For each of the following lists of premises, derive the conclusion and supply the justification for it. There is only one possible answer for each problem.1. R ⊃ D2. E ⊃ R3. ________ ____

All the inequalities that represent the graph include the following:

B. y > -5/4(x) + 5

E. y + 5 > -1.25(x - 8)

In Mathematics and Geometry, the point-slope form of a straight line can be calculated by using the following mathematical expression:

y - y₁ = m(x - x₁)

Where:

First of all, we would determine the slope of this line;

Slope (m) = (y₂ - y₁)/(x₂ - x₁)

Slope (m) = (5 - 0)/(0 - 4)

Slope (m) = 5/-4

Slope (m) = -5/4

At data point (0, 5) and a slope of -5/4, a linear equation for this line can be calculated by using the point-slope form as follows:

y - y₁ = m(x - x₁)

y - 5 = -5/4(x - 0)  

y - 5 = -1.25(x - 0)

y = -5x/4 + 5

y > -5x/4 + 5 (shaded above the dashed line).

At data point (8, -5) and a slope of -5/4, a linear equation for this line can be calculated by using the point-slope form as follows:

y - y₁ = m(x - x₁)

y - (-5) = -5/4(x - 8)  

y + 5 = -1.25(x - 8)

y + 5 > -1.25(x - 8)

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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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Answer:

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

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.

The solution of the differential equation is  U=x+2y. (A)

To solve the differential equation (x + 2y)dx + ydy = 0 using substitution, you can use the substitution U = x + 2y.


1. Substitute U for x+2y: dU = (dx + 2dy)

2. Replace (x + 2y)dx + ydy with dU - 2ydy + ydy: dU - ydy = 0

3. Factor out dy: dU - ydy = dy(U - y) = 0

4. Separate variables: (1/dU) dU = dy/y

5. Integrate both sides: ∫(1/dU) dU = ∫(dy/y)

6. Obtain the solution: ln|U| = ln|y| + C

7. Replace U with x+2y: ln|x+2y| = ln|y| + C

8. Exponentiate both sides: x+2y = k*y, where k = e^C

Thus, the differential equation (x + 2y)dx + ydy = 0 can be solved using the substitution U = x + 2y.(A)

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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 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 absolute maximum value of f(t) is approximately -0.1213 at t = 9/4, and the absolute minimum value is -2 at t = 4.

To find the absolute maximum and absolute minimum values of f on the given interval [−1, 4], we first need to find the critical points of the function f(t) = t − 3√t.
Taking the derivative of f(t) with respect to t, we get:
f'(t) = 1 - (3/2)t^(-1/2)
Setting f'(t) = 0 to find critical points, we get:
0 = 1 - (3/2)t^(-1/2)
(3/2)t^(-1/2) = 1
t^(-1/2) = 2/3
t = (2/3)^(-2) = 2.25
So the only critical point of f(t) on the interval [−1, 4] is t = 2.25.
Now we need to evaluate f(t) at the endpoints of the interval and at the critical point to determine the absolute maximum and minimum values of f on the interval:
f(-1) = -1 - 3√(-1) = -1 - 3i
f(4) = 4 - 3√4 = 4 - 6 = -2
f(2.25) = 2.25 - 3√2.25 = 2.25 - 3(1.5) = -2.25
Therefore, the absolute maximum value of f on the interval [−1, 4] is f(-1) = -1 - 3i, and the absolute minimum value of f on the interval is f(4) = -2.
To find the absolute maximum and minimum values of f(t) = t - 3√t on the interval [-1, 4], we need to evaluate the function at its critical points and endpoints.
First, we find the critical points by taking the derivative of the function and setting it to zero:
f'(t) = 1 - (3/2)t^(-1/2)
To solve for critical points, set f'(t) = 0:
0 = 1 - (3/2)t^(-1/2)
(3/2)t^(-1/2) = 1
t^(-1/2) = 2/3
t = (2/3)^(-2) = 9/4
Since 9/4 is within the interval [-1, 4], it is a valid critical point.
Now, evaluate the function at the critical point and the endpoints:
f(-1) = -1 - 3√(-1)

(Note: This value is complex, and we're looking for absolute max/min in the real domain, so we'll ignore this endpoint)
f(9/4) = (9/4) - 3√(9/4) ≈ -0.1213
f(4) = 4 - 3√4 = -2
So, the absolute maximum value of f(t) is approximately -0.1213 at t = 9/4, and the absolute minimum value is -2 at t = 4.

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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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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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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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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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Answer:

(the answer is y ≤ X + 1).....

The following parts can be answered by the concept of combination of functions.

20) fg: Since f is even and g is odd, the product (fg) will be an odd function.

21) fg: The answer is the same as #20. The product (fg) will be an odd function.

22) g∘f: For a composition of functions, the even/odd properties depend on the functions themselves. Since g is odd and f is even, the composition g∘f will also be an odd function.

23) f∘f: Since both functions are even, the composition of two even functions, f∘f, will result in an even function.

24) g∘g: Similarly, since both functions are odd, the composition of two odd functions, g∘g, will result in an even function.
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Answer:

The correct answer is: Yes; it is a reflection over the y-axis.

To see why, imagine folding the graph along the y-axis. Points on the right-hand side of the y-axis remain on the right-hand side, while points on the left-hand side move to the right. This transformation is equivalent to reflecting the original figure across the y-axis.

Step-by-step explanation:

The average value of P=f(t)=2.04(1.03)^t for 0≤t≤30 is approximately 3.236. The average value of P=f(t)=2.04(1.03) for 0≤t≤30 is 2.10.

To find the average value of the function P=f(t)=2.04(1.03)^t for 0≤t≤30, you'll need to evaluate the appropriate integral and use the formula for the average value of a function.
The formula for the average value of a function is:
Average value = (1/(b-a)) * ∫[f(t) dt] from a to b
In this case, a = 0, b = 30, and f(t) = 2.04(1.03)^t.
Step 1: Evaluate the integral.
∫[2.04(1.03)^t dt] from 0 to 30
Step 2: Use a calculator to find the definite integral value.
We should find that the integral value is approximately 97.091.
Step 3: Substitute the integral value, a, and b into the average value formula.
Average value = (1/(30-0)) * 97.091
Step 4: Calculate the average value.
Average value ≈ (1/30) * 97.091 ≈ 3.236
So, the average value of P=f(t)=2.04(1.03)^t for 0≤t≤30 is approximately 3.236.

To find the average value of P=f(t)=2.04(1.03) for 0≤t≤30, we need to first evaluate the integral of the function over the given interval.
∫(0 to 30) 2.04(1.03) dt
Using a calculator, we can simplify and solve this integral as follows:
2.04(1.03)∫(0 to 30) dt
= 2.10t |(0 to 30)
= 2.10(30) - 2.10(0)
= 63.00
So, the integral of P=f(t) over the interval 0≤t≤30 is 63.00.
To find the average value of P over this interval, we divide this integral by the length of the interval:
Average value of P = (1/30-0) * 63.00
= 2.10
Therefore, the average value of P=f(t)=2.04(1.03) for 0≤t≤30 is 2.10.

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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]

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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To obtain a feasible solution, you would need to revise the assignment limits or add additional constraints that do not violate the given constraint.

Based on the constraint you provided, x21 + x22 + x23 + x24 ≤ 3, it means that the sum of variables x21, x22, x23, and x24, representing the number of tasks assigned to agents 1, 2, 3, and 4 respectively, cannot exceed 3.

This constraint implies that agent 3 can be assigned to a maximum of 2 tasks (since x23 ≤ 2), and agent 2 can be assigned to a maximum of 3 tasks (since x22 ≤ 3).

However, there seems to be a contradiction with the statement that "agent 3 can be assigned to 2 tasks" and "agent 2 can be assigned to 3 tasks" because the sum of these maximum assignments would already exceed 3, which is not feasible according to the constraint.

Therefore, To obtain a feasible solution, you would need to revise the assignment limits or add additional constraints that do not violate the given constraint, such as reducing the maximum number of tasks that can be assigned to agent 2 or agent 3, or adjusting the total number of tasks available for assignment.

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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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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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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 correct answer is option C i.e. 0.15 - 0.02 to 0.15 + 0.02

The correct calculation for a 95% confidence interval for the true proportion of all US adults who feel a gym membership is a necessity is:

Margin of error = z√(p(1-p)/n)

where z is the z-score corresponding to the desired level of confidence (95% in this case), p is the sample proportion (0.15), and n is the sample size (1522).

From a standard normal distribution table, the z-score for a 95% confidence level is approximately 1.96.

Substituting these values into the formula, we get:

Margin of error = 1.96 * √(0.15*(1-0.15)/1522) ≈ 0.02

Therefore, the 95% confidence interval is:

0.15 - 0.02 to 0.15 + 0.02

which simplifies to: [0.13, 0.17]

So, the correct answer is option C.

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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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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.

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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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.

The stress (σ) for n = 100 is approximately 333.33 MPa.

The stress (σ) for n = 103 is also approximately 333.33 MPa.

To calculate the stress (σ) for n = 100 and 103 using the given Sut (ultimate tensile strength) and Fsut (factor of safety for ultimate tensile strength), we can use the formula:

σ = Sut / Fsut

Let's assume the given values of Sut and Fsut are as follows:

Sut = 500 MPa (megapascals)

Fsut = 1.5 (dimensionless)

For n = 100:

Plugging in the values into the formula, we get:

σ = Sut / Fsut

= 500 MPa / 1.5

≈ 333.33 MPa

So, the stress (σ) for n = 100 is approximately 333.33 MPa.

Similarly, for n = 103:

Using the same formula with the given values of Sut and Fsut:

σ = Sut / Fsut

= 500 MPa / 1.5

≈ 333.33 MPa

So, the stress (σ) for n = 103 is also approximately 333.33 MPa.

Please note that these calculations are based on the given values of Sut and Fsut, and the units are assumed to be in megapascals (MPa) as per the given formula.

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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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