Consider the differential equation d y d x = ( y − 1 ) x 2 where x ≠ 0 . A) Find the particular solution y = f ( x ) to the differential equation with the initial condition f ( 2 ) = 1 .B) For the particular solution y = f ( x ) described in part A) find lim x → [infinity] f ( x )

The missing side lengths and the missing angles of the parallelogram are computed below

Given that we have

The opposite sides and angles of a paralleogram are equal

So, we have

CD = 17

AC = 9
CB = 17.5

TD = 10.5

Also, we have

ACD = 75

CDB = 105

CAB = 105

DBC = 45

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The latest time Jane and her family can start hiking is 3:19 p.m.

The latest time Jane and her family can start hiking is as follows:

1. First, find the total time needed for hiking and setting up camp: 1 hour and 36 minutes for hiking, plus 57 minutes for setting up camp.

2. Convert the hours and minutes to minutes only: 1 hour = 60 minutes, so 1 hour and 36 minutes = 60 + 36 = 96 minutes.

3. Add the hiking time (96 minutes) to the camp setup time (57 minutes): 96 + 57 = 153 minutes.

4. Convert 153 minutes back to hours and minutes: 153 minutes = 2 hours and 33 minutes.

5. Subtract the total time needed (2 hours and 33 minutes) from the sunset time (5:52 p.m.): 5:52 p.m. - 2 hours and 33 minutes.

6. The latest time Jane and her family can start hiking is 3:19 p.m.

Your answer: The latest time Jane and her family can start hiking is 3:19 p.m.

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To write the iterated integral for the shaded region r in exercises 21-26, we need to use the concept of double integrals. A double integral is a type of iterated integral that allows us to integrate over a two-dimensional region.

The iterated integral for the shaded region r can be written as:

∫∫r f(x,y) dA

where f(x,y) is the function we are integrating and dA represents the infinitesimal area element.

To evaluate the double integral, we can use either the row-first or column-first method. The row-first method involves fixing the value of y and integrating with respect to x first, while the column-first method involves fixing the value of x and integrating with respect to y first.

For example, in exercise 21, the shaded region r is the rectangle with vertices (0,0), (0,2), (3,2), and (3,0). If we want to integrate the function f(x,y) over this region, we can write the iterated integral as:

∫0^3 ∫0^2 f(x,y) dy dx

This means we first integrate f(x,y) with respect to y from y=0 to y=2, and then integrate the resulting expression with respect to x from x=0 to x=3.

Similarly, we can write the iterated integral for the shaded region r in exercises 22-26 using the same concept of double integrals.

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Vivian paid $17.36 for 3.5 pounds.

Given that, one pound of sirloin steak costs $4.95, Vivian bought 3.5 pounds of the sirloin steak for a church cookout,

We need to find the total she paid for 3.5 pounds,

So,

if 1 pound = 4.96

so, 3.5 = 4.96×3.5

= 17.36

Hence, Vivian paid $17.36 for 3.5 pounds.

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The measure of the angle x will be 57°.

The angle of an arc in a circle is defined as the angle subtended by the arc at the centre of the circle. The angle of the arc is measured in degrees or radians.

There is a theorem related to the angle of an arc in a circle called the central angle theorem. This theorem states that the angle subtended by an arc at the centre of a circle is equal to double the angle subtended by the same arc at any point on the circumference of the circle.

The measure of the angle x is,

x = ( 360 - 123 - 123 ) / 2

x = 57°

Hence, the value of an angle x is 57°.

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The exponential distribution is a continuous probability distribution that describes the time between events in a Poisson process, where events occur continuously and independently at a constant rate. The probability density function of the exponential distribution is given by f(x) = λe^(-λx), where λ is the rate parameter.

a. To show that EX" = EXn-1, we need to use the memoryless property of the exponential distribution. This property states that the conditional probability of X > t+s given that X > s is equal to the unconditional probability of X > t, for any s,t > 0. Using this property, we can write:

EX" = E(X|X > n-1) + (n-1) = E(X) + n-1 = 1/λ + n-1

EXn-1 = E(X|X > 1) + (n-2) = E(X) + n-2 = 1/λ + n-2

Since EX" = EXn-1, we have shown that the memoryless property holds.

b. To find EX" = n!, we can use the moment generating function (MGF) of the exponential distribution, which is given by M(t) = λ/(λ-t). The nth moment of X is defined as E(X^n) = (-1)^n d^n M(t)/dt^n at t=0. Differentiating the MGF n times, we get:

E(X^n) = n!λ^n/(λ-t)^n+1 at t=0

Setting n=1, we get E(X) = 1/λ, which is the mean of the exponential distribution. Setting n=2, we get E(X^2) = 2/λ^2, which is the variance of the exponential distribution.

For n>2, we can use the formula above to find the nth moment:

E(X^n) = n!λ^n/(-1)^{n+1} = n!/(λ^n)

Therefore, for n = 1,2,3,..., we have:

EX" = E(X^n|X > n-1) = (n-1)!/(λ^n-1) = n!/λ^n

Thus, we have shown that EX" = n! for n = 1,2,3,...

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To evaluate the principal value of the integral 00 dx / (x^2 + 2), we need to find the limit of the integral as the limits of integration approach 0 from both the positive and negative sides. The Principal value of the integral = arctan(x / sqrt(2)) + C.

Evaluate the principal value of the integral ∫(1 / (x^2 + 2)) dx
Here is a step-by-step explanation to solve this integral:
Step 1: Identify the function to integrate
The given function is f(x) = 1 / (x^2 + 2).
Step 2: Perform substitution
We can perform a trigonometric substitution to make integration easier. Let x = sqrt(2) * tan(u), so dx = sqrt(2) * sec^2(u) du.
Step 3: Rewrite the integral with substitution
The integral becomes ∫(sqrt(2) * sec^2(u) du / (2*tan^2(u) + 2)).
Step 4: Simplify the integrand
Simplify the expression inside the integral to get ∫(sec^2(u) du / (sec^2(u))).
Step 5: Integrate the simplified expression
Since the numerator and denominator are the same, the integrand simplifies to 1. Now, we just need to integrate ∫1 du. The result is the integral of 1 with respect to u, which is simply u + C, where C is the constant of integration.
Step 6: Replace u with the original variable
Recall that we set x = sqrt(2) * tan(u), so u = arctan(x / sqrt(2)). Therefore, the final answer is:
Principal value of the integral = arctan(x / sqrt(2)) + C

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To find aw/as and aw/at using the Chain Rule:

(a) Using the

Chain Rule

, we have:

aw/as = (dw/ds) * (ds/as) = ((dw/dx) * (dx/ds) + (dw/dy) * (dy/ds) + (dw/dz) * (dz/ds)) * (ds/as)

Substituting

the given expressions for x, y, and z, we get:

x = 2t, y = s - 2t, z = st

dx/ds = 0, dy/ds = 1, dz/ds = t

dx/dt = 2, dy/dt = -2, dz/dt = s

Therefore:

aw/as = ((z/x) * 2 + (z/y) * (-2) + (x*y)) * (1/s) = (2st/x - 2st/y + 2st) * (1/s)

= 2st * (y/x - y/y + 1) * (1/s) = 2st * (1 - 2/s) * (1/s)

= 2t * (1 - 2/s)

Similarly, we can find

aw/at

:

aw/at = (dw/dx) * (dx/dt) + (dw/dy) * (dy/dt) + (dw/dz) * (dz/dt)

= z * 2 + (-z) * 2 + xy

= 2st - 2st + 2ts

= 2ts

Therefore, aw/as = 2t * (1 - 2/s) and aw/at = 2ts.

(b) To find aw/as and aw/at by converting w to a function of s, we substitute the expressions for x and y in terms of s:

x = 2t, y = s - 2t

into the expression for w:

w = xyz = (2t)(s - 2t)(st) = 2st^2 - 4t^2s

Then we differentiate w with respect to s and t to get:

dw/ds = 4t^2 - 4t

dw/dt = 4st - 8ts

Using the Chain Rule, we can find aw/as and aw/at:

aw/as = dw/ds = 4t^2 - 4t

aw/at = dw/dt = 4st - 8ts

Therefore,

aw/as = 4t^2 - 4t

and aw/at = 4st - 8ts.

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If numerator of fraction is 3 less than denominator which is equivalent to "9/10", then the fraction is 27/30.

A "Fraction" is a mathematical representation of a part of a whole, expressed as one number (the numerator) divided by another (the denominator), separated by a horizontal line.

Let us assume the denominator of the fraction be = x.

According to the problem, the numerator of the fraction is 3 less than the denominator.

So, numerator of fraction can be represented as :  x - 3,

We also know that the fraction is equivalent to 9/10.

So, the equation is :

⇒ (x - 3)/x = 9/10,

Next, we cross-multiply,

⇒ 10(x - 3) = 9x,

⇒ 10x - 30 = 9x,

⇒ x = 30,

Now, we substitute it in the expression for the numerator:

We get,

⇒ x - 3 = 30 - 3 = 27,

Therefore, the fraction is 27/30, which can be simplified by dividing both the numerator and denominator by 3 to get : 9/10.

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Different function model used in to model different situations of real life.l, for example Linear, quadratic and polynomial function model. The slope of a hill, roller coaster designers are real life example of polynomial model.

Various functions can be used to test real-world situations. We have a linear business model associated with the product or the main features of the business that makes them ascending or descending. A quadratic function simulates an increase followed by a decrease.

Polynomial functions simulate many changes in direction. Multinomial models can now be used to investigate situations where the relationship between variable and estimator is curvilinear. Sometimes nonlinear relationships at the small scale of the description can also be modeled with polynomials. For example, roller coaster designers may use polynomial model to describe the bends of their rides. Other examples include the continuation of slopes, curved bridges or mountains which are based on polynomial function modelling.

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The derivative of the natural logarithm function, ln(x), is simply the reciprocal of x i.e. d/dx ln(x) = 1/x.

The derivative of ln(x) with respect to x, denoted as d/dx ln(x), can be found using the logarithmic differentiation technique. First, we express ln(x) as the natural logarithm of e raised to the power of ln(x):
ln(x) = ln(e^(ln(x)))
Using the chain rule, we can then find the derivative of ln(x) as:
d/dx ln(x) = d/dx ln(e^(ln(x)))
          = 1/x

Therefore, the derivative of ln(x) with respect to x is simply 1/x. This result can be useful in solving problems involving exponential and logarithmic functions, particularly when finding the slopes of tangent lines or rates of change.

In the context of exponentials and logs, "d/dx ln(x)" refers to finding the derivative of the natural logarithm function, ln(x), with respect to x. The natural logarithm is the inverse of the exponential function with base e (approximately equal to 2.718).
The derivative of ln(x) with respect to x is given by:
d/dx ln(x) = 1/x

So, the derivative of the natural logarithm function, ln(x), is simply the reciprocal of x.

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L<1, the series is absolutely convergent by the Ratio Test.

What is ratio test?

The ratio test is a convergence test used to determine whether an infinite series converges or diverges. It is based on the idea that if the ratio of successive terms in a series approaches a limit L as n approaches infinity, then the series converges absolutely if L < 1, diverges if L > 1, and inconclusive if L = 1.

More formally, given a series [tex]\sum_{n=1}^\infty a_n[/tex], we consider the limit

[tex]L = \lim_{n\to\infty} \left|\frac{a_{n+1}}{a_n}\right|[/tex]

If L < 1, then the series converges absolutely. If L > 1, then the series diverges. If L = 1, then the ratio test is inconclusive and we may need to use other tests to determine convergence or divergence.

Using the ratio test, we have:

[tex]\lim_{n\to\infty} \left|\frac{a_{n+1}}{a_n}\right| &= \lim_{n\to\infty} \left|\frac{(-1)^{n+1}\frac{(n+1)^3 3^{n+1}}{(n+1)!}}{(-1)^n\frac{n^3 3^n}{n!}}\right|\\\&=\lim_{n\to\infty} \left|\frac{(n+1)^3 3}{n^3}\right|\&=\lim_{n\to\infty} \left|\frac{n^3+3n^2+3n+1}{n^3}\cdot 3\right|\\&=\lim_{n\to\infty} \left(3+\frac{3}{n}+\frac{3}{n^2}+\frac{1}{n^3}\right)\\&=3[/tex]

Since L<1, the series is absolutely convergent by the Ratio Test.

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To sketch the curve, we first need to plot the points where the equations intersect with the y-axis. For f(y) = y^2 + 2, when y = 0, f(y) = 2. So the point (0, 2) is on the curve. For g(y) = 0, the equation intersects with the y-axis at y = 0.

To sketch the curve for the given equations, follow these steps:

1. Identify the equations: We have f(y) = y^2 + 2, g(y) = 0, y = -1, and y = 2.
2. Plot the functions: f(y) is a parabolic curve with a vertex at (0, 2). g(y) is a horizontal line along the y-axis (y = 0). y = -1 and y = 2 are two horizontal lines at y = -1 and y = 2 respectively.
3. Sketch the curve: Draw the parabola f(y) = y^2 + 2 with its vertex at (0,

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The surface area of revolution about the y-axis of y = 36 - 4x² over the interval 0 is approximately 2261.63 square units.

To find the surface area of revolution about the y-axis of y = 36 - 4x² over the interval 0, we can use the formula:

SA = 2π ∫[a,b] x √(1 + (dy/dx)²) dx In this case, a = 0 and b = 6 (since y = 0 when x = ±3), so we have: SA = 2π ∫[0,6] x √(1 + (dy/dx)²) dx

To find dy/dx, we can take the derivative of y with respect to x: dy/dx = -8x Substituting this into the formula, we get: SA = 2π ∫[0,6] x √(1 + (-8x)²) dx

Integrating this function is not trivial, but we can use a substitution u = 1 + (-8x)² to simplify it: SA = π ∫[1,1+(-8*6)²] (u-1)/16 √u du

Now we can use the power rule to integrate: SA = π [(2/3)(u-1)^(3/2)]|[1,1+(-8*6)²] SA = π [(2/3)(1+(-8*6)²-1)^(3/2)-(2/3)(1-1)^(3/2)] SA = π (2/3)(1+(-8*6)²)^(3/2) SA ≈ 2261.63

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Answer:

[tex] \sqrt{72 {m}^{5} {n}^{2} } [/tex]

[tex] \sqrt{2 \times 36 \times {m}^{4} \times m \times {n}^{2} } [/tex]

[tex]6 {m}^{2} n \sqrt{2m} [/tex]

C is the correct answer.

The probability that the sweet is a cherry sweet is given as follows:

p = 4/11.

A probability is calculated as the division of the desired number of outcomes by the total number of outcomes in the context of a problem/experiment.

The probability that it is an apple sweet is 2/11, and the ratio of apple sweets to pear sweets is 2: 5, hence the probability of a pear sweet is given as follows:

p = 5/11.

Then the probability that the sweet is a cherry sweet is given as follows:

p = 1 - (5/11 + 2/11)

p = 1 - 7/11

p = 11/11 - 7/11.

p = 4/11.

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To solve the equation 2y^(4) +11y^(3) + 18y" + 4ť' - 8y=0, we can first factor out a common factor of 2y: 2y(y^3 + 5y^2 + 9y + 2ť' - 4) = 0 Next, we can focus on the expression inside the parentheses, which can be written as:

y^3 + 5y^2 + 9y + 2ť' - 4

We can recognize this expression as the polynomial 2r^4 +11r^3 + 18r^2 + 4r – 8 evaluated at r=y-1/2.

So, 2r^4 +11r^3 + 18r^2 + 4r – 8 = (2r - 1)(r + 2)^3.

Finally, we can factor out a common factor of 2 and use the factored form of the polynomial:

2(y-1)(y+2)^3(2y+1) = 0

Therefore, the solutions to the original equation are:

y = 1, -2, -1/2 (multiplicity 3)
To solve the given polynomial equation, we will first rewrite it in a more accurate and coherent form by removing the irrelevant terms:

2y^4 + 11y^3 + 18y^2 + 4y - 8 = 0

We are given a hint that the equation can be factored as:

(2r - 1)(r + 2)^3

Now, we will replace r with y to match the variable in the given equation:

(2y - 1)(y + 2)^3 = 0

Now that we have the factored form of the equation, we can set each factor equal to zero and solve for y:

1) 2y - 1 = 0
2y = 1
y = 1/2

2) (y + 2)^3 = 0
y + 2 = 0
y = -2

So the solutions to the given equation are y = 1/2 and y = -2.

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Answer:

64

Hope this helps!

Step-by-step explanation:

1, 4 ( 1 × 4 ), 16 ( 4 × 4 ), 64 ( 16 × 4 ), 256 ( 64 × 4 ), etc.

(a) The electric dipole moment (p) is zero in this case due to equal and opposite charges.

(b) The magnetic field in the radiation zone is given by a complex formula involving various parameters and coordinates.

(c) The angular distribution of the radiation is proportional to a simplified expression involving cosines of angles.

The total time-averaged power radiated is calculated using a formula involving parameters such as speed of light, dipole moment, and distance.

(a) The lowest nonvanishing multipole moments are the electric dipole moment (p) and the magnetic dipole moment (m), which are given by:

p = qd

m = (1/c) ∫r x j dV

where q is the charge, d is the displacement vector, j is the current density, and V is the volume. In this case, the two electric dipoles have equal and opposite charges, so their net charge is zero and the electric dipole moment is:

p = qd = 0

(b) The magnetic field in the radiation zone is given by the formula:

H = (cpa/2πk³) [(x+iy)cos∅ - zsin∅e^(jo∅)]cos∅e^(ikr)/r
where c is the speed of light, p is the dipole moment, a is the distance between the dipoles, k is the wave number, r is the distance from the source, x, y, and z are the coordinates of the observation point, ∅ is the angle between the observation point and the axis of rotation, and e is the base of the natural logarithm. The overall phase factor is not important for the purposes of this problem.

(c) The angular distribution of the radiation is proportional to (cos²∅ + cos⁴∅), which can be simplified as follows:
cos²∅ + cos⁴∅ = (1/2)(1 + cos²∅ + 2cos⁴∅/2)
= (1/2)(1 + cos²∅ + (1/2)(1 + cos 2∅ + cos 4∅))
= (3/4) + (1/4)cos 2∅ + (1/8)cos 4∅
The total time-averaged power radiated is given by the formula:
p = (4/15π€₀)ck⁶p²a²
where €₀ is the vacuum permittivity.

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Answer:

Step-by-step explanation:

Kylie has to try at most 3,024 different combinations in order to call Katerina if she does not dial repetitive phone numbers.

Kylie only recalls the first six numbers of Katerina's phone number, so she must guess the last four. Because she cannot repeat any of the numbers she has already successfully predicted, the first remaining digit has just nine viable alternatives (all digits from 0 to 9 except the one she already knows). Similarly, the second remaining digit has just eight options, the third has seven, and the fourth has six.

Therefore, the total number of possible combinations is:

9 x 8 x 7 x 6 = 3,024

This implies Kylie will have to try at most 3,024 different combinations to reach Katerina, provided she gets it right on the last try.

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Answer:

A) 7

B) (47.5,54.5)

Step-by-step explanation:

a) The margin of error is half the length of the confidence interval, so the length of the confidence interval is 2 times the margin of error:

Length of confidence interval = 2 x 3.5 = 7.

b) If the sample mean is 51 and the margin of error is 3.5, then the confidence interval can be calculated as follows:

Lower bound = sample mean - margin of error = 51 - 3.5 = 47.5

Upper bound = sample mean + margin of error = 51 + 3.5 = 54.5

Therefore, the 95% confidence interval for the population mean is (47.5, 54.5).

The area of the region inside the cardioid and outside the circle is 3π/2 square units.

The area of the region inside the cardioid r=1+cosθ and outside the circle r=3cosθ using a double integral, follow these steps:

1. Determine the bounds of integration for θ: Find where the cardioid and circle intersect by setting r equal for both equations: (1+cosθ) = 3cosθ. Solve for θ, which results in θ = 0 and θ = π.

2. Set up the double integral: The area of the region can be found using the double integral of the difference between the two polar functions with respect to r and θ: Area = ∬(1+cosθ - 3cosθ) rdrdθ.

3. Determine the bounds of integration for r: The lower bound for r is the circle r=3cosθ, and the upper bound is the cardioid r=1+cosθ.

4. Integrate with respect to r: ∫[∫(1+cosθ - 3cosθ) rdr]dθ from r=3cosθ to r=1+cosθ. This results in: [1/2(r^2)] evaluated from r=3cosθ to r=1+cosθ.

5. Plug in the limits of integration for r: [(1/2)((1+cosθ)^2) - (1/2)(3cosθ)^2]dθ.

6. Integrate with respect to θ: ∫[(1/2)((1+cosθ)^2) - (1/2)(3cosθ)^2] dθ from θ=0 to θ=π.

7. Evaluate the integral: After integrating and evaluating the limits, you will find that the area of the region inside the cardioid and outside the circle is 3π/2 square units.

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To graph the function d = 8 tan 2xt on the interval (0,2), we can plot several points by selecting different values of t and calculating the corresponding value of d. The function d = 8 tan 2xt is undefined whenever the tangent function is undefined, which occurs when the angle of the tangent is equal to (2k + 1)π/2, where k is an integer.

a) For example, when t = 0.5 seconds, we have d = 8 tan(2(0.5)x) = 8 tan(x), and we can evaluate this expression for different values of x to get points on the graph. Repeat this process for different values of t to obtain more points and connect them to form the graph.

b) Since tan(x) is undefined when x = (2k + 1)π/2, we have 2xt = (2k + 1)π/2, which implies t = (2k + 1)π/4x for k ∈ Z. Therefore, the function is undefined at times t = (2k + 1)π/4x for k = -4, -3, -2, -1, 0, 1, 2, 3, 4. Plugging in x = 10.2, we get t = ±0.3877, ±1.162, ±1.936, ±2.71. Therefore, the function is undefined at these values of t.

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The estimated temperature at the point (2.04, 0.98) on the unevenly heated metal plate is approximately 138.66 degrees Celsius.

To estimate the temperature at the point (2.04, 0.98), we can use the first-order Taylor approximation, which includes the partial derivatives of the temperature T(x,y) with respect to x and y.

Given the information T(2,1) = 138, Tx(2,1) = 12, and Ty(2,1) = -9, we can estimate T(2.04, 0.98) as follows:

T(2.04, 0.98) ≈ T(2,1) + Tx(2,1) * (2.04 - 2) + Ty(2,1) * (0.98 - 1) T(2.04, 0.98) ≈ 138 + 12 * (0.04) - 9 * (-0.02) T(2.04, 0.98) ≈ 138 + 0.48 + 0.18 T(2.04, 0.98) ≈ 138.66.

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For i) we get: a = -1/3 and b = -20/3 and for ii) we get the polynomial satisfying all the given properties is: P(x) = x³ - (1/3)x² - (20/3) and for iii) we get c=0

Explanation:

To satisfy property iii) P(0)=0, we know that c must be equal to 0.

Let's now use the first two properties to find the values of a and b.

i) At x = -3, P'(x) = 0 and P''(x) < 0 for a local maximum.

P'(x) = 3x² + 2ax + b

P''(x) = 6x + 2a

Substituting x = -3 in the above equations, we get:

9a - 9 + b = 0

-18 + 2a < 0

Solving the above two equations simultaneously, we get:

a = -1/3 and b = -20/3

ii) At x = 7, P'(x) = 0 and P''(x) > 0 for a local minimum.

Using the same approach as above, we get:

a = -2/3 and b = 532/9

Therefore, the polynomial satisfying all the given properties is:

P(x) = x³ - (1/3)x² - (20/3)x

Note that property iii) is satisfied because we set c=0 earlier.

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Since all three conditions are satisfied, we can conclude that S is a subspace of M2(R). To prove that s is a subspace of (m2(r), , .), we need to show that s satisfies the three axioms of a subspace:

1. s contains the zero vector:
The zero vector in m2(r) is the 2x2 matrix with all entries equal to zero. We can verify that this matrix satisfies ab = ba for any matrix b, so the zero vector is in s.

2. s is closed under vector addition:
Let b1 and b2 be matrices in s. We need to show that their sum, b1 + b2, is also in s.

(ab1 + ab2) = a(b1 + b2) = ab1 + ab2 (using the distributive property of matrix multiplication)

Similarly,

(b1a + b2a) = (b1 + b2)a = b1a + b2a

So b1 + b2 satisfies the condition ab = ba and is therefore in s.

3. s is closed under scalar multiplication:
Let b be a matrix in s, and let c be a scalar. We need to show that the product cb is also in s.

(acb) = a(cb) = a(bc) = (ab)c = (ba)c = b(ac)

So cb satisfies the condition ab = ba and is therefore in s.

Since s satisfies all three axioms of a subspace, we can conclude that s is indeed a subspace of (m2(r), , .).
To determine if the set S is a subspace of the vector space M2(R), we need to check if it satisfies three conditions: closure under addition, closure under scalar multiplication, and the existence of the zero vector.

1. Closure under addition:
Let B1 and B2 be two matrices in S such that AB1 = B1A and AB2 = B2A. We need to check if the sum B1 + B2 is also in S.
A(B1 + B2) = AB1 + AB2 = B1A + B2A = (B1 + B2)A, which shows that the sum B1 + B2 is in S.

2. Closure under scalar multiplication:
Let B be a matrix in S such that AB = BA, and let c be a scalar in R. We need to check if cB is also in S.
A(cB) = c(AB) = c(BA) = (cB)A, which shows that the product cB is in S.

3. Existence of the zero vector:
The zero matrix 0 satisfies A0 = 0A = 0, so the zero matrix is in S.

Since all three conditions are satisfied, we can conclude that S is a subspace of M2(R).

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The highest free throw of 82% has an average bowling score of 108

This ordered pair represents the ordered pair of the highest x value in the graph

From the graph, the ordered pair is (82, 108)

It means that the student with the highest free throw of 82% has an average bowling score of 108

This ordered pair represents the ordered pair of the highest y value in the graph

From the graph, the ordered pair is (80, 112)

It means that the student with the highest free throw of 80% has an average bowling score of 112

From the graph, we can see that the association is a positive linear association

This is because as the free throw percents increases, the average bowling score is also expected to increase

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Complete question

The members of a school basketball team went bowling for a season ending party. They made this scatter plot to compare their free throw percents for the season and their average bowling score for 3 games.

Part A:

What ordered pair represents the student with the highest free throw percent? Explain the meaning of each coordinate in the ordered pair.

Part B:

What ordered pair represents the student with the highest bowling average? Explain the meaning of each coordinate in the ordered pair.

Part C:

Is the association between free throw percent and bowling average linear or nonlinear? If it is linear, is the relationship positive, negative, or neither? State the association, if any, in terms of the variables.

Larry has a 0.12 chance of hitting the inner bullseye and thus, his probability of winning is also 0.12.

His probability of hitting the outer bullseye is 0.31, thereby resulting in a winning likelihood of 0.19 when subtracting 0.12.

Pia holds a 0.13 likelihood for hitting the inner bullseye, estimating her overall probability of securing victory as 0.01 - being 0.13 after deducting 0.12. Moreover, her potential of hitting the outer bullseye stands at 0.35, rendering a probability of success to be 0.22 when considering the 0.13 deduction.

Lastly, Carina's chances of making it to the inner bullseye stand at 0.25, indicating a probability of attaining triumph at 0.13 - following a subtraction of 0.12 from 0.25. On top of that, her possibility of striking the outer bullseye rests at 0.49, resulting in an odds of conquering the game as 0.24 after subtracting 0.25.

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