not 10%.

So the Davenports can only deduct

$5,250 - ($65,000 x 7.5%) = $375

if they decided to itemize their deductions.

The threshold will increase back to 10% starting 2020 (due 2021) tax returns.

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Answer: y = 4(x² − 1)(x² − 4).

Step-by-step explanation:

We need to find the equation of the curve that passes through the point (2, 0).

We start by separating the variables dy/dx and y and integrating both sides:

(2x² - 5) dy/dx = 8x(y + 9)

dy/(y + 9) = (4x/(2x² - 5)) dx

Integrating both sides:

ln|y + 9| = 2ln|2x² - 5| + C

where C is the constant of integration.

Rewriting in exponential form:

|y + 9| = e^(2ln|2x² - 5| + C)

|y + 9| = e^(ln|2x² - 5|² + C)

|y + 9| = k(2x² - 5)²

where k is the constant of integration.

Since the curve passes through the point (2, 0), we can substitute these values into the equation above to find k:

|0 + 9| = k(2(2)² - 5)²

9 = k(36)

k = 1/4

Substituting this value of k back into the equation, we get:

|y + 9| = (1/4)(2x² - 5)²

y + 9 = (1/4)(2x² - 5)² or y + 9 = -(1/4)(2x² - 5)²

Simplifying the right-hand side of each equation, we get:

y + 9 = (1/4)(4x⁴ - 20x² + 25)

or

y + 9 = -(1/4)(4x⁴ - 20x² + 25)

Expanding and simplifying, we get:

y = 4x⁴/4 - 5x²/2 + 25/4 - 9 or y = -4x⁴/4 + 5x²/2 - 25/4 - 9

y = x⁴ - 5x² + 19/4 or y = -x⁴/4 + 5x²/2 - 41/4

Thus, the equation of the curve passing through the point (2, 0) with the given gradient is y = 4(x² − 1)(x² − 4).

(a) The interval (-∞, ∞).

(b) The interval (-∞, ∞).

(c) The interval (-∞, ∞).

(d) The interval (-π/2, π/2) \ {0}.

(a) The longest interval in which the given initial value problem is certain to have a unique twice-differentiable solution is the interval where the coefficient function, 3t, is continuous and bounded. Since 3t is a continuous and bounded function for all t in the interval (-∞, ∞), the given initial value problem is certain to have a unique twice-differentiable solution for all t in (-∞, ∞).

(b) The longest interval in which the given initial value problem is certain to have a unique twice-differentiable solution is the interval where the coefficient functions, t(t - 4), 3t, and 4, are continuous and bounded. Since t(t - 4), 3t, and 4 are continuous and bounded functions for all t in the interval (-∞, ∞), the given initial value problem is certain to have a unique twice-differentiable solution for all t in (-∞, ∞).

(c) The longest interval in which the given initial value problem is certain to have a unique twice-differentiable solution is the interval where the coefficient functions, cost and In|t|, are continuous and bounded. Since cost and In|t| are continuous and bounded functions for all t in the interval (-∞, ∞), the given initial value problem is certain to have a unique twice-differentiable solution for all t in (-∞, ∞).

(d) The longest interval in which the given initial value problem is certain to have a unique twice-differentiable solution is the interval where the coefficient functions, x - 2, 1, and (x - 2)tanx, are continuous and bounded. Since x - 2, 1, and (x - 2)tanx are continuous and bounded functions for all x in the interval (-π/2, π/2) \ {0} , the given initial value problem is certain to have a unique twice-differentiable solution for all x in (-π/2, π/2) \ {0}.

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The given question is incomplete, the complete question is:

determine the longest interval in which the given initial value problem is certain to have a unique twice- differentiable solution. Do not attempt to find the solution. (a) ty" + 3y = 1, y(1) = 1, y'(1) = 2   (b)   t(t – 4)y" + 3ty' + 4y = 2, y(3) = 0, y'(3) = -1 (c)   y" + (cost)y' + 3( In |t|) y = 0, y(2) = 3, y'(2) = 1 (d) (x - 2)y"+y' +(x - 2)(tan x) y = 0, y(3) = 1, y'(3) = 2

To solve the equation -12x + 24 = 0, we want to get x by itself on one side of the equation.

First, we can subtract 24 from both sides:

- 12x + 24 - 24 = 0 - 24

This simplifies to:

- 12x = -24

Next, we can divide both sides by -12:

- 12x / -12 = -24 / -12

This simplifies to:

x = 2

Therefore, the solution to the equation -12x + 24 = 0 is x = 2.

Answer:

From the given information, we know that the area to the left of c under the standard normal distribution curve is 0.0166.

Using a standard normal distribution table or calculator, we can find the corresponding z-score for this area.

A z-score represents the number of standard deviations away from the mean. For a standard normal distribution, the mean is 0 and the standard deviation is 1.

Looking up the area of 0.0166 in the z-table, we find that the corresponding z-score is approximately -2.06.

Therefore, we have:

P(z < c) = 0.0166

P(z < -2.06) = 0.0166

So, c = -2.06.

Answer:

Using a standard normal distribution table, we can find the z-score corresponding to a probability of 0.0166:

z = -2.07

Therefore, c = -2.07.

Step-by-step explanation:

The Monotone Convergence Theorem can be demonstrated by considering the decreasing sequence {a_n} = 1/n, which is bounded below by zero and converges to zero.

Consider the sequence of real numbers {a_n} defined as a_n = 1/n. We want to show that the sequence converges to zero.

First, notice that the sequence is decreasing since a_n+1 = 1/(n+1) < 1/n = a_n for all n ≥ 1. Moreover, the sequence is bounded below by zero since a_n > 0 for all n. Thus, the sequence {a_n} is a decreasing bounded sequence and by the Monotone Convergence Theorem, it must converge to some limit L.

Let's now calculate the limit L. Since the sequence is decreasing and bounded below by zero, its limit L must be greater than or equal to zero. Furthermore, for any ε > 0, there exists an N such that 1/n < ε for all n > N, since the sequence converges to zero. Therefore, we have

|a_n - 0| = |1/n - 0| = 1/n < ε for all n > N.

This shows that the limit of the sequence is zero, i.e., lim (n → ∞) 1/n = 0.

Thus, we have demonstrated that the Monotone Convergence Theorem applies to the sequence {a_n}, which is decreasing and converges to zero.

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I have solved the question in general, as the given question is incomplete.

The complete question is:

Give an example to show that the Monotone Convergence Theorem?

The percentage of students earned a B in the statistics course is 22.5%. So, the correct option is B).

To find the percentage of students who earned a B in the course, we need to determine the total number of students who took the course and the number of students who earned a B.

Using the information given in the bar chart, we can determine that there were a total of 40 students who took the statistics course. The number of students who earned a B is given as 9 in the bar chart. Therefore, the percentage of students who earned a B is (9/40) x 100%, which simplifies to 22.5%.

The total number of students who took the statistics course is:

Y = A + B + C + D + F = 5 + 9 + 11 + 8 + 7 = 40

The percentage of students who earned a B is:

(B/Y) x 100% = (9/40) x 100% = 22.5%

Therefore, the correct answer is (B) 22.5%.

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Step-by-step explanation:

The balance will be the initial deposit ( p = 3000)  plus the interest earned for two years  p r t     where r = decimal interest per year   t = 2 years

Balance = $   3000  +    3000 * .025 * 2  = $ 3150.00

Answer: 54, 1 2/3, 1, 0.5, 0.03, 0, -1/4, -1, -4, -103

Step-by-step explanation:

54, 1, 1 2/3, 0.5, 0.03, 0, -1/4, -1, -4, -103.

First, we order the numbers by their sign: 54, 1, 1 2/3, 0.5, 0.03, 0, -1/4, -1, -4, -103.

Then we order the positive numbers in decreasing order: 54, 1 2/3, 1, 0.5, 0.03, 0.

Finally, we order the negative numbers in increasing order: -103, -4, -1, -1/4.

Putting it all together, we have: 54, 1 2/3, 1, 0.5, 0.03, 0, -1/4, -1, -4, -103.

Continuous numerical values make up the data type for the variable "Age of a tennis player selected at random in the Wimbledon tennis tournament."

Discrete numerical, continuous numerical, and categorical data are the three basic types that can be identified.

- Non-numerical categorical variables, such as gender or eye colour, represent categories or groups.

- Discrete numerical data, such as the number of siblings or pets, are numerical data that can only take on specified values.

Continuous numerical data, like age or weight, are numerical data that can have any value within a range.

Because age can have any value within a range, the data for the variable "Age of a randomly chosen tennis player in the Wimbledon tennis competition" is continuous numerical (for example, a player could be 18.5 years old or 25.2 years old). Hence, continuous numerical data is the right response.

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The probability of obtaining a reading between 0.59°C and 0.88°C is 0.7224 and 0.8106.

The sum of all possible values, weighted by the chance of each value, is equal to the mean of a discrete probability distribution of the random variable X. Each possible number of X must be multiplied by its probability P(x) before being added as a whole to determine the mean. In statistics, the mean is one measure of central trend in addition to the mode and median. The mean is simply the average of the numbers in the specified collection. It suggests that values in a specific data gathering are evenly distributed. In order to find the mean, the total values given in a datasheet must be added, and the result must be divided by the total number of values.

In this question, using the formula,

z-score = (x – μ) / σ

where:

x: individual data value

μ: population mean

σ: population standard deviation

for x=0.59

μ= 0

σ= 1

z-score= 0.59

Probability=0.7224

for x=0.88

z-score= 0.88

Probability=0.8106

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Answer:

Yes. If you are asking if the duration between those two times is a total of 12 hours, the answer is yes.

Step-by-step explanation:

9:45am is 12 hours away from 9:45pm. This applies to all times and their am/pm counterparts such as 12am/12pm.

B. We fail to reject the null hypothesis. In hypothesis testing, the significance level, denoted by a, is the probability of rejecting the null hypothesis when it is true.

If the p-value is less than the significance level, we reject the null hypothesis. In this case, the p-value is 0.05, which is greater than the significance level of 0.04. Option C is not necessarily true as the term "unusual" is subjective and can vary depending on the context. Option D is not necessarily true as the critical region may be in the other tail of the normal curve.

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In response to the stated question, we may state that Hence the chances  probability of Baxter's Shelties not having a Tri-Color puppy this year are 0.45, or 45%.

Probabilistic theory is a branch of mathematics that calculates the likelihood of an event or proposition occurring or being true. A risk is a number between 0 and 1, with 1 indicating certainty and a probability of around 0 indicating how probable an event appears to be to occur. Probability is a mathematical term for the likelihood or likelihood that a certain event will occur. Probabilities can also be expressed as numbers ranging from 0 to 1 or as percentages ranging from 0% to 100%. In relation to all other outcomes, the ratio of occurrences among equally likely alternatives that result in a certain event.

To determine the likelihood that Baxter's Shelties will not have a Tri-Color puppy this year, add the probabilities of all other potential colour combinations and subtract them from one (since the sum of all probabilities must be 1).

White and Sable: 0.18 + 0.12 = 0.3

White and Blue Merle: 0.1 + 0.05 = 0.15

0.05 Bi-Black

Bi-Blue: 0.02 Sable Merle: 0.03

As a result, the overall likelihood of NOT getting a Tri-Color puppy is:

1 - (0.3 + 0.15 + 0.05 + 0.03 + 0.02) = 1 - 0.55 = 0.45

Hence the chances of Baxter's Shelties not having a Tri-Color puppy this year are 0.45, or 45%.

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The probability of obtaining a reading less than 0.35° C is approximately 35%.

Accοrding tο the prοbability fοrmula, the likelihοοd οf an event οccurring is equal tο the ratiο οf the number οf favοurable οutcοmes tο the tοtal number οf οutcοmes. Prοbability οf an event οccurring P(E) = The number οf favοurable οutcοmes divided by the tοtal number οf οutcοmes.

The readings at freezing οn a set οf thermοmeters are nοrmally distributed, with a mean (x) οf 0°C and a standard deviatiοn (μ) οf 1.00°C. We want tο knοw hοw likely it is that we will get a reading that is less than 0.35°C.

To solve this problem, we must use the z-score formula to standardise the value:

[tex]$Z = \frac{x - \mu}{\sigma}[/tex]

Z = standard score

x = observed value

[tex]\mu[/tex] = mean of the sample

[tex]\sigma[/tex] = standard deviation of the sample

Here

x = 0.35° C

[tex]\mu[/tex] = 0° C

[tex]\sigma[/tex] = 1.00°C

Using the values on the formula:

[tex]$Z = \frac{0.35 - 0}{1}[/tex]

Z = 0.35

The probability of obtaining a reading less than 0.35° C is approximately 35%.

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The Born exponent or interatomic potential energy for Rbl is 11 ( approximately). The correct option is D).

The Born exponent for RbI can be calculated using the relationship between the Born exponent and the interionic distance. The Born exponent is defined as the ratio of the repulsive to attractive contributions to the interatomic potential energy, and it depends on the charges and sizes of the ions.

For Rb+ and I-, the Born exponents are 10 and 12, respectively. This means that the repulsive interaction between Rb+ and I- is weaker than the attractive interaction, as the repulsion is proportional to Rb+^10 and the attraction is proportional to I^-12. Therefore, the attractive interaction dominates.

For RbI, we can use the relationship between the Born exponent and the interionic distance to calculate the Born exponent. This relationship is given by:

B = (1/d) * ln[(l1 + l2)/|l1 - l2|]

where B is the Born exponent, d is the interionic distance, and l1 and l2 are the ionic radii of the cation and anion, respectively.

Assuming the ionic radii of Rb+ and I- are additive, we have:

l1 + l2 = l(RbI) = l(Rb+) + l(I-) = 1.52 + 1.81 = 3.33 Å

|l1 - l2| = |l(Rb+) - l(I-)| = |1.52 - 1.81| = 0.29 Å

Substituting these values into the equation for B, we get:

B = (1/d) * ln[(l1 + l2)/|l1 - l2|] = (1/d) * ln[3.33/0.29] ≈ 11.02

Therefore, the Born exponent for RbI is approximately 11.02.

The correct answer is D).

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The values or variables listed in the function declaration are called formal parameters to the function.

They are used to store the data that is passed into the function when it is called. The formal parameters are local variables, meaning that the values stored in them are only available within the function.

The arguments are the values passed to the function when it is called. These values are then assigned to the formal parameters and are used within the function to perform the desired task.

Formal arguments are produced at function entry and removed at function exit, behaving similarly to other local variables inside the function.

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The cylindrical has a surface area of 414 square units due to its 9-unit radius and 14-unit height.

A cylinder is a three-dimensional geometric form made up of two circular bases that are parallel to one another and are joined by a curved lateral surface. It can be pictured as a solid item with a constant circular cross-section along its entire length. The measurements of a cylinder, such as the radius and height of the circular bases, affect its characteristics. The surface area, volume, and horizontal surface area of a cylinder are some of its typical characteristics. Mathematical formulas can be used to determine these properties.

given

The following algorithm determines a cylinder's surface area:

[tex]A = 2\pi r^2 + 2\pi rh[/tex]

where r is the cylinder's base's radius, h is the cylinder's height, and (pi) is a mathematical constant roughly equivalent to 3.14.

Inputting the numbers provided yields:

[tex]A = 2\pi (9)^2 + 2\pi (9)(14)\\[/tex]

A = 2π(81) + 2π(126) 

A = 162π + 252π

A = 414π

The cylindrical has a surface area of 414 square units due to its 9-unit radius and 14-unit height.

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A = {u + 2v, -3} is the basis for subspace U given that the set A is now linearly independent and that U = span(A).

Since U = span(S), and the set S is linearly independent, let S = {u, v} be the basis of the subspace U.

Now determine whether or not the set A = {u + 2v, -3v} is linearly independent.

A set of vectors must all have linear combinations that add up to zero in order for them to be considered linearly independent. Let a and b represent any scalars so that,

a(u + 2v) + b(-3v) = 0

Simplify the obtained equation.

au + 2av - 3bv = 0

au + v(2a - 3b) = 0

Make 2a - 3b = A.

Rewrite the equation that was found using this.

Now because u and v are linearly independent, a and A must be zero, and as a result, the constant b is also zero.

Set A is hence linearly independent.

Also, au + Av ∈ U, so, U = span(A).

Considering that the set A is now linearly independent and that U = span(A), the basis for subspace U is A = {u + 2v, -3}.

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The complete question is:

If {u, v} is a basis for the subspace U, show that {u + 2v, −3v} is also a basis for U.

There is a vertical asymptote at x = 2 and the slope and intercept of the oblique asymptote are 2 and - 1, respectively.

In this problem we find the definition of a rational function:

f(x) = (2 · x² - 5 · x + 3) / (x - 2)

The vertical asympote correspond to the vertical line at the x-value where the function is undefined. And the oblique asymptote is defined by a equation of the form:

y = m · x + b

Where:

And the slope and the intercept of the asymptote can be found by means of the following equation:

Slope

[tex]m = \lim_{x \to \pm \infty} \left[\frac{f(x)}{x}\right][/tex]

Intercept

[tex]b = \lim_{x \to \pm \infty} [f(x) - m \cdot x][/tex]

First, factor and simplify the rational equation to determine whether any zero is evitable:

f(x) = (2 · x² - 5 · x + 3) / (x - 2)

f(x) = (2 · x - 3) · (x - 1) / (x - 2)

The discontinuity at x = 2 is not evitable. Then, the equation for the vertical asymptote is x = 2.

Second, determine the slope and the intercept of the oblique asymptote:

[tex]m = \lim_{x \to \pm \infty} \left[\frac{2\cdot x^{2}-5\cdot x + 3}{x^{2} - 2\cdot x} \right][/tex]

m = 2

[tex]b = \lim_{x \to \pm \infty} \left[\frac{2\cdot x^{2}-5\cdot x + 3}{x - 2} - 2 \cdot x\right][/tex]

[tex]b = \lim_{x \to \pm \infty} \left[\frac{2\cdot x^{2}-5\cdot x + 3-2 \cdot x^{2}+4\cdot x}{x-2}\right][/tex]

[tex]b = \lim_{x \to \pm \infty} \left[\frac{3 - x}{x-2} \right][/tex]

b = - 1

The slope and the intercept of the oblique asymptote are 2 and - 1, respectively.

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Step-by-step explanation:

(a) Experimental probability of rolling an even number = (number of rolls for 2, 4, and 6) / (total number of rolls) = (4 + 5 + 6) / 40 = 0.375

(b) Theoretical probability of rolling an even number = number of even outcomes / total number of outcomes = 3 / 6 = 0.5

(c) Statement 2 is true: With a small number of rolls, it is not surprising when the experimental probability is much greater than the theoretical probability.

Answer: it represents half of the students in 1 class

Step-by-step explanation:

1/2 divided by s

Answer:

1/2s would then represent one half (or 50%) of the students in the singular class stated.

The set of ordered pair that does not represent a function is option 1 {(4,0), (8, -8), (4,1), (5,8)}.

A function in mathematics is a relationship between two sets in which every element of the first set (referred to as the domain) is connected to exactly one element of the second set (called the range). A function is typically represented by the symbol f(x), where x is a domain element and f(x) is a corresponding range element.

We know that, a set of ordered pairs represents a function if each input is associated with only one output.

From the given options we observe that, {(4,0), (8, -8), (4,1), (5,8)}, does not represent a function.

Hence, the set of ordered pair that does not represent a function is option 1 {(4,0), (8, -8), (4,1), (5,8)}.

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When [tex]x[/tex] approaches infinity, the function's graph moves closer to the x-axis and horizontal equilibrium point at [tex]y = 3[/tex]. For [tex]x -2[/tex], which is denoted by such an open ring at [tex](-2, 3)[/tex] on the graph, the function is undefined.

A graphs is a pictorial display or diagram that displays facts or numbers in an organized way in math. The relationships between multiple things are frequently represented by the points on a graph.

The graph is a mathematics structure made up of a collection of points Coordinates and a set of lines connecting some pairs of VERTICES that may or may not be empty. There is a chance that the edges will be directed, or orientated.

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Answer: Area: 460.48 ft^2         Perimeter: 90.12 ft

Step-by-step explanation:

The area is 1/2 * 3.14 * (16 / 2)^2 (area of semicircle)

+ 10 * 12 / 2 (area of triangle)

+ 20 * 15 (area of rectangle)

= 460.48

The perimeter is 1/2 * 16 * 3.14 (perimeter of semicircle)

+ 10 (perimeter of triangle)

+ 20 + 15 + 20 (perimeter of rectangle)

= 90.12

The value of error term E(x,y) = 8x^2 - 8x - 56.

The definition of differentiability states that a function f(x,y) is differentiable at a point (a,b) if there exists a linear function L(x,y) such that:

f(x,y) - f(a,b) = L(x,y) + E(x,y)

where E(x,y) is an error term that approaches 0 as (x,y) approaches (a,b).

In the case of the function f(x,y) = 8x^2 - 8y, we want to find E(x,y) at the point (-1,-7).

First, we need to calculate f(-1,-7):

f(-1,-7) = 8(-1)^2 - 8(-7) = 56

Next, we need to find the linear function L(x,y) that approximates f(x,y) near (-1,-7). To do this, we can use the gradient of f(x,y) at (-1,-7):

∇f(-1,-7) = (16,-8)

The linear function L(x,y) is given by:

L(x,y) = f(-1,-7) + ∇f(-1,-7) · (x+1, y+7)

where · denotes the dot product.

Substituting the values, we get:

L(x,y) = 56 + (16,-8) · (x+1, y+7)

= 56 + 16(x+1) - 8(y+7)

= 8x - 8y

Finally, we can calculate the error term E(x,y) as:

E(x,y) = f(x,y) - L(x,y) - f(-1,-7)

= 8x^2 - 8y - (8x - 8y) - 56

= 8x^2 - 8x - 56

Therefore, the error term E(x,y) for the function f(x,y) = 8x^2 - 8y at the point (-1,-7) is E(x,y) = 8x^2 - 8x - 56.

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Answer: 10 is two units from 0 on the number line, so there are six numbers that are 10 units from 0.

Step-by-step explanation:

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