How can we reduce bias in an estimator: OA. use nonrandom sampling. B. use random sampling. C. increase the number of items included in the sample. D. decrease the number of items included in the sample.

Answer:

  x ∈ {-4, 0, 2}

Step-by-step explanation:

You want the solutions to the cubic 2x³ +4x² -16x = 0.

We observe that x and 2 are factors of all terms, so this can be written ...

  2x(x² +2x -8) = 0

The quadratic will have binomial factors with constants that are factors of -8 that have a sum of 2.

  2x(x +4)(x -2) = 0

Solutions are the values of x that make the factors zero:

  x = 0

  x +4 = 0   ⇒   x = -4

  x -2 = 0   ⇒   x = 2

The solutions are x = -4, 0, 2.

Answer:

Step-by-step explanation:

step 1;

  49  = 1 x 49

1 + 49 = 50

step 2;

49   = 7 x 7

7+7 = 14

*Hence 7 and 7 are the two positive numbers whose product is 49 and whose sum is minimum.

The slope of the tangent line to the curve x = 4t₂, y = 2 24t − 14t₂ is 1 at the point (2,10).

To find the slope of the tangent line, we need to take the derivative of y with respect to x, which is dy/dx = (dy/dt) / (dx/dt).

Substituting the given values, we get dy/dx = (48t - 14) / 4, which simplifies to dy/dx = 12t - 3.

To find the point where the slope is 1, we set dy/dx = 1 and solve for t. This gives us t = (1+3)/12 = 1/3.

Substituting t = 1/3 into the equations for x and y, we get x = 4(1/3)² = 4/9 and y = 2(24/3) - 14(1/3) = 16.

Therefore, the point where the tangent line has slope 1 is (4/9, 16).

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In the two functions as the value of V(x) increases, the value of W(x) also increases.

The value of functions, V(x) and W(x) is determined as follows;

for h(-2, 1/4); the value of the functions is calculated as follows;

v(x) = 2ˣ ⁺ ³ = 2⁻²⁺³ = 2¹ = 2

w(x) = 2ˣ ⁻ ³ = 2⁻²⁻³ = 2⁻⁵ = 1/32

for h(-1, 1/2); the value of the functions is calculated as follows;

v(x) = 2ˣ ⁺ ³ = 2² = 4

w(x) = 2ˣ ⁻ ³ = 2⁻⁴ = 1/16

for h(0, 1); the value of the functions is calculated as follows;

v(x) = 2ˣ ⁺ ³ = 2³ = 8

w(x) = 2ˣ ⁻ ³ = 2⁻³ = 1/8

for h(1, 2); the value of the functions is calculated as follows;

v(x) = 2ˣ ⁺ ³ = 2⁴ = 16

w(x) = 2ˣ ⁻ ³ = 2⁻² = 1/4

for h(2, 4); the value of the functions is calculated as follows;

v(x) = 2ˣ ⁺ ³ = 2⁵ = 32

w(x) = 2ˣ ⁻ ³ = 2⁻¹ = 1/2

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The surface described by the equation [tex]r^2 + z^2 = 4[/tex]is a right circular cylinder with a radius of 2 units, centered along the z-axis.

The surface whose equation is given is a cylinder with a radius of 2 units and a height of 4 units, centered on the z-axis.
Hi! I'd be happy to help you identify the surface with the given equation. The equation provided is:

[tex]r^2 + z^2 = 4[/tex]

This equation represents a right circular cylinder with a radius of 2 units, centered along the z-axis. Here's why:

1. Notice that the equation contains r^2 and [tex]z^2[/tex] terms, which suggests a cylindrical coordinate system.
2. The equation does not contain the θ term, which implies that the surface is symmetric about the z-axis.
3. The equation is in the form [tex]r^2 + z^2[/tex] = constant, which is the equation of a right circular cylinder in cylindrical coordinates.

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The two parallel sides, given the perimeter of the trapezium, can be found to be 9 cm and 18 cm.

The two parallel sides of the second trapezium can be found to be 10 cm and 20 cm.

Let's denote the parallel sides of the trapezium MUST as TS (shorter side) and MU (longer side), and the non-parallel sides as MS and UT.

TS + 2 * TS + MS + UT = 48

3 x TS + MS + UT = 48

Since MS + UT = 21, we can substitute this into the above equation:

3 x TS + 21 = 48

Now, solve for TS:

3 x TS = 27

TS = 9 cm

Now, find MU using equation (3):

MU = 2 x TS = 2 x 9 = 18 cm

So, the two parallel sides are 9 cm and 18 cm.

Let's denote the parallel sides of the trapezium as a (shorter side) and b (longer side), and the height as h.

The area of a trapezium can be calculated using the formula:

Area = (1/2) x (a + b) x h

180 = (1/2) x (a + 2a) x 12

180 = 3a x 6

180 = 18a

a = 10 cm

Now, find b using equation (3):

b = 2 * a = 2 * 10 = 20 cm

So, the two parallel sides are 10 cm and 20 cm.

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The acceleration at time t = 2 is zero. So, the correct answer is c).

To find the acceleration at time t=2, we need to take the second derivative of the position function s(t).

s(t) = t^3 - 6t^2 + 4t + 5

Taking the first derivative

s'(t) = 3t^2 - 12t + 4

Taking the second derivative

s''(t) = 6t - 12

Now, plugging in t=2

s''(2) = 6(2) - 12 = 0

Therefore, the acceleration at time t=2 is 0, so the correct option is (c) 0. This means that the object is not accelerating at t=2, but rather it is either at rest or moving at a constant velocity.

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Answer:

sorry can't understand what your trying to say but i can help a lil if you translate it into english

Without more information, we cannot provide a more specific answer.

Without a diagram or more information about the plate and the forces acting on it, it is difficult to give a definitive answer to this question. However, we can make some general observations about the forces involved.

The ball-and-socket joint at a allows the plate to rotate freely around that point, while the roller point at b allows the plate to move horizontally without resistance. The cable at c is likely providing some sort of tension or support to the plate.

Assuming that the plate is in equilibrium, the sum of the forces acting on it must be zero. The force at point Bz would be the reaction force of the roller point at b. This force would be perpendicular to the surface of the roller and would depend on the weight of the plate and any other forces acting on it.

If we knew the weight of the plate and the angle at which the cable at c is pulling, we could use trigonometry and the principles of statics to calculate the magnitude and direction of the force at Bz. However, without more information, we cannot provide a more specific answer.

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The rate of change of photosynthesis with respect to intensity of light is 230 when x = 1. The rate found in part (a) is decreasing at a rate of 470 when x = 3. The rate of change of photosynthesis with respect to the intensity of light is 230 when x = 1 and -2190 when x = 3.

(a) To find the rate of change of photosynthesis with respect to the intensity of light, we need to take the derivative of the equation f(x) = 175x^2 - 40x^3 with respect to x.

f'(x) = 350x - 120x^2

(b) To find the rate of change when x = 1, we substitute x = 1 into the derivative equation:

f'(1) = 350(1) - 120(1)^2 = 230


To find the rate of change when x = 3, we substitute x = 3 into the derivative equation:

f'(3) = 350(3) - 120(3)^2 = -630

Therefore, the rate of change of photosynthesis with respect to intensity of light is -630 when x = 3.

To find how fast the rate found in part (a) is changing when x = 1, we need to take the second derivative of the equation:

f''(x) = 350 - 240x

Then we substitute x = 1 into the second derivative equation:

f''(1) = 350 - 240(1) = 110

Therefore, the rate found in part (a) is increasing at a rate of 110 when x = 1.

To find how fast the rate found in part (a) is changing when x = 3, we also need to take the second derivative of the equation:

f''(x) = 350 - 240x

Then we substitute x = 3 into the second derivative equation:

f''(3) = 350 - 240(3) = -470


(a) To find the rate of change of photosynthesis with respect to the intensity of light, we need to find the derivative of the given function f(x) = 175x^2 - 40x^3.

f'(x) = d/dx (175x^2 - 40x^3) = 350x - 120x^2

(b) To find the rate of change when x = 1 and x = 3, simply plug these values into the derivative:

f'(1) = 350(1) - 120(1^2) = 350 - 120 = 230
f'(3) = 350(3) - 120(3^2) = 1050 - 3240 = -2190

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If from total of 28 students, 12 has clock, 16 have calculator and 4 do not have these items, then the number of students that have clock and a calculator are (b) 4.

The total number of students are 28 students,

The number of students who has a clock is = 12 students,

The number of students who has a calculator is = 16 students,

The number of students who do not have any of these is = 4 students,

Let the number of students who has-both clock and calculator be = x,

So, the number of students having only clock are = 12-x,

the number of students having only calculator are = 16-x,

The equation for total students is written as :

⇒ 12-x + x + 16-x + 4 = 28,

⇒ 32 - x = 28,

⇒ x = 32 - 28,

⇒ x = 4,

Therefore, 4 students have a clock and calculator, Option(b) is correct.

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The cost of 1 kg orange is 66.67% less than the cost of 1 kg apple.

"Rate" usually refers to the cost per unit of a certain item or service. Rates can vary depending on factors such as quantity, time, and discounts. For instance, a product may have a lower rate if purchased in bulk, or a service may have a discounted rate for repeat customers.

In mathematics, percent is a way of expressing a number as a fraction of 100.

Let's start by finding the cost of 1 kg orange and 1 kg apple using the given information.

Let the cost of 1 kg orange be x and the cost of 1 kg apple be y.

From the first statement, we know that:

3x + 5y = 1300

Simplifying this equation, we get:

x + (5/3)y = 1300/3

x = (1300/3) - (5/3)y

From the second statement, we know that:

2(1.1x) + 3(0.8y) = 700

Simplifying this equation, we get:

2.2x + 2.4y = 700

Substituting the value of x from the first equation, we get:

2.2[(1300/3) - (5/3)y] + 2.4y = 700

Simplifying this equation, we get:

y = 200

Substituting the value of y in the first equation, we get:

x = 100

Therefore, the cost of 1 kg orange is Rs. 100 and the cost of 1 kg apple is Rs. 200.

To find the percentage by which the cost of 1 kg orange is more or less than the cost of 1 kg apple, we can use the following formula:

Percentage difference = [(|x - y|)/((x + y)/2)] * 100

Substituting the values of x and y, we get:

Percentage difference = [(|100 - 200|)/((100 + 200)/2)] * 100

Percentage difference = [100/150] * 100

Percentage difference = 66.67%

Therefore, the cost of 1 kg orange is 66.67% less than the cost of 1 kg apple.

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The vector w can be expressed as:
w = 5v1 - 5v2 - v3

Given the orthogonal set of vectors v1, v2, and v3 in R5, and the vector w in the span of these vectors, you have provided the following information:

- (v1,v1) = 35
- (v2,v2) = 334
- (v3,v3) = 1
- (w,v1) = 175
- (w,v2) = -1670
- (w,v3) = -1

To find w in terms of v1, v2, and v3, use the following formula for orthogonal projections:

w = (w,v1)/||(v1)||^2 * v1 + (w,v2)/||(v2)||^2 * v2 + (w,v3)/||(v3)||^2 * v3

Since (v1,v1) = 35, ||(v1)||^2 = 35.
Since (v2,v2) = 334, ||(v2)||^2 = 334.
Since (v3,v3) = 1, ||(v3)||^2 = 1.

Substitute the given values:

w = (175/35) * v1 + (-1670/334) * v2 + (-1/1) * v3

Simplify the coefficients:

w = 5 * v1 - 5 * v2 - v3

So, the vector w can be expressed as:

w = 5v1 - 5v2 - v3

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The possible value of p in the given equation is -3/2.

A plane's equation is a linear expression made up of the constants a, b, c, and d as well as the variables x, y, and z. The direction numbers of a vector perpendicular to the plane are represented by the coefficients a, b, and c. The constant d can be thought of as the distance along the normal vector of the plane from the origin. The formula a(x1, y1, z1) = 0 can be used to get the equation of a plane, where (x1, y1, z1) is a specified point in the plane.

The acute angle between a line and a plane is given by:

cos(theta) = |n . d| / (|n| |d|)

where n is the normal vector to the plane, d is the direction vector of the line, and |.| denotes the magnitude.

From the equation of the plane, we can read off the normal vector n as (2, 1, 2).

From the vector equation of the line, we can read off the direction vector d as (-2, p, 0).

Substituting these values into the formula for the cosine of the acute angle, we get:

cos(30°) = |(2, 1, 2) . (-2, p, 0)| / (|(2, 1, 2)| |(-2, p, 0)|)

Simplifying and solving for p, we get:

cos(30°) = |(2, 1, 2) . (-2, p, 0)| / (|(2, 1, 2)| |(-2, p, 0)|)

1/2 = |-4 + 2p| / (√9 |p² + 4|)

Squaring both sides and simplifying, we get:

4(p² + 4) = 4(p² - 2p + 1)

4p² + 16 = 4p² - 8p + 4

8p = -12

p = -3/2

Therefore, the possible value of p is -3/2.

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The possible value of p in the given equation is -3/2.

A plane's equation is a linear expression made up of the constants a, b, c, and d as well as the variables x, y, and z. The direction numbers of a vector perpendicular to the plane are represented by the coefficients a, b, and c. The constant d can be thought of as the distance along the normal vector of the plane from the origin. The formula a(x1, y1, z1) = 0 can be used to get the equation of a plane, where (x1, y1, z1) is a specified point in the plane.

The acute angle between a line and a plane is given by:

cos(theta) = |n . d| / (|n| |d|)

where n is the normal vector to the plane, d is the direction vector of the line, and |.| denotes the magnitude.

From the equation of the plane, we can read off the normal vector n as (2, 1, 2).

From the vector equation of the line, we can read off the direction vector d as (-2, p, 0).

Substituting these values into the formula for the cosine of the acute angle, we get:

cos(30°) = |(2, 1, 2) . (-2, p, 0)| / (|(2, 1, 2)| |(-2, p, 0)|)

Simplifying and solving for p, we get:

cos(30°) = |(2, 1, 2) . (-2, p, 0)| / (|(2, 1, 2)| |(-2, p, 0)|)

1/2 = |-4 + 2p| / (√9 |p² + 4|)

Squaring both sides and simplifying, we get:

4(p² + 4) = 4(p² - 2p + 1)

4p² + 16 = 4p² - 8p + 4

8p = -12

p = -3/2

Therefore, the possible value of p is -3/2.

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The trigonometric identity for this problem is proven, as tan(x)/csc(x) = 1/cos(x) - 1/sec(x).

The trigonometric expression for this problem is defined as follows:

tan(x)/csc(x).

The definitions for the tangent and for the cossecant are given as follows:

When two fractions are divided, we multiply the numerator by the inverse of the denominator, hence:

tan(x)/csc(x) = sin(x)/cos(x) x sin(x) = sin²(x)/cos(x).

The sine squared can be given as follows:

sin²(x) = 1 - cos²(x).

Hence the simplified expression is given as follows:

(1 - cos²(x))/cos(x) = 1/cos(x) - cos(x).

The secant is one divided by the cosine, hence:

1/sec(x) = 1/1/cos(x) = cos(x).

Thus we can prove the identity, as:

tan(x)/csc(x) = 1/cos(x) - 1/sec(x).

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By answering the presented question, we may conclude that As a result, exponential about 739 banana plants in the plantation will be affected with the blight after 9 years.

The word "exponential growth" refers to the process of increasing quantity through time. When the instantaneous rate of change of a quantity with respect to time is proportional to the quantity, this is said to be proportional to the quantity. Exponential growth is a statistical pattern in which bigger gains are seen with time. Compound interest delivers exponential rewards in the world of finance. Savings accounts with compound interest can occasionally experience exponential growth. characterised by a rapid increase in the exponential growth rate (in size or extent). exponentially. The exponential function formula is f(x)=abx, where a and b are positive real values. Draw exponential functions for various values of a and b using the tools provided below.

To tackle this problem, we may apply the exponential growth formula:

[tex]N = N0 * (1 + r)^t[/tex]

Where N0 is the initial number of infected plants (476)

r = rate of increase (5% = 0.05)

t = time span (9 years)

When we plug in the values, we get:

[tex]N = 476 * (1 + 0.05)^9 \sN = 476 * 1.55128 \sN = 738.94[/tex]

When we round to the next full number, we get:

N ≈ 739

As a result, about 739 banana plants in the plantation will be affected with the blight after 9 years.

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The equation of the osculating circle to the function [tex]f(x) = \frac{1}{16} x^4 - \frac{1}{4} x^2[/tex]at the origin is [tex]x^2 + (y - 4/3)^2 = 16/9[/tex].

The radius of the circle is 4/3, and its center is at (0, 4/3).

To find the equation of the osculating circle to the function [tex]f(x) = \frac{1}{16} x^4 - \frac{1}{4} x^2[/tex] at the origin, we need to find the radius and center of the circle.

The osculating circle at a point (a, f(a)) has the same curvature as the graph of the function at that point, so we can use the formula for curvature:

[tex]k = |f''(a)| / [1 + (f'(a))^2]^{(3/2)[/tex]

where f''(a) and f'(a) are the second and first derivatives of f(x) evaluated at x = a.

At the origin (a = 0), we have:

f(a) = f(0) = -0.0625

f'(a) = f'(0) = 0

f''(a) = f''(0) = 3/4

Substituting these values into the formula for curvature, we get:

[tex]k = |f''(0)| / [1 + (f'(0))^2]^{(3/2)}\\= (3/4) / [1 + 0^2]^{(3/2)[/tex]

= 3/4

Since the radius of the osculating circle is 1/k, the radius of the circle at the origin is:

r = 1 / (3/4) = 4/3

To find the center of the circle, we note that it must lie on the normal line to the graph of f(x) at the origin.

Since the slope of the tangent line at the origin is f'(0) = 0, the slope of the normal line is undefined (i.e., it is vertical).

Therefore, the center of the osculating circle is at (0, r), or (0, 4/3).

The equation of the osculating circle is therefore:

[tex](x - 0)^2 + (y - 4/3)^2 = (4/3)^2\\x^2 + (y - 4/3)^2 = 16/9[/tex]

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The proper basis step used in strong induction is showing that P(n) is true for n = 30, 31, 32, and 33.

To prove the statement P(n) that n cents of postage can be formed with just 4-cent and 11-cent stamps using strong induction where n ≥ 30:

We need to first identify the proper basis step used in strong induction.
Step 1:  Basis step
We need to show that P(n) is true for the initial values of n.

Since n ≥ 30, we will check for n = 30, 31, 32, and 33.
For n = 30: 3 * 4-cent stamps + 2 * 11-cent stamps = 12 + 22 = 30 cents
For n = 31: 1 * 4-cent stamps + 3 * 11-cent stamps = 4 + 33 = 31 cents
For n = 32: 8 * 4-cent stamps = 32 cents
For n = 33: 5 * 4-cent stamps + 2 * 11-cent stamps = 20 + 22 = 33 cents
Since P(n) is true for n = 30, 31, 32, and 33, the basis step is complete.
The proper basis step used in strong induction is showing that P(n) is true for n = 30, 31, 32, and 33.

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We will have to pay back $206 after one year for the loan of $200 with a 3% interest rate.

Explain interest rate

The interest rate is the amount of money charged by a lender to a borrower for the use of funds or the return earned on an investment. It is usually expressed as a percentage of the principal amount and is paid or earned annually. Interest rates are influenced by various factors such as inflation, economic conditions, and government policies. Higher interest rates lead to increased borrowing costs and higher returns on investments, while lower interest rates encourage borrowing and spending but reduce returns. Interest rates play a crucial role in shaping the economy and financial markets.

According to the given information

The simple interest formula is:

Interest = Principal x Rate x Time

In this case:

So, the interest charged on the loan after one year is:

Interest = $200 x 0.03 x 1 = $6

Therefore, the total amount paid back after one year will be:

Total amount = Principal + Interest = $200 + $6 = $206

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When measured in straight line, the distance of the cars apart would be = 10 miles.

To calculate the distance of the cars apart in a straight line, the Pythagorean formula should be used. That is;

C² = a²+b²

c² = 8²+6²

= 64+36

c² = 100

c = √100

= 10 miles

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The evaluation of -2/3+1/6-5/12 is -11/12

A fraction has two parts, the numerator and the denominator.

In a simple fraction, both are integers. Examples are; 2/5 , 3/5. A complex fraction has a fraction in the numerator or denominator. In a proper fraction, the numerator is less than the denominator.

Solving, -2/3 +1/6 -5/12

1/6 -2/3 -5/12

= (2-8-5)/12

= (2-13)/12

= -11/12

therefore the evaluation of -2/3+1/6-5/12 is -11/12

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The measure of the smallest angle between the beams and the roof is 35⁰.

The measure of the smallest angle between the beams and the roof is calculated by determining the value of x.

The value of x is calculated as follows;

3x + 10 + 90 + 2x + 5 = 180

5x + 105 = 180

5x = 75

x = 75/5

x = 15

The measure of the angles is calculated as follows;

3x + 10

= 3 x 15 + 10

= 55⁰

2x + 5

= 2 x 15 + 5

= 35⁰

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The equation y² - x⁵ = -7 at x = 2 and y = 5, the value of dy/dx is 8.

To find dy/dx for the equation y² - x⁵ = -7 at x = 2, y = 5, follow these steps:

1. Differentiate both sides of the equation with respect to x using implicit differentiation.

Differentiating y² with respect to x, we get 2y(dy/dx).
Differentiating -x⁵ with respect to x, we get -5x⁴.

So, we have: 2y(dy/dx) - 5x⁴ = 0.

2. Plug in the given values of x and y into the differentiated equation.

Substitute x = 2 and y = 5: 2(5)(dy/dx) - 5(2⁴) = 0.

3. Solve for dy/dx.

First, simplify the equation: 10(dy/dx) - 80 = 0.

Next, add 80 to both sides: 10(dy/dx) = 80.

Finally, divide both sides by 10 to get: dy/dx = 8.

So, for the equation y² - x⁵ = -7 at x = 2 and y = 5, the value of dy/dx is 8.

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Answer:

3x10^3

Step-by-step explanation:

Move the decimal so there is one non-zero digit to the left of the decimal point. The number of decimal places you move will be the exponent on the

10

The sign of the exponent will depend on the direction you are moving the decimal.

Answer:

3x10^3

Step-by-step explanation:

Move the decimal so there is one non-zero digit to the left of the decimal point. The number of decimal places you move will be the exponent on the

10

The sign of the exponent will depend on the direction you are moving the decimal.

A snowmobile manufacturer produces three models, the XJ6, the XJ7, and the XJ8. The final testing bay is available for 40 hours in any given production planning week.

For each snowmobile model, a certain amount of testing time is required. The XJ6 requires 1 hour of testing, the XJ7 requires 1.5 hours of testing, and the XJ8 requires 2 hours of testing. To determine how many of each model can be produced in a week, linear programming techniques. We need to set up an equation using the available testing time. Let x, y, and z be the number of XJ6, XJ7, and XJ8 models produced in a week, respectively. Then, we have the following equation:[tex]1x + 1.5y + 2z[/tex][tex]\leq 40[/tex]

This equation represents the constraint that the total testing time required by all the snowmobile models cannot exceed the available testing time of 40 hours. We can use linear programming techniques to find the optimal values of x, y, and z that maximize the company's profits or minimize its costs, subject to this constraint and any other constraints that may be relevant.

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A snowmobile manufacturer produces three models, the XJ6, the XJ7, and the XJ8. In any given production planning week, the company has 40 hours available in its final testing bay. Each XJ6 requires 1 hour of testing, each XJ7 requires 1.5 hours, and each XJ8. Complete the final equation.

Check the picture below.

[tex]\cfrac{2^3}{8^3}=\cfrac{V}{96\pi }\implies \cfrac{1}{64}=\cfrac{V}{96\pi }\implies \cfrac{96\pi }{64}=V\implies 4.71\approx V[/tex]

a. Using backwards finite difference approximations for both partial derivatives, we get:

∂Q/∂x ≈ (Q(i,j) - Q(i-1,j))/a

∂Q/∂y ≈ (Q(i,j) - Q(i,j-1))/a

Therefore,

∂Q/∂x + ∂Q/∂y ≈ (Q(i,j) - Q(i-1,j))/a + (Q(i,j) - Q(i,j-1))/a

≈ Q(i,j)/a - [Q(i-1,j) + Q(i,j-1)]/a

b. Using backwards finite difference approximations for both partial derivatives twice, we get:

∂^2Q/∂x^2 ≈ (Q(i,j) - 2Q(i-1,j) + Q(i-2,j))/a^2

∂^2Q/∂y^2 ≈ (Q(i,j) - 2Q(i,j-1) + Q(i,j-2))/a^2

Therefore,

∂^2Q/∂x^2 + ∂^2Q/∂y^2 ≈ (Q(i,j) - 2Q(i-1,j) + Q(i-2,j))/a^2 + (Q(i,j) - 2Q(i,j-1) + Q(i,j-2))/a^2

≈ 2Q(i,j)/a^2 - [Q(i-1,j) + Q(i-2,j) + Q(i,j-1) + Q(i,j-2)]/a^2

c. Using backwards finite difference approximations for both partial derivatives thrice, we get:

∂^3Q/∂x^3 ≈ (Q(i,j) - 3Q(i-1,j) + 3Q(i-2,j) - Q(i-3,j))/a^3

∂^3Q/∂y^3 ≈ (Q(i,j) - 3Q(i,j-1) + 3Q(i,j-2) - Q(i,j-3))/a^3

Therefore,

∂^3Q/∂x^3 + ∂^3Q/∂y^3 ≈ (Q(i,j) - 3Q(i-1,j) + 3Q(i-2,j) - Q(i-3,j))/a^3 + (Q(i,j) - 3Q(i,j-1) + 3Q(i,j-2) - Q(i,j-3))/a^3

≈ 3Q(i,j)/a^3 - [Q(i-1,j) + Q(i-2,j) + Q(i-3,j) + Q(i,j-1) + Q(i,j-2) + Q(i,j-3)]/a^3

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There do not exist positive integers a and n such that a^2+3=3^n.

The given proof is not complete. The statement to be proven is that there do not exist positive integers a and n such that a^2+3=3.

The proof starts by assuming the opposite, i.e., assuming that there exist positive integers a and n such that a^2+3=3. However, the proof then only considers the case where n=1, which is not the most general case.

The proof correctly shows that if we put n=1, we get a^2+3=3, which simplifies to a^2=0. However, the conclusion that this is impossible is not explained. The reason this is impossible is that a is a positive integer, so a^2 must also be a positive integer. But a^2=0 implies that a=0, which contradicts the assumption that a is a positive integer.

To complete the proof, we need to consider the case where n>=2. In this case, we have:

a^2 + 3 = 3^n

Subtracting 3 from both sides, we get:

a^2 = 3^n - 3

We can factor the right-hand side as:

a^2 = 3(3^(n-1) - 1)

Since a is a positive integer, a^2 must be a multiple of 3. But 3^(n-1) - 1 is never a multiple of 3 for n>=2, so a^2 cannot be equal to 3(3^(n-1) - 1). Therefore, there do not exist positive integers a and n such that a^2+3=3^n.

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Answer:

71

Step-by-step explanation:

Replace x and y with their values:

7(2)(4)+5(2)-4(2)+2(2)(4)-3

14(4)+10-8+4(4)-3

56+10-8+16-3

66-8+16-3

58+16-3

74-3

71

The value of the above polynomial is 71

The given algebraic expressions is a polynomial in x and y.on seeing this expression carefully we found that their are like terms of coefficient 'xy' and 'x'. So we will simplify the expression as :

7xy+5x-4x+2xy-3=9xy+x-3.....(i)

Now, we will substitute the values of x and y in this expression to derive it's value.

9x2x4+2-3=71

Hence the value of expression is 71.

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