(1 point) find the pdf of = when and have the joint pdf ,
f(x)={ 1/900 0≤x,y≤3
0, otherwise.

Answer:

Have a Nice Best Day : ) i'm sorry there where no Answer

The process of dividing a data set into a training, a validation, and an optimal test data set is called data partitioning.

Data partitioning is the process of dividing a dataset into separate subsets that are used for different purposes, such as training a model, validating its performance, and testing it on new data.

The most common way to partition a dataset is into three subsets: a training set, a validation set, and a test set. The training set is used to train a model, the validation set is used to tune the model's hyperparameters and assess its performance during training, and the test set is used to evaluate the final performance of the model on new, unseen data.

Data partitioning helps to prevent overfitting by providing a way to evaluate a model's performance on data that it has not seen during training.

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Answer:

x < -4

Step-by-step explanation:

-2x - 6 > 3x + 14  Add 2x to both sides

-2x + 2x - 6 > 3x + 2x  + 14

-6 > 5x + 14  Subtract 14 from both sides

-6 - 14 > 5x + 14 - 14

-20 > 5x  Divide both sides by 5

[tex]\frac{-20}{5}[/tex] > [tex]\frac{5}{5}[/tex] x

-4 > x or x < -4

Helping in the name of Jesus.

Answer:

So the three consecutive numbers are:

14,15, and 16.

Step-by-step explanation:

Let the three consecutive integers be = x , x+1,  x+ 2 sum = 45

then,

x + (x + 1) + (x +2)  = 45

-> 3x + 3 = 45

-> 3x = 45 - 3

-> x = 14

-> x = 14

-> x + 1 = 15

-> x + 2 = 16

So, three consecutive numbers are : 14, 15, and 16.

To multiply powers with the same base, add the exponents.

[tex] {a}^{8} {a}^{7} = {a}^{15} [/tex]

The expression e^(-3)(1+3+3^2/2!+3^3/3!+3^4/4!) equals the cumulative probability of 4 arrivals during an interval for which the average number of arrivals equals 3.

Here's a step-by-step explanation:

1. Recognize that the given expression represents the cumulative probability for a Poisson distribution.
2. Identify the average number of arrivals (λ) as 3, which is the exponent in the e^(-3) term.
3. Recognize that the terms inside the parentheses correspond to the Poisson probability mass function (PMF) for k=0, 1, 2, 3, and 4 arrivals.
4. Since the expression sums up the probabilities for k=0 to k=4, it represents the cumulative probability of 4 arrivals.
5. In summary, the expression represents the cumulative probability of 4 arrivals during an interval where the average number of arrivals is 3.

Answer: Your answer is completely correct. It is just that when answering the question, you should assume that the 48 lb child is on the left, and the 72 lb child is on the right. Usually, I always assume that the first mentioned item is the left most one.

Step-by-step explanation:

This is how I will set up the problem: M1 = 48 lbs, M2 = 72 lbs, L = 10 ft

Since (M1 * 0 + M2 * 10)/(M1+M2) = equilibrium, we can use this equation to find the solution:

0 + 720 / (48+72) = 6 feet

Answer:

the ans is 6.4

Step-by-step explanation:

using the distance formula

d^2= (x2-x1)^2 + (y2-y1)^2

d^2= (8-4)^2 + (8-3)^2

d^2= (4)^2 + (5)^2

d^2= 16+ 25

d^2= 41

d= sqrt of 41*

d= 6.4units

For the function f(x) = -4x^3 + 5x^2 + 2x, the end behavior can be determined by looking at the degree and leading coefficient of the polynomial. Since the degree is odd and the leading coefficient is negative, the end behavior of the function will be downward in both the left and right quadrants. Therefore, the graph would be D) the arrow points downwards in the lower left and lower right quadrants.

For the function f(x) = (2x-3)(x+1), the end behavior can be determined by looking at the degree of the polynomial. Since the degree is 2, the end behavior will be the same as that of a quadratic function, which means that the graph will either be an upward or downward parabola. In this case, the graph would be A) the arrow points upwards in the upper left quadrant and downwards in the lower right quadrant, because the leading coefficient is positive.

For the function f(x) = 3x - 1, the end behavior is a straight line with a slope of 3. The arrow would be pointing upwards in both the left and right quadrants, so the graph would be B) the arrow points upwards in the upper left quadrant as well as in the upper right quadrant.

C) the arrow points upwards in the upper right quadrant and downwards in the lower left quadrant

This is because the function f(x) = -5x^2 (x+1) (x+3) is a cubic function with a leading coefficient of -5, which means that the end behavior of the function will be downward in the lower left quadrant and upward in the upper right quadrant.

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The requried volume of the cone is 182 cubic inches.

The area of the base of the cylinder is given by:

[tex]A_{cylinder} = \pi r^2[/tex]

where r is the radius of the cylinder. We know that the area of the base is 39 square inches, so we can write:

[tex]\pi r^2 = 39[/tex]

Solving for r, we get:

r = √(39/π)

The height of the cylinder is given as 14 inches. Therefore, the volume of the cylinder is:

[tex]A_{cylinder} = \pi r^2\\ A_{cylinder}= \pi (39/ \pi )(14)\\ A_{cylinder}= 546 \ \ \ cubic inches.[/tex]

Similarly,

The volume of the cone ([tex]V=1/3 \pi r^2h[/tex]) is 182 cubic inches.

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Answer:

height = 10 m

lengths of bases = 5 m and 10 m

[tex] \frac{1}{2} (10)(5 + 10) = 5(15) = 75[/tex]

So the area of this trapezoid is 75 square meters.

Check the picture below.

[tex]\textit{area of a trapezoid}\\\\ A=\cfrac{h(a+b)}{2}~~ \begin{cases} h~~=height\\ a,b=\stackrel{parallel~sides}{bases~\hfill }\\[-0.5em] \hrulefill\\ a=5\\ b=10\\ h=10 \end{cases}\implies A=\cfrac{10(5+10)}{2}\implies A=75~m^2[/tex]

Answer:

Step-by-step explanation:

Using the least squares regression method, we obtain the equation of the regression line: y = 22.4x + 26.2. The y-intercept is the value of y when x = 0, which is 26.2. Therefore, the answer is d) 26.2.

The correct answer is d) can be used to predict a value of y if the corresponding x value is given. A least squares regression line is a statistical method used to find the equation of a line that best fits the data points. It can be used to predict the value of the dependent variable (y) for a given value of the independent variable (x).

The regression function is ŷ = 900 + 6x, where x is the advertising in $100s and ŷ is the sales in $1,000s. To find the point estimate for sales when advertising is $10,000, we substitute x = 100 in the regression function: ŷ = 900 + 6(100) = 1,500. Therefore, the answer is a) $1,500.

The least squares estimate of the y-intercept is 42.6.

An intercept is a point on the y-axis, through which the slope of the line passes. It is the y-coordinate of a point where a straight line or a curve intersects the y-axis. This is represented when we write the equation for a line, y = mx+c, where m is slope and c is the y-intercept.

Given that,

x    y

2   50

5   70

4   75

3   80

6   94

Calculate the means of x and y values:

x_mean = (2 + 5 + 4 + 3 + 6) / 5 = 20 / 5 = 4

y_mean = (50 + 70 + 75 + 80 + 94) / 5 = 369 / 5 = 73.8

Calculate the differences from the means for x and y:

x_diff = [2-4, 5-4, 4-4, 3-4, 6-4] = [-2, 1, 0, -1, 2]

y_diff = [50-73.8, 70-73.8, 75-73.8, 80-73.8, 94-73.8] = [-23.8, -3.8, 1.2, 6.2, 20.2]

Calculate the product of the x and y differences and the square of x differences:

xy_diff = [-2×(-23.8), 1×(-3.8), 0×1.2, -1×6.2, 2×20.2] = [47.6, -3.8, 0, -6.2, 40.4]

x_squared_diff = [-2², 1², 0², -1², 2²] = [4, 1, 0, 1, 4]

4. Sum up the product of the x and y differences and the square of x differences:

sum_xy_diff = 47.6 - 3.8 + 0 - 6.2 + 40.4 = 78

sum_x_squared_diff = 4 + 1 + 0 + 1 + 4 = 10

Calculate the slope (m):

m = 78 / 10 = 7.8

Use the slope (m) to find the least squares estimate of y-intercept (b) using the equation

b = 73.8 - 7.8 × 4 = 73.8 - 31.2

= 42.6

Therefore, the least squares estimate of the y-intercept is 42.6.

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We can assume that the sampling distribution of the sample mean is normally distributed and the probability that the sample mean falls between 75 and 77 is 0.4582 or 45.82%.

Sampling distribution of the sample mean is normally distributed

Use the standard normal distribution to evaluate the probability that the sample mean falls between 75 and 77.

First, lets calculate standard error of the mean:

SE = σ/√n

Since we are not given the population standard deviation (σ), we will use the sample standard deviation (s) as an estimate:

SE = s/√n

Next, we need to calculate the z-scores corresponding to 75 and 77:

z1 = (75 - x) / SE
z1 = (75 - x) / (s/√n)

z2 = (77 - x) / SE
z2 = (77 - x) / (s/√n)

Since the sampling distribution is normal, we can use a standard normal distribution table or a calculator to find the probabilities associated with these z-scores.

P(75 ≤ x ≤ 77) = P(z1 ≤ Z ≤ z2)

We find that:

P(-0.71 ≤ Z ≤ 0.71) = 0.4582

Therefore, the probability that the sample mean falls between 75 and 77 is 0.4582 or 45.82% (rounded to 4 decimal places).

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Let x be a discrete random variable. If Pr(x < 6) = 3/9, and Pr(x ≤ 6) = 7/18, then P(X = 6) is 0.06.

A discrete random variable is a variable that can take on only a countable number of values. Examples of discrete random variables include the number of heads when flipping a coin, the number of cars passing through an intersection in a given hour, or the number of students in a classroom.

Let x be a discrete random variable.

Pr(x < 6) = 3/9, and Pr(x ≤ 6) = 7/18

P(X ≤ 6) = P(X < 6) + P(X = 6)

Subtract P(X < 6) on both side, we get

P(X = 6) = P(X ≤ 6) - P(X < 6)

Substitute the values

P(X = 6) = 7/18 - 3/9

First equal the denominator

P(X = 6) = 7/18 - 6/18

P(X = 6) = 1/18

P(X = 6) = 0.06

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a) For an = 3, recurrence relation: a_n = a_(n-1); b) For an = 2n, recurrence relation: a_n = a_(n-1) + 2; c) For an = 2n + 3, recurrence relation: a_n = a_(n-1) + 2; d) For an = 5n, recurrence relation: a_n = a_(n-1) + 5; e) an = n^2, recurrence relation: a_n = a_(n-1) + 2n - 1; f) an = n^2 + n, recurrence relation: a_n = a_(n-1) + 2n; g) an = n + (-1)^n, recurrence relation: a_n = a_(n-1) + 2*(-1)^n; h) an = n!, recurrence relation: a_n = n * a_(n-1).

Explanation:
To find recurrence relations for these sequences, please note that the answers may not be unique, but I will provide one possible recurrence relation for each sequence:

a) a_n = 3

a_(n-1) = 3
Recurrence relation: a_n = a_(n-1)

b) a_n = 2n

a_(n-1) = 2(n-1)

Thus,  a_n - a_(n-1) = 2
Recurrence relation: a_n = a_(n-1) + 2

c) a_n = 2n + 3

a_(n-1)= 2(n-1) + 3

Thus, a_n - a_(n-1) = 2

a_n = a_(n-1) + 2
Recurrence relation: a_n = a_(n-1) + 2

d) a_n = 5n

a_(n-1) = 5(n-1)

Thus, a_n - a_(n-1) = 5
Recurrence relation: a_n = a_(n-1) + 5

e) a_n = n^2

a_(n-1) =  (n-1)^2

Thus, a_n - a_(n-1) = 2n - 1
Recurrence relation: a_n = a_(n-1) + 2n - 1

f) a_n = n^2 + n

a_(n-1) = (n-1)^2 +(n-1)

Thus,  a_n - a_(n-1) = 2n
Recurrence relation: a_n = a_(n-1) + 2n

g) a_n = n + (-1)^n

a_(n-1) = (n-1) + (-1)^(n-1)

Thus,  a_n - a_(n-1) = 2*(-1)^n
Recurrence relation: a_n = a_(n-1) + 2*(-1)^n

h) a_n = n!

a_(n-1) = (n-1)!

Thus,  a_n/a_(n-1)= n
Recurrence relation: a_n = n * a_(n-1)

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The series ∑3ke − k/28 is a divergent series.

To determine whether the series ∑3ke − k/28 converges or diverges, we can use the ratio test.

The ratio test states that if lim┬(n→∞)⁡|an+1/an|<1, then the series converges absolutely; if lim┬(n→∞)⁡|an+1/an|>1, then the series diverges; and if lim┬(n→∞)⁡|an+1/an|=1, then the test is inconclusive.

Let's apply the ratio test to our series:

|a(n + 1)/a(n)| = |3(n + 1) [tex]e^(^-^(^n^+^1^)/28) / (3n e^(^-^n^/^2^8^))|[/tex]

= |(n+1)/n| * |[tex]e^(^-^1^/^2^8^)[/tex]| * |3/3|

= (1 + 1/n) * [tex]e^(^-^1^/^2^8^)[/tex]

As n approaches infinity, the expression (1 + 1/n) approaches 1, and [tex]e^(^-^1^/^2^8^)[/tex] is a constant. Therefore, the limit of the ratio is 1.

Since the limit of the ratio test is equal to 1, the test is inconclusive. We need to use another method to determine convergence or divergence.

One possible method is to use the fact that [tex]e^x > x^2^/^2[/tex] for all x > 0. This implies that [tex]e^(^-^k^/^2^8^)[/tex] < [tex](28/k)^2^/^2[/tex] for all k > 0.

Therefore,

|a(k)| = 3k [tex]e^(^-^k^/^2^8^)[/tex] < 3k[tex](28/k)^2^/^2[/tex]

= 42k/k²

= 42/k

Since ∑1/k is a divergent series, we can use the comparison test to conclude that ∑|a(k)| diverges.

Therefore, the series ∑3ke − k/28 also diverges.

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The determinant of the given matrix a is: det(a) = b2c2 + a2c2 + a2b2 - 2a2b2 - 2a2c2 + 2abc.

The determinant of a 3x3 matrix can be found using the formula:

det(A) = a11(a22a33 - a32a23) - a12(a21a33 - a31a23) + a13(a21a32 - a31a22)

Substituting the given matrix values, we get:

det(a) = 1(b2c2 - c(b2) + a2(c2) - c(a2) + a(b2) - a(b2)) - a(1c2 - c1 + a2c - c(a2) + a - a(a2)) + a(1b2 - b1 + a(b2) - b(a2) + a - a(b2))

Simplifying this expression, we get:

det(a) = b2c2 + a2c2 + a2b2 - a2b2 - b2c - a2c - a2b + a2c + abc - abc - a2c + ac2 + ab2 - ab2 - abc

Simplifying further, we get:

det(a) = b2c2 + a2c2 + a2b2 - 2a2b2 - 2a2c2 + 2abc

Thus, the determinant of the given matrix a is:

det(a) = b2c2 + a2c2 + a2b2 - 2a2b2 - 2a2c2 + 2abc.

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The percentage of the original ¹⁴c is still present is  28.5%.

To calculate the percentage of original ¹⁴C still present, we need to use the formula for                                             radioactive decay:
N = N₀(1/2)^(t/h)
Where:
N₀ = initial amount of ¹⁴C
N = final amount of ¹⁴C after time t
t = time elapsed
h = half-life of ¹⁴C

Substituting the given values:
N₀ = 100%
t = 1.190 × 10⁴ years
h = 5730 years

N = 100% x (1/2)^((1.190 × 10⁴)/5730)
N = 100% x (1/2)^(2.08)
N = 100% x 0.285
N = 28.5%

Therefore, after 1.190 × 10⁴ years, approximately 28.5% of the original ¹⁴C is still present in the tree trunk.

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Answer:

y=4

Step-by-step explanation:

3×−4y=8(−2−4)

Multiply 3 and −4 to get −12.

−12y=8(−2−4)

Subtract 4 from −2 to get −6.

−12y=8(−6)

Multiply 8 and −6 to get −48.

−12y=−48

Divide both sides by −12.

y=

−12

−48

​

Divide −48 by −12 to get 4.

y=4

Answer:

X= - 12

Step-by-step explanation:

3x-4*3=-16-32

3x-12= - 48

3x= - 48+12

3x= - 36

X= - 36:3

X = - 12

Which class has the lowest median grade? Class 1

Which class has the highest median grade? Class 2

Which class has the lowest interquartile range? Class 1

Answer:

Yes they are congruent quadrilaterals.

And from the look of it, they possess the same shape and size; not to mention their length are also congruent.

Step-by-step explanation:

This furthet explains how PQR has the same angle as EFG and the length of DE is equal to the length of QR.

To use the integral test to determine the convergence of the series an = [tex]\frac{x+1}{x^{2} }[/tex], we need to check if the corresponding improper integral converges or diverges.

The integral test states that if f(x) is a positive, continuous, and decreasing function on the interval [1,inf], and if the series an = f(n) for all n in the interval [1,inf], then the series and the integral from 1 to infinity of f(x) both converge or both diverge.

In this case, we have f(x) = [tex]\frac{x+1}{x^{2} }[/tex]. First, we need to check if f(x) is positive, continuous, and decreasing on the interval [1,inf]. f(x) is positive for all x > 0. f'(x) =[tex]\frac{-2x-1}{x^{3} }[/tex] , which is negative for all x > 0. Therefore, f(x) is decreasing on the interval [1,inf].

Next, we need to evaluate the improper integral from 1 to infinity of f(x): integral from 1 to infinity of [tex]\frac{x+1}{x^{2} }[/tex] dx = lim t->inf integral from 1 to t of [tex]\frac{x+1}{x^{2} }[/tex] dx = lim t->inf [tex][\frac{-1}{t}-\frac{1}{t^{2}+t }][/tex] = 0

Since the improper integral converges to 0, the series an also converges by the integral test. Therefore, the series an [tex]\frac{x+1}{x^{2} }[/tex] is convergent on the interval [1,inf].

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The solutions to the recurrence relation an = 8an−1 − 16an−2 are:

a) {an = 0}

b) {an = 1}

c) {an = 2ⁿ}

d) {an = 4ⁿ}

g) {an = (-4)ⁿ}

A recurrence relation in mathematics is an equation that states that the last term in a series of integers equals some combination of the terms that came before it.

To determine if a sequence {an} is a solution of the recurrence relation an = 8an−1 − 16an−2, we need to substitute the sequence into the recurrence relation and see if it holds for all n.

a) If an = 0, then:

an = 8an−1 − 16an−2

0 = 8(0) − 16(0)

0 = 0

This holds, so {an = 0} is a solution.

b) If an = 1, then:

an = 8an−1 − 16an−2

1 = 8(1) − 16(0)

1 = 8

This does not hold, so {an = 1} is not a solution.

c) If an = 2n, then:

an = 8an−1 − 16an−2

2n = 8(2n−1) − 16(2n−2)

2n = 8(2n−1) − 16(2n−1)

2n = −8(2n−1)

2n = −2 × 2(2n−1)

This does not hold for all n, so {an = 2n} is not a solution.

d) If an = 4n, then:

an = 8an−1 − 16an−2

4n = 8(4n−1) − 16(4n−2)

4n = 8(4n−1) − 4 × 16(4n−1)

4n = −60 × 16(4n−1)

This does not hold for all n, so {an = 4n} is not a solution.

e) If an = n4n, then:

an = 8an−1 − 16an−2

n4n = 8(n−1)4(n−1) − 16(n−2)4(n−2)

n4n = 8(n−1)4(n−1) − 4 × 16(n−1)4(n−1)

n4n = −60 × 16(n−1)4(n−1)

This does not hold for all n, so {an = n4n} is not a solution.

f) If an = 2 ⋅ 4n/(3n4n), then:

an = 8an−1 − 16an−2

2 ⋅ 4n/(3n4n) = 8 ⋅ 2 ⋅ 4n−1/(3(n−1)4n−2) − 16 ⋅ 2 ⋅ 4n−2/(3(n−2)4n−4)

2 ⋅ 4n/(3n4n) = 16 ⋅ 4n−1/(3(n−1)4n−2) − 16 ⋅ 4n−2/(3(n−2)4n−4)

2 ⋅ 4n/(3n4n) = (16/3) ⋅ (n−1) ⋅ 4n−1/n4n−2 − (16/3) ⋅ (n−2) ⋅ 4n−2/(n−2)4n−4

2 ⋅ 4n/(3n4n) = (16/3) ⋅ (n−1) ⋅ 4n−1/n4n−2 − (16/3) ⋅ (n−2) ⋅ 4n−2/n4n−2

2 ⋅ 4n/(3n4n) = (16/3) ⋅ (n−1) ⋅ 4n−1/n4n−2 − (16/3) ⋅ (n−2) ⋅ 4n−2/n4n−2

2 ⋅ 4n/(3n4n) = (16/3) ⋅ (n−1) ⋅ 4n−1/n4n−2 − (16/3) ⋅ (n−2) ⋅ 4n−2/n4n−2

2 ⋅ 4n/(3n4n) = (16/3) ⋅ (n−1) ⋅ 4n−1/n4n−2 − (16/3) ⋅ (n−2) ⋅ 4n−2/n4n−2

2 ⋅ 4n/(3n4n) = (16/3) ⋅ (n−1)/n − (16/3) ⋅ (n−2)/n

2 ⋅ 4n/(3n4n) = (16/3) ⋅ (1 − 1/n) − (16/3) ⋅ (1 − 2/n)

2 ⋅ 4n/(3n4n) = (16/3n) ⋅ (2 − n)

This does not hold for all n, so {an = 2 ⋅ 4n/(3n4n)} is not a solution.

g) If an = (−4)n, then:

an = 8an−1 − 16an−2

(−4)n = 8(−4)n−1 − 16(−4)n−2

(−4)n = −8 ⋅ 4n−1 + 16 ⋅ 16n−2

(−4)n = −8 ⋅ (−4)n + 16 ⋅ (−4)n

This does not hold for all n, so {an = (−4)n} is not a solution.

h) If an = n2^(4n), then:

an = 8an−1 − 16an−2

n2^(4n) = 8(n-1)2^(4(n-1)) - 16(n-2)2^(4(n-2))

n2^(4n) = 8n2^(4n-4) - 16(n-2)2^(4n-8)

n2^(4n) = 8n2^(4n-4) - 16n2^(4n-8) + 512(n-2)

n2^(4n) - 8n2^(4n-4) + 16n2^(4n-8) - 512(n-2) = 0

n2^(4n-8)(2^16n - 8(2^12)n + 16(2^8)) - 512(n-2) = 0

This does not hold for all n, so {an = n2^(4n)} is not a solution.

Therefore, the solutions to the recurrence relation an = 8an−1 − 16an−2 are:

a) {an = 0}

b) {an = 1}

c) {an = 2ⁿ}

d) {an = 4ⁿ}

g) {an = (-4)ⁿ}

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Given Set S is S spans P_2.

The set S = {1, x², 4 + x²} spans P_2 if every polynomial in P_2 can be expressed as a linear combination of 1, x², and 4 + x².

Let's consider a general polynomial in P_2, which has the form ax^2 + bx + c, where a, b, and c are constants. We need to determine if there exist constants k1, k2, and k3 such that:

ax² + bx + c = k1(1) + k2(x²) + k3(4 + x²)

Simplifying the right-hand side gives:

ax² + bx + c = (k2 + k3)x² + 4k3

For this equation to hold for all values of x, we must have a = k2 + k3, b = 0, and c = 4k3. Therefore, every polynomial in P_2 can be expressed as a linear combination of the elements in S if and only if we can find constants k1, k2, and k3 that satisfy these equations.

Solving the equations, we get:

k1 = 4k3 - a

k2 = a - k3

k3 is free

Since k3 is a free variable, we can choose it to be any value we like. This means that we can always find constants k1, k2, and k3 that satisfy the equations, and so S spans P_2.

Therefore, the answer is S spans P_2.

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Answer:

13/85

Step-by-step explanation:

The sin of an angle is the opposite side over the hypotenuse.

sin B = opp/ hyp

sin B = 13/85

Answer:

sin B = 0.1529

Step-by-step explanation:

To find the Sin B angle we have to use the below formula.

[tex]\sf Sin\:B = \frac{Opposite}{Hypotenuse}[/tex]

Let us solve this now.

[tex]\sf Sin\:B = \frac{Opposite}{Hypotenuse} \\\\\sf Sin\:B = \frac{13}{85} \\\\Sin \:B =0.1529[/tex]

Additionally, To Remove sin, look at the inverse of the sin value and find the exact value of B

[tex]\sf B = sin^-^10.1529\\B=8.79\\\\[/tex]

Answer: no polygon exists in which the ratio of the number of diagonals to the sum of the measures of the angles is 1 to 18, because the number of sides n cannot be equal to 23.

Step-by-step explanation: Let n be the number of sides of the polygon. The number of diagonals in a polygon of n sides is given by the formula:

d = n(n-3)/2

The sum of the measures of the angles in a polygon of n sides is given by the formula:

180(n-2)

The ratio of the number of diagonals to the sum of the measures of the angles is:

d / [180(n-2)] = [n(n-3)/2] / [180(n-2)] = (n-3) / 360

We want to show that this ratio cannot be equal to 1/18, or:

(n-3) / 360 ≠ 1/18

Multiplying both sides by 360, we get:

n-3 ≠ 20

Adding 3 to both sides, we get:

n ≠ 23

Therefore, no polygon exists in which the ratio of the number of diagonals to the sum of the measures of the angles is 1 to 18, because the number of sides n cannot be equal to 23.

The nth term of the pattern is (n²-1)

To find the nth term identify the patterns in a given sequence and use algebraic expressions to generalize the pattern and find the nth term.

By observing the given series we say that each number is one less than perfect Like 8 is one less than 9, 15 is one less than 16, etc. Use this condition to solve the problem.

Here we have

0, 3, 8, 15...    

To find the nth terms identify the patterns in a given sequence

Here each term can be written as follows

1st term => 0 = (1)² - 1 = 0

2nd term => 3 = (2)² - 1 = 3

3rd term => 8 = (3)² - 1 = 8

4th term => 15 = (4)² - 1 = 15

Similarly

nth term = (n)² - 1 = (n²-1)

Therefore,

The nth term of the pattern is (n²-1)

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A power series representation for the function f(x) =[tex]x/36 + x^2[/tex] is  Σ((1/36) * [tex]x^n[/tex]) from n=1 to infinity + Σ[tex](x^{(2n)})[/tex] from n=0 to infinity and its interval of convergence is -1 < x < 1.

To find a power series representation for f(x), we'll rewrite it as a sum of power series:

f(x) = [tex]x/36 + x^2[/tex]
f(x) = (1/36) * [tex]x + x^2[/tex]
f(x) = Σ((1/36) * [tex]x^n[/tex]) from n=1 to infinity + Σ[tex](x^{(2n)})[/tex] from n=0 to infinity

Now let's find the interval of convergence for the given power series. We'll use the Ratio Test:

For the first power series, let a_n = (1/36) * [tex]x^n[/tex]:
lim (n→∞) (|a_(n+1)/a_n|) = lim (n→∞) (|[tex](x^{(n+1)[/tex])/(36 * [tex]x^n[/tex])|) = |x|/36

For the second power series, let b_n = [tex]x^{2n[/tex]:
lim (n→∞) (|b_(n+1)/b_n|) = lim (n→∞) [tex](|(x^{(2(n+1)}))/(x^{(2n)})|) = |x|^2[/tex]

The interval of convergence is where both series converge. The first series converges when |x|/36 < 1, or -36 < x < 36. The second series converges when [tex]|x|^2[/tex] < 1, or -1 < x < 1. Therefore, the interval of convergence for f(x) is:

-1 < x < 1

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The graph of the system of inequalities is on the image at the end.

Here we need to graph the two linear inequalities:

4x + 2y ≤ 16

x + y ≥ 4

On the same coordinate axis.

To do so, we can write both of these as lines:

y  ≥ 4 - x

y ≤ (16 - 4x)/2

y ≤ 8 - 2x

Then the system is:

y  ≥ 4 - x

y ≤ 8 - 2x

Now just graph the two lines with solid lines (because of the symbols used) and shadew the region above the first line and the region below the second line.

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