Use cylindrical coordinates. ∭_e x^2 dv .where E … solid thati x2 dV, where E is the solid that lies within the cylinder x^2 + y^2=4, above the plane z = 0, and below the cone z^2 4x^2 + 4y^2

The correct order of integration and associated limits for the given triple integral I = ∫∫∫[f(x, y, z)] dx dy dz with limits: 0 ≤ x ≤ 1 - 2y, 0 ≤ y, 0 ≤ z ≤ 2y²

The correct order of integration and associated limits for the given triple integral, let's first examine the region of integration R and its slices in different planes.

Region R is defined by 0 ≤ z ≤ 2y² and 0 ≤ x ≤ 1 - 2y.

1.Slices in the zy-plane: In the zy-plane, z is restricted to 0 ≤ z ≤ 2y², and y is unrestricted. Therefore, the integral can be written as:

I = ∫∫∫ f(x, y, z) dV = ∫∫∫ f(x, y, z) dz dy dx

2.Slices in the zx-plane: In the zx-plane, z is unrestricted, and x is restricted to 0 ≤ x ≤ 1 - 2y. Therefore, the integral can be written as:

I = ∫∫∫ f(x, y, z) dV = ∫∫∫ f(x, y, z) dx dz dy

3.Slices in the yz-plane: In the yz-plane, y is unrestricted, and z is restricted to 0 ≤ z ≤ 2y². Therefore, the integral can be written as:

I = ∫∫∫ f(x, y, z) dV = ∫∫∫ f(x, y, z) dy dz dx

Considering the given hint, we can choose any of the above orders of integration as all of them are correct ways to write the integral. However, for simplicity, let's choose the order: I = ∫∫∫ f(x, y, z) dz dy dx.

Now, let's determine the limits of integration for each variable in this order:

∫∫∫ f(x, y, z) dz dy dx = ∫∫ [∫[f(x, y, z) dz] from z=0 to z=2y²] dy dx

The innermost integral with respect to z is evaluated from 0 to 2y². The next integral with respect to y is evaluated from 0 to a certain limit determined by the region R. Finally, the outermost integral with respect to x is evaluated from 0 to 1 - 2y.

Therefore, the order of integration and the associated limits for the triple integral are:

I = ∫∫∫ f(x, y, z) dz dy dx

I = ∫∫ [∫[f(x, y, z) dz] from z=0 to z=2y²] dy dx

I = ∫∫∫[f(x, y, z)] dx dy dz with limits: 0 ≤ x ≤ 1 - 2y, 0 ≤ y, 0 ≤ z ≤ 2y²

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The function y = sin(kt) satisfies the differential equation y'' - 64y = 0 for pospositiveypospositiveyitiveitive values of k that are multiples of 8.

To determine the values of k for which the function y = sin(kt) satisfies the given differential equation, we need to substitute y into the equation and solve for k. Let's start by finding the first and second derivatives of y with respect to t.
The first derivative of y with respect to t is y' = kcos(kt), and the second derivative is y'' = -k^2sin(kt). Substituting these derivatives into the differential equation gives us:
(-k^2sin(kt)) - 64sin(kt) = 0Simplifying the equation, we get:
sin(kt) = -64*sin(kt)/k^2
We can divide both sides of the equation by sin(kt) (assuming sin(kt) is not zero) to get:
1 = -64/k^2
Solving for k^2, we find k^2 = -64. Since k must be positive, there are no positive values of k that satisfy this equation. Therefore, there are no positive values of k for which the function y = sin(kt) satisfies the given differential equation y'' - 64y = 0.

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To prove that for any real number x, if  x²+ 7x < 0, then x < 0, we can use the properties of quadratic functions and inequalities.

By analyzing the quadratic expression, we can determine the conditions under which it is negative. This analysis shows that the inequality x²+ 7x < 0 holds true when x is less than 0. Consider the quadratic expression x² + 7x. To determine when this expression is negative, we can factor it as x(x + 7). According to the zero product property, this expression is equal to zero when either x or (x + 7) is equal to zero. Thus, the two critical points are x = 0 and x = -7.

Now, let's analyze the behavior of the quadratic expression in the intervals (-∞, -7), (-7, 0), and (0, +∞). Choose a test point from each interval, such as -8, -3, and 1, respectively. Evaluating the expression x²⁺7x for these test points, we find that for -8 and -3, the expression is positive, and for 1, it is positive as well.

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False, the null distribution is the distribution of the test statistic assuming the null hypothesis is true; it is mound-shaped and symmetric about the null mean.

The null distribution is the distribution of the test statistic under the assumption that the null hypothesis is true. However, its shape and symmetry are not necessarily predetermined.

The null distribution can take various forms depending on the specific test and the underlying data. It may or may not be mound shaped or symmetric about the null mean. The shape and characteristics of the null distribution are determined by the specific hypothesis being tested, the sample size, and other factors.

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The instantaneous rate of change for the function f(x) at x = 3 is 16.9068.

To determine the instantaneous rate of change for the function;  f(x) = 5x^ln(x), where x = 3, we can use the formula for the instantaneous rate of change, approximating the limit by using smaller and smaller values of h.

The instantaneous rate of change of the function f(x) at x = 3 can be found as follows:

Let h be a small increment of x that approaches zero. Then the formula for the instantaneous rate of change is given by:

f'(3) = lim[h→0] {(5(3+h)^(ln(3+h))-5(3^(ln3)))/h}

For the above formula, we have: Let f(x) = 5x^ln(x)

Then, f'(x) = 5x^ln(x) * [(d/dx) ln(x)] + 5*ln(x)*x^(ln(x) - 1)

Now, for the given problem, we can substitute 3 for x, and solve as follows: f'(3) = 5(3^ln(3)) * [(d/dx) ln(x)] + 5*ln(3)*3^(ln(3) - 1)

f'(3) = 5(3^ln(3)) * [(1/x)] + 5*ln(3)*3^(ln(3) - 1)

f'(3) = 5(3^ln(3)) * [(1/3)] + 5*ln(3)*3^(ln(3) - 1)

f'(3) = 5(3^ln(3) / 3) + 5*ln(3)*3^(ln(3) - 1)

Then, f'(3) = 5e ln(3) + 5 ln(3) / 3= (5e ln(3) + 5 ln(3) / 3)= 16.9068 (rounded to four decimal places).

Therefore, the instantaneous rate of change for the function f(x) at x = 3 is 16.9068.

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Answer:

$27.50

Step-by-step explanation:

First, find the cost with the 15% off sale

29.95(0.85)

= 25.46

Find the price with the sales tax:

25.46(1.08)

= 27.5

So, the sale price of the game with tax is approximately $27.50

Answer:

the bottom triangle is scaled down version of above one

then

[tex] \frac{18}{6} = \frac{x}{2} [/tex]

[tex]x = 6[/tex]

and

[tex] \frac{18}{6} = \frac{y}{5} [/tex]

[tex]y = 15[/tex]

a. The relation R defined on the set S = {1, 2, 3, ..., 18, 19} is an equivalence relation. It is reflexive, symmetric, and transitive, b. The equivalence class of 1, denoted [1], consists of the perfect squares in S: {1, 4, 9, 16}, c. The equivalence classes with more than one element are [1], [2], [3], ..., [18], and [19]. Each equivalence class represents a set of numbers that are squares of integers.

a. To show that the relation R is an equivalence relation, we need to demonstrate that it is reflexive, symmetric, and transitive.

i. Reflexive: For R to be reflexive, every element in S must be related to itself. Since the square of any integer is still an integer, xRx holds for all x in S, satisfying reflexivity.

ii. Symmetric: For R to be symmetric, if xRy holds, then yRx must also hold. Since multiplication is commutative, if xy is a square of an integer, then yx is also a square of an integer. Hence, R is symmetric.

iii. Transitive: For R to be transitive, if xRy and yRz hold, then xRz must also hold. Since the product of two squares of integers is itself a square of an integer, xz is also a square of an integer. Thus, R is transitive.

b. To find the equivalence class of 1, denoted [1], we determine all elements in S that are related to 1 under R. In this case, [1] consists of the perfect squares in S: {1, 4, 9, 16}.

c. The equivalence classes with more than one element are [1], [2], [3], ..., [18], and [19]. Each equivalence class represents a set of numbers that are squares of integers. The equivalence class [1] includes all perfect squares in S, while the other equivalence classes consist of a single element, which are non-square integers.

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Answer: 40

Step-by-step explanation:

To plot an image of your own choosing, you can follow these steps. First, create a list containing a set of ordered pairs that represent the points needed to create your image.

To create the image, you need to define the set of ordered pairs that form the outline of the desired shape. The points in the list should be ordered in a way that connects them together to form the shape. Including (0,0) at the beginning and end ensures that the last point connects back to the first point.

Once you have the list of ordered pairs, you can convert it into a matrix using the "HGMatrix" command. This step is necessary for plotting the image using the "Plotimage" command. The matrix, named "mylistmat," will represent the points in a format suitable for plotting.

Finally, using the "Plotimage" command, you can plot the image based on the points in "mylistmat." The command will take the ordered pairs and connect them together to create the desired image.

By following these steps, you can plot your own image using the specified list, matrix conversion, and image plotting commands.

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Answer:

x²+11x+30

Step-by-step explanation:

for this question, all we need to do is multiply (x+5) by (x+6)

we can use the distributive property.

(x+5)(x+6)

x² + 5x + 6x + 30

x² + 11x + 30

Answer:

True

Step-by-step explanation:

Brainliest?

Answer:

I Believe the answer is True

Step-by-step explanation:

Answer

Answer:

d

Step-by-step explanation:

Note that

> means greater than

< means less than

for example : 2 < 3 means 2 is less than 3

3 >2 means 3 is greater than 2

To determine which comparison is better we have to convert litre to millimetre

1 litre = 1000mm

A 5 mm > (50 x 1000)

5 mm > 50,000

5 is not greater than 50,000

B. (2 x 1000) < 200 mm

2000 < 200

2000 is not less than 200

C. (100 x 1000) < 1,000

100,000 < 1000

100,000 is not less than 1000

D. 3200 > (3 X 1000)

3200 > 3000

3200 is greater than 3000

Answer:

55

Step-by-step explanation:

To find this, simply plug 10 in for x.

6(10)-5

6*10=60

60-5=55

Answer:

Step-by-step explanation:

1

Answer:

56

Step-by-step explanation:

i just is smart

yhhhhhhhhhhh

Answer:

Step-by-step explanation:

Rewrite this quadratic in standard form:  3x^2 + 7x - 1.

The coefficients of x are {3, 7, -1}, and so the discriminant is b^2 - 4ac, or

7^2 - 4(3)(-1), or 49 + 12, or 61.  Because the discriminant is positive, this quadratic has two real, unequal roots

(a) The probability of a computer failure from Company A is 0.001; from Company B is 0.002; and from Company C is 0.005.

Therefore, the probability that a computer will experience a hard drive failure within one year is:(0.50 x 0.001) + (0.30 x 0.002) + (0.20 x 0.005)= 0.0012. The probability of a randomly selected computer experiencing a hard drive failure within one year is 0.0012 or 0.12%.
(b) Bayes' theorem will be used to calculate this probability:Let A be the event that the computer's hard drive was manufactured by Company C. Let B be the event that the computer experienced a hard drive failure. P(A|B) is the probability that the hard drive was manufactured by Company C given that a hard drive failure was experienced.

P(A|B) = P(B|A) P(A) / P(B) Where: P(B|A) = 0.005 (the probability of failure if the hard drive was manufactured by Company C)P(A) = 0.20 (the proportion of hard drives that the computer manufacturer gets from Company C)P(B) = (0.50 x 0.001) + (0.30 x 0.002) + (0.20 x 0.005) = 0.0012 (as in part a)

Therefore: P(A|B) = (0.005 x 0.20) / 0.0012 = 0.0833 or 8.33%.
(c)Let A be the event that both hard drives were manufactured by Company A; B be the event that both hard drives were manufactured by Company B; and C be the event that both hard drives were manufactured by Company C. Then we need to find the probability of event A or B or C, given that a hard drive failure was experienced:P(A U B U C|F) = P(F|A U B U C) P(A U B U C) / P(F)where F is the event that the hard drive in the replacement computer fails.P(F|A U B U C) = P(F) = (0.50 x 0.001) + (0.30 x 0.002) + (0.20 x 0.005) = 0.0012P(A U B U C) = (0.50)^2 + (0.30)^2 + (0.20)^2 = 0.46P(F) = P(A U B U C) P(F|A U B U C) + P(A' n B n C) P(F|A' n B n C)= 0.46 x 0.0012 + 0.04 x 0.3 = 0.000552P(A U B U C|F) = P(F|A U B U C) P(A U B U C) / P(F)= (0.0012 x 0.46) / 0.000552 = 1.00or 100%. Therefore, the probability that the original and replacement computers were produced by the same company is 100%.
(d) Bayes' theorem will be used to calculate this probability:Let A be the event that the hard drive was manufactured by Company C. Let B be the event that the computer did not experience a hard drive failure. P(A|B) is the probability that the hard drive was manufactured by Company C given that no hard drive failure was experienced.P(A|B) = P(B|A) P(A) / P(B)Where:P(B|A) = 1 - 0.005 = 0.995 (the probability that the hard drive did not fail if it was manufactured by Company C)P(A) = 0.20 (as in part b)P(B) = 1 - (0.50 x 0.001) - (0.30 x 0.002) - (0.20 x 0.005) = 0.9988

Therefore:P(A|B) = (0.995 x 0.20) / 0.9988 = 0.1989 or 19.89%. Therefore, the probability that the hard drive was manufactured by Company C given that it did not fail is 19.89%.

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The probability that the hard drive was manufactured by company C given that a failure was not experienced by the computer within one year is approximately 0.256.

(a)Probability that a randomly chosen computer purchased from this manufacturer will experience a hard drive failure within one year = 0.5 x 0.001 + 0.3 x 0.002 + 0.2 x 0.005 = 0.0016

(b)Let's denote the event that a computer failure is experienced within one year by F and the event that the hard drive is made by company C by C.

Then we are required to calculate P(C | F), which is the probability that the hard drive was manufactured by company C given that a failure was experienced by the computer within one year. This can be found by using the Bayes' rule as follows:

[tex]$$P(C|F) = \frac{P(F|C)P(C)}{P(F|A)P(A) + P(F|B)P(B) + P(F|C)P(C)}$$[/tex]
where P(C) = 0.2, P(A) = 0.5 and P(B) = 0.3.$$P(F|A) = 0.001, P(F|B) = 0.002, P(F|C) = 0.005$$

Thus, we have:[tex]$$P(C|F) = \frac{0.005 \times 0.2}{0.001 \times 0.5 + 0.002 \times 0.3 + 0.005 \times 0.2} \approx 0.476$$[/tex]

Therefore, the probability that the hard drive was manufactured by company C given that a failure was experienced by the computer within one year is approximately 0.476.

(c)Let's denote the event that the original hard drive is manufactured by company A, B and C by A, B, and C respectively.

Similarly, let's denote the event that the replacement hard drive is manufactured by company A, B, and C by A', B', and C' respectively.

We are required to calculate P(A = A', B = B', C = C' | F), which is the probability that the hard drives in the original and replacement computers were manufactured by the same company given that a failure was experienced by both computers within one year.

This can be found by using the Bayes' rule as follows:

[tex]$$P(A = A', B = B', C = C'|F) = \frac{P(F|A = A', B = B', C = C')P(A = A')P(B = B')P(C = C')}{P(F)}$$[/tex]

where: [tex]$$P(F) = P(F|A = A', B = B', C = C')P(A = A')P(B = B')P(C = C') + P(F|A \ne A', B \ne B', C \ne C')P(A \ne A')P(B \ne B')P(C \ne C')$$[/tex]

Here, we are assuming that the probabilities of computer failure are independent of each other and the company that manufactured the hard drives of the two computers are independent of each other. Therefore, we have:

[tex]$$P(F|A = A', B = B', C = C') = P(F|A)P(F|B)P(F|C) = 0.001 \times 0.002 \times 0.005$$[/tex]

[tex]$$P(F|A \ne A', B \ne B', C \ne C') = 0$$[/tex]

Also, we have:$$P(A = A') = P(B = B') = P(C = C') = \frac{1}{3}$$

[tex]$$P(A \ne A', B \ne B', C \ne C') = \frac{2}{3} \times \frac{2}{3} \times \frac{2}{3} = \frac{8}{27}$$[/tex]

Thus, we have:$$P(A = A', B = B', C = C'|F) = \frac{0.001 \times 0.002 \times 0.005 \times (\frac{1}{3})^3}{P(F)}$$

[tex]$$P(A \ne A', B \ne B', C \ne C'|F) = \frac{P(F) - 0.001 \times 0.002 \times 0.005 \times (\frac{1}{3})^3}{\frac{8}{27}}$$[/tex]

Now, we need to find P(F). This can be done as follows:

[tex]$$P(F) = P(F|A = A', B = B', C = C')P(A = A')P(B = B')P(C = C') + P(F|A \ne A', B \ne B', C \ne C')P(A \ne A')P(B \ne B')P(C \ne C')$$$$= 0.001 \times 0.002 \times 0.005 \times (\frac{1}{3})^3 + 0 = 4.6296 \times 10^{-8}$$Thus, we have:$$P(A = A', B = B', C = C'|F) = 0.0296$$[/tex]

[tex]$$P(A \ne A', B \ne B', C \ne C'|F) = 0.9704$$[/tex]

Therefore, the probability that the hard drives in the original and replacement computers were manufactured by the same company given that a failure was experienced by both computers within one year is 0.0296.(d)Let's denote the event that the hard drive is made by company C by C and the event that a computer failure is not experienced within one year by F'. We are required to calculate P(C | F'), which is the probability that the hard drive was manufactured by company C given that a failure was not experienced by the computer within one year. This can be found by using the Bayes' rule as follows:

[tex]$$P(C|F') = \frac{P(F'|C)P(C)}{P(F'|A)P(A) + P(F'|B)P(B) + P(F'|C)P(C)}$$[/tex]

where P(C) = 0.2, P(A) = 0.5 and P(B) = 0.3.$$P(F'|A) = 0.999, P(F'|B) = 0.998, P(F'|C) = 0.995$$

Thus, we have: [tex]$$P(C|F') = \frac{0.995 \times 0.2}{0.999 \times 0.5 + 0.998 \times 0.3 + 0.995 \times 0.2} \approx 0.256$$[/tex]

Therefore, the probability that the hard drive was manufactured by company C given that a failure was not experienced by the computer within one year is approximately 0.256.

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By filling in the elements accordingly and comparing the Venn diagram with the calculated results, the accuracy of the union and intersection can be confirmed i.e. AUB = {1, 2, 3, 4, 5, 6, 7, 8, 9} and An B = {4, 8}.

To find the union (AUB) and intersection (An B) of the sets A = {1, 4, 6, 8, 9} and B = {2, 3, 4, 5, 7, 8}, we compare the elements in the sets. The union (AUB) is the combination of all unique elements from both sets, while the intersection (An B) consists of the elements common to both sets.

(a) The union (AUB) is {1, 2, 3, 4, 5, 6, 7, 8, 9}, which includes all the distinct elements present in both sets.

(b) The intersection (An B) is {4, 8}, representing the elements that are common to both sets A and B.

To verify these results, you can draw a Venn diagram. Draw two overlapping circles representing sets A and B, and fill in the elements accordingly. The overlapping region represents the intersection, and the combination of all elements in both circles represents the union. Comparing the Venn diagram with the calculated results confirms the accuracy of the union and intersection.

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14. (1112 - 6vw - 3wa)-(-702 + vw + 13w) = 1814 - 7vw - 3wa - 13w

15. The equivalent of the expression (5a + 26 - 4c) is 25a2 + 10ab - 20ac + 482 - 86c + 1602 + c.

16.  (b + b)(4 - 5)(25 - 8) = -34

14. Simplify (1112 - 6vw - 3wa)-(-702 + vw + 13w).

Given expression is (1112 - 6vw - 3wa)-(-702 + vw + 13w)

⇒ 1112 - 6vw - 3wa + 702 - vw - 13w

⇒ 1814 - 7vw - 3wa - 13w

15. We are to find the equivalent of the expression (5a + 26 - 4c).

a. 25a2 + 20ab - 40ac +482 - 16bc + 1602

b. 25a2 + 10ab - 20ac + 482 - 86C + 1602

c. 25a2 + 482 + 1602

d. 10a + 4b-8c5a + 26 - 4c

= 5a - 4c + 26 = 25a2 - 20ac +482 - 4c2 + 52 - 8ac

= 25a2 - 20ac + 482 - 4c2 + 10a - 8c = Option (b)

⇒ 25a2 + 10ab - 20ac + 482 - 86c + 16c2 + c.

16. Expand and simplify. (b + b)(4 - 5)(25 - 8)

Given expression is (b + b)(4 - 5)(25 - 8) = 2b(-1)(17) = -34

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Answer:

3=20

4=15

5=12

Step-by-step explanation:

a. System in the matrix form is x' = Ax where A = [tex]\left[\begin{array}{ccc}0&1\\-5&6\end{array}\right][/tex] and x = [y, u].

b. The eigenvalues of the system are λ₁ = 5 and λ₂ = 1 and eigenvector are v₁ and v₂ = v₁, and v₁ is any non-zero value.

c. The general solution is equal to y(x) = c₁ × [tex]e^{(5x)[/tex] × [v₁] + c₂× [tex]e^{(x)[/tex]× [v₂].

a. Solution to the equation. y' - 5y + y = 32 is y(x) = c₁ × [tex]e^{(5x)[/tex] + c₂ × [tex]e^{(x)[/tex].

(a) To write the second order linear ODE as a system of two first order ODEs,

Introduce a new variable u = y'.

Then, we have,

u' = y'' - 5y + 6y

   = -5y + 6u

Now, write this as a system of two first order ODEs,

y' = u

u' = -5y + 6u

To express this system in matrix form,

Define the vector x = [y, u] and the matrix A = [tex]\left[\begin{array}{ccc}0&1\\-5&6\end{array}\right][/tex]

The system can then be written as,

x' = Ax

(b) To find the eigenvalues and eigenvectors of matrix A, solve the characteristic equation,

|A - λI| = 0

where I is the identity matrix.

Substituting the values of A, we have,

[tex]|\left[\begin{array}{ccc}0&1\\-5&6\end{array}\right][/tex] [tex]-\lambda\left[\begin{array}{ccc}1&0\\0&1\end{array}\right]|[/tex] = 0

[tex]\left[\begin{array}{ccc}-\lambda&1\\-5&6-\lambda\end{array}\right][/tex] = 0

(-λ)(6-λ) - (-5)(1) = 0

λ²- 6λ + 5 = 0

Factoring the quadratic equation, we get,

(λ - 5)(λ - 1) = 0

So the eigenvalues are λ₁ = 5 and λ₂ = 1.

To find the corresponding eigenvectors,

solve the equation (A - λI)v = 0 for each eigenvalue.

Let us start with λ = 5

(A - 5I)v = 0

[tex]|\left[\begin{array}{ccc}1&1\\-5&6\end{array}\right]|[/tex] v = 0

v₁ + v₂ = 0

-5v₁ + v₂ = 0

From the first equation, we get v₂ = -v₁.

Substituting this into the second equation, we have -5v₁ - v₁ = 0,

which simplifies to -6v₁ = 0.

This implies v₁ = 0, and consequently, v₂ = 0.

So, for λ = 5, the eigenvector is v₁ = 0 and v₂ = 0.

Now, let us find the eigenvector for λ = 1.

(A - I)v = 0

[tex]|\left[\begin{array}{ccc}-1&1\\-5&5\end{array}\right][/tex] v = 0

-v₁ + v₂ = 0

-5v₁ + 5v₂ = 0

From the first equation, we get v₂ = v₁.

Substituting this into the second equation, we have -5v₁ + 5v₁ = 0,

which simplifies to 0 = 0.

This implies that v₁ can be any non-zero value.

So, for λ = 1, the eigenvector is v₁ and v₂ = v₁, where v₁ is any non-zero value.

(e) The general solution to the second order ODE can be expressed using the eigenvalues and eigenvectors as follows,

y(x) = c₁ ×[tex]e^{(\lambda_{1} x)[/tex] × v₁ + c₂ × [tex]e^{(\lambda_{2} x)[/tex]× v₂

Plugging in the values we found earlier, the general solution becomes,

y(x) = c₁ × [tex]e^{(5x)[/tex] × [v₁] + c₂× [tex]e^{(x)[/tex]× [v₂]

where [v₁] and [v₂] are the eigenvectors corresponding to the eigenvalues λ₁ = 5 and λ₂ = 1 respectively.

(a) To find the solution to the equation y' - 5y + y = 32,

Use the general solution obtained above.

Comparing the equation with the standard form y' - 5y + 6y = 0,

The equation corresponds to the case where λ₂ = 1.

Substitute λ = 1, v₁ = 1, and v₂ = 1 into the general solution.

y(x) = c₁ × [tex]e^{(5x)[/tex] × [1] + c₂ × [tex]e^{(x)[/tex] × [1]

Simplifying this expression, we have,

y(x) = c₁ × [tex]e^{(5x)[/tex] + c₂ × [tex]e^{(x)[/tex]

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Answer:

There is a 1/3 (or 0.33%) probability of rolling a number greater than or equal to 5.

Step-by-step explanation:

First, find what numbers are greater than or equal to 5:

5 and 6

Find what options you can get on a number cube:

1, 2, 3, 4, 5, and 6

Out of the 6 possible outcomes, there are only 2 that will get a number greater than or equal to 5. Write this as a fraction:

2/6

Simplify:

1/3

In a river bank

you can put the answers from the end

1.) 1/12

2.) 2/13

3.) 8/41

4.) 1/4

5.) 11/74

6.) 2/35

7.) 15/58

8.) 5/18

9.) 1/7

10.) 1/13

Answer:

hi sh Sheet sh I monorailg Jericho improve Odom Ybor

It cant be written any way else its already in its simplest form

Given:

The end points of the line segment AB are A(-2,-3) and B(4,-1).

The rule of translation is:

[tex](x,y)\to (x+4,y-3)[/tex]

To find:

The coordinates of the end points of the line segment A'B'.

Solution:

It is given that the end points of the line segment AB are A(-2,-3) and B(4,-1).

We have,

[tex](x,y)\to (x+4,y-3)[/tex]

By using the above translation rule, we get

[tex]A(-2,-3)\to A'(-2+4,-3-3)[/tex]

[tex]A(-2,-3)\to A'(2,-6)[/tex]

And

[tex]B(4,-1)\to B'(4+4,-1-3)[/tex]

[tex]B(4,-1)\to B'(8,-4)[/tex]

Hence, the endpoint of the line segment A'B' are A'(2,-6) and B'(8,-4).

Answer

c

Step-by-step explanation:

It is not d because of the way that it is formed

Answer: the second one/ \/36/6

Step-by-step explanation:

Answer:

a=-2

Step-by-step explanation:

3a−6=−12

Step 1: Add 6 to both sides.

3a−6+6=−12+6

3a=−6

Step 2: Divide both sides by 3.

3a/ 3 = −6/ 3

a=−2