Use the laplace transform to solve the given initial-value problem. y'' − 6y' 13y = 0, y(0) = 0, y'(0) = −5

[tex]\displaystyle\\Answer:\theta=\frac{\pi }{2}+\pi n.[/tex]

Step-by-step explanation:

[tex]\displaystyle\\7*cosec^2\theta-9*cot^2\theta=7\\7-7*cosec^2\theta+9*cot^2\theta=0\\7-\frac{7}{sin^2\theta}+9*\frac{cos^2\theta}{sin^2\theta} } =0\\\frac{7*sin^2\theta-7+9*cos^2\theta}{sin^2\theta} =0\\\frac{-7*(1-sin^2\theta)+9*cos^2\theta}{sin^2\theta} =0\\\frac{-7*cos^2\theta+9*cos^2\theta}{sin^2\theta} =0\\\frac{2*cos^2\theta}{sin^2\theta}=0\\[/tex]

[tex]2*cot^2\theta=0\\Divide\ the\ right\ and\ initial\ parts\ by\ 2:\\cot^2\theta=0\\cot\theta=0\\\theta=\frac{\pi }{2}+\pi n\ \ \ (n=0,\ 1,\ 2,\ 3\ ...).[/tex]

Answer:

B

Step-by-step explanation:

if an angle is congruent, it is the same measurement

The median of this box and whisker plot is 73

A box and whisker plot is defined as a graphical method of displaying variation in a set of data. In most cases, a histogram analysis provides a sufficient display, but a box and whisker plot can provide additional detail while allowing multiple sets of data to be displayed in the same graph. Box and whisker plots are very effective and easy to read, as they can summarize data from multiple sources and display the results in a single graph. Box and whisker plots allow for comparison of data from different categories for easier, more effective decision-making.

From the given data, following can be inferred

which is demonstrated in the box and whisker plot below

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Answer:

680

Step-by-step explanation:

x = total number

.6x  = bus riders = 408              (.6 is decimal for 60%)

.6x = 408

x = 408/.6 = 680

Answer:

Allana is 2 1/2 and her mother is 15 years old.

Step-by-step explanation:

Let Allana's age be x years then,

her mother's age is 6x.

Using the given information in ten years time we have the equation:-

6x + 10 = 2(x + 10)

6x + 10 = 2x + 20

6x - 2x = 20 - 10

4x = 10

x = 2 1/2

So 6x = 6 * 2 1/2 = 15.

Taylor series is [tex]f(x) = ln2 + \sum_{n=1)^{\infty}(-1)^{n-1} \frac{(n-1)!}{n!(9)^{n}(x9)^{2} }[/tex]

To find the Taylor series for f(x) = ln(x) centering at 9, we need to observe the pattern for the first four derivatives of f(x). From there, we can create a general equation for f(n). Starting with f(x), we have

f(x) = ln(x)

[tex]f^{1}(x)= \frac{1}{x} \\f^{2}(x)= -\frac{1}{x^{2} }\\f^{3}(x)= -\frac{2}{x^{3} }\\f^{4}(x)= \frac{-6}{x^{4} }[/tex]

.

.

.

Since we need to have it centered at 9, we must take the value of f(9), and so on.

f(9) = ln(9)

[tex]f^{1}(9)= \frac{1}{9} \\f^{2}(9)= -\frac{1}{9^{2} }\\f^{3}(x)= -\frac{1(2)}{9^{3} }\\f^{4}(x)= \frac{-1(2)(3)}{9^{4} }[/tex]

.

.

.

Following the pattern, we can see that for [tex]f^{n}(x)[/tex],

[tex]f^{n}(x)=(-1)^{n-1}\frac{1.2.3.4.5...........(n-1)}{9^{n} } \\f^{n}(x)=(-1)^{n-1}\frac{(n-1)!}{9^{n}}[/tex]

This applies for n ≥ 1, Expressing f(x) in summation, we have

[tex]\sum_{n=0}^{\infinite} \frac{f^{n}(9) }{n!} (x-9)^{2}[/tex]

Combining ln2 with the rest of series, we have

[tex]f(x) = ln2 + \sum_{n=1)^{\infty}(-1)^{n-1} \frac{(n-1)!}{n!(9)^{n}(x9)^{2} }[/tex]

Taylor series is [tex]f(x) = ln2 + \sum_{n=1)^{\infty}(-1)^{n-1} \frac{(n-1)!}{n!(9)^{n}(x9)^{2} }[/tex]

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Answer:

C. 4≤x≤6

Step-by-step explanation:

First things first, let’s learn the difference between domain and range. Domain is the spread of the x values and range is the spread of the y values. This means that in this question, you can rule out options A and B because those could potentially show range. Now we are left with options C and D. When we look at the first x coordinate, we see that it’s a 4. The last x coordinate is a 6. This means that the domain is greater than or equal to 4, but less than or equal to 6. You can also represent it as 4≤x≤6.

Hope that helps! Have a fantastic day!

(A) Algorithms are mathematical formulas placed in software that performs an analysis on a data set.

Therefore, (A) algorithms are mathematical formulas placed in software that performs an analysis on a data set.

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Step-by-step explanation:

it is a line, so that is 1 point, where it cuts the x-axis.

in general that means y = 0.

so,

0 = 4x - 3

4x = 3

x = 3/4

the point, where the line cuts the x-axis is

(3/4, 0)

The measure of the angle between the vectors

[tex]$\arccos[ (-\sqrt{13 } i )/ (\sqrt{5 }) ]\\\sqrt{5 }[/tex].

What is the measure of the angle between the vectors?

Given:

[tex]$\mathrm{u}=\langle -3,2\rangle$[/tex] and [tex]$v=\langle 2,1\rangle$[/tex]

Computing the angle between the vectors, we get

[tex]$\quad \cos (\theta)=\frac{\vec{a} \cdot \vec{b}}{|\vec{a}| \cdot|\vec{b}|}$[/tex]

To estimate the lengths of the vectors, we get

Computing the Euclidean Length of a vector,

[tex]$\left|\left(x_{1}, \ldots, x_{n}\right)\right|=\sqrt{\sum_{i=1}^{n}\left|x_{i}\right|^{2}}$[/tex]

Let, [tex]$\mathrm{u} &=\langle -3,2\rangle \\[/tex] and [tex]$\mathrm{v} &=\langle 2,1\rangle \\[/tex]

If [tex]$\mathrm{u} &=\langle -3,2\rangle \\[/tex]

[tex]$|u| &=\sqrt{-3^{2}+(2)^{2}} \\[/tex]

[tex]$&=\sqrt{5}i \\[/tex] and

[tex]$\mathrm{v} &=\langle 2,1\rangle \\[/tex]

[tex]$|v| &=\sqrt{2^{2}+(1)^{2}} \\[/tex]

[tex]$&=\sqrt{5}[/tex]

Finally, the angle is given by:

Computing the angle between the vectors, we get

[tex]$ $\cos (\theta)=\frac{\vec{a} \cdot \vec{b}}{|\vec{a}| \cdot|\vec{b}|}$[/tex]

[tex]$&\cos (\Phi)=-\sqrt{13 } i/ \sqrt{5 } \\[/tex]

simplifying the above equation, we get

[tex]$&\Phi=\arccos (\cos (\Phi))[/tex]

[tex]$=\arccos[ (-\sqrt{13 } i )/ (\sqrt{5 }) ]\\\sqrt{5 }[/tex]

Therefore, the measure of the angle between the vectors

[tex]$\arccos[ (-\sqrt{13 } i )/ (\sqrt{5 }) ]\\\sqrt{5 }[/tex].

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The point where the line intersect the parabola are at (1, 8) and (4, 5) which gives the solution to the graph shown.

Quadratic equation are equation that has a leading degree of 2 while a linear equation has a leading degree of 1.

From the given graph, the curve shown has a two solutions which is the point where the curve intersect the x-axis.

For the graph shows a system consisting of a linear equation and a quadratic equation, the solution will be the points where the line intersects the parabola.

The point where the line intersect the parabola are at (1, 8) and (4, 5) which gives the solution to the graph shown

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The inequality that describes this graph is y ≥ x+5

From the given graph, we have the following highlights:

The first highlight above implies, the inequality can be any of ≥ and ≤

While the second highlight above implies, the inequality is ≥

Hence, the inequality that describes this graph is y ≥ x+5

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The 0 = 0 in the solution to the system implies that the system of equations involving the equations 6x - 5y = -8 and 6x - 5y = -8 have infinitely many solutions

Linear equations are equations that have constant average rates of change, slope or gradient

A system of linear equations is a collection of at least two linear equations.

In this case, the system of equations is given as

6x - 5y = -8

6x - 5y = -8

Subtract the equation 6x - 5y = -8 from the equation 6x - 5y = -8

6x - 5y = -8

- 6x - 5y = -8

----------------------

0 = 0

This in other words means that the difference between both equations is 0

When the solution to a system of equation is 0 = 0, it means that the system of equations have infinitely many solutions

Hence, the 0 = 0 in the solution to the system implies that the system of equations involving the equations 6x - 5y = -8 and 6x - 5y = -8 have infinitely many solutions

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Answer: The answer is D for those who don't like to read extra long expert answers (no offense to experts, just shorten answers to the direct answer pls)

Step-by-step explanation: Edge 2022

Answer:

The ribbon is 5m long.

The length of the piece = 3 ^ 4 / 5 m.

The length of the remaining piece.

We can simply solve this mathematical problem by using the following mathematical process.

First of all, we have to convert the length of the cut piece into simple decimal value.

Length of cut piece =3^ 4 / 5 = (3 * 5 + 4) / 5 = 19/5 = 3.8m

Length of the remaining ribbon = Intital length cut length = 5 - 3.8 = 1.2m = 12/10 - m = 6/5 * m = 1 1/5 m

Answer: 1 1/5

Step-by-step explanation:

5 - 3 = 2 - 4/5 = 1 1/5

The Area of the kite with the given diagonals is: 125 square meters.

A kite can be described as a quadrilateral having four (4) sides, which can be categorized into two pairs of congruent sides, each of them is adjacent to the other. A kite has two diagonals that intersect, where by the longer diagonal divides the smaller diagonal into equal halves.

An example of a kite is shown in the image attached below, where:

JL and KM are the diagonals of the kite.

The formula for finding the area of a kite is given as:

Area = 1/2 × (product of the lengths of the diagonals)

Given the following:

Length of diagonal 1 = 10 meters

length of diagonal 2 = 25 meters

Area = 1/2 × (product of the lengths of the diagonals) = 1/2 × (10 × 25)

Area of kite = 1/2 × (250)

Area of kite = 1/2 × (250)

Area of kite = 125 square meters.

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Step-by-step explanation:

If the value of determinant a is 3, determinant b is 5 then the value of determinant ab is 15.

Given that the value of determinant a is 3, determinant b is 5.

We are required to find the value of determinant ab.

Determinant of a matrix is the scalar value computed for a given square matrix.It means the matrix which is not square does not have a determinant value.  It is denoted by IAI.

We know that,

IaI*IbI=IabI

So we have to just put the values in the above formula to get the determinant of ab.

IabI=3*5

=15

Hence if the value of determinant a is 3, determinant b is 5 then the value of determinant ab is 15.

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[tex]\cos(x)[/tex] is an even function, while [tex]\sin(x)[/tex] is odd. This means

[tex]\cos(-x) = \cos(x) \text{ and } \sin(-x) = -\sin(x)[/tex]

[tex]\cot(x)[/tex] is defined by

[tex]\cot(x) = \dfrac{\cos(x)}{\sin(x)}[/tex]

so it is an odd function, since

[tex]\cot(-x) = \dfrac{\cos(-x)}{\sin(-x)} = \dfrac{\cos(x)}{-\sin(x)} = -\cot(x)[/tex]

Putting everything together, it follows that

[tex]\cot(-x) \cos(-x) \sin(-x) = (-\cot(x)) \cos(x) (-\sin(x)) \\\\~~~~~~~~= \cot(x) \cos(x) \sin(x) \\\\ ~~~~~~~~= \cos^2(x)[/tex]

The 32° measure of arc CB, and m<BCD which is 52° gives m<BAC and m<ACB as 16° and 90° respectively, from which we have;

First part;

Given;

Angle subtended by arc CB = 32°

m<BCD = 52°

Based on circle theory, we have;

Therefore;

Arc CB = 2 × m<BAC

Which gives;

Arc CB = 32° = 2 × m<BAC

m<BAC = 32° ÷ 2 = 16°

Angle subtended by the diameter at the circumference is 90°.

Therefore;

m<ACB = 90°

In triangle ∆ABC, we have;

m<ACB + m<BAC + m<CBA = 180°

m<CBA = 180° - (m<ACB + m<BAC)

Therefore;

m<CBA = 180° - (90° + 16°)

m<CBA = 180° - (90° + 16°) = 74°

Second part;

The arc subtending m<ACD is AB which is also the diameter.

Angle formed by the diameter, which is a straight line = 180°

Therefore;

Angle subtended at the center by arc AB = 180°

Angle subtended at the circumference, m<ACB is therefore;

m<ACB = 180° ÷ 2 = 90°

A

m<ACB = m<ACD + m<BCD (Angle addition property)

Therefore;

90° = m<ACD + 52°

m<ACD = 90° - 52° = 38°

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Answer:

6%- 457 Million

Step-by-step explanation:

w=241(1.06)^11=457 million

The pH of the solution is 12.2.

Molarity (kon) = No. of moles of FoH x 1000

Mckor, 7:12 × 10-4 moles x Looo 50.0 mL.

Volume of solution (m²), McKong = 1.424x102M, [on] =1.424×102M POH = -log [on] pOH =

-log [1.424x102] pon = 1.85.

pH + pOH = 14

PH = 14-pOH = 14-1.85 pH = 12.15.)

12.2.

pH is a measure of the concentration of hydrogen ions in water. Specifically, pH is the negative logarithm of the hydrogen ion (H+) concentration (mol/L) in an aqueous solution. pH = -log10(H+) This term indicates the basicity or acidity of a solution from 0 to 14 where pH 7 is neutral.

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The general terms for the given arithmetic sequences are given as follows:

1. [tex]a_n = -2 -3(n - 1)[/tex]

2. [tex]a_n = 22 - 2(n - 1)[/tex]

In an arithmetic sequence, the difference between consecutive terms is always the same, called common difference d.

The general term of an arithmetic sequence is given by:

[tex]a_n = a_1 + (n - 1)d[/tex]

For sequence 1, given by -2, -5, -8, -11, the first term and the common ratio are given as follows:

[tex]a_1 = -2, d = -3[/tex]

Hence the general term that defines sequence 1 is presented below:

[tex]a_n = -2 -3(n - 1)[/tex]

For sequence 2, represented by 22, 20, 18, 16, the first term and the common ratio are given as follows:

[tex]a_1 = 22, d = -2[/tex]

Hence the general term that defines sequence 2 is presented below:

[tex]a_n = 22 - 2(n - 1)[/tex]

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The coordinates for P and Q are as follows,

P = (a, a)

Q = (0, a)

Finding the Missing Coordinates:

Triangle OPQ is an isosceles triangle, hence two of its legs are equal.

Since the coordinates of the end point of the leg OQ lies on the y-axis, and OP is parallel to x-axis, OQ ⊥ QP ............ (1)

Also, it indicates that the x- coordinate of point Q is 0.

⇒ The coordinates of point Q are (0, a)

From (1), we can infer that,

OP = OQ [∵ OP is the hypotenuse]

⇒ The distance of point P from x-axis = a

The distance of point P from y-axis =a

Hence, the coordinates of point P are given as (a, a).

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The y-intercept exists reduced by 2 units. Hence there exists a translation of 2 units down.

Let the function be g(z) = 3 sin (TZ).

A vertical stretch by a factor of k indicates that the point (x, y) on the graph of f (x) exists transformed to the point (x, ky) on the graph of g(x), where k < 1.

If k = 1, then the same graph we get, and if k > 1 we get a vertically shrink graph.

In our question, there exists a vertical stretch of 2. This means the new graph would have points as (x, y/2)

i.e. instead of f(x) = y, we have now f(x) = y/2

So transformation is g(x) = 3f(x)

The y-intercept exists reduced by 2 units. Hence there exists a translation of 2 units down.

Therefore, the correct answer is option C.

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The identified vertical pair of angles in the given figure is: angle LOP and angle MOE.

A pair of angles are regarded as vertical angles that are formed when two straight line intersect each other. The vertical angles formed are non-adjacent angles. They share a common point or vertex but do not have any shared common side.

According to the vertical angles theorem, this vertical angles pair that are non-adjacent angles have angle measure that is congruent to each other.

In the figure shown below, the shared vertex O. The two non-adjacent angles that share this vertex but share no common sides are angle LOP and angle MOE. These vertical angles that are non-adjacent angles are congruent.

Therefore, the identified vertical pair of angles in the given figure is: angle LOP and angle MOE.

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Answer:

D

Step-by-step explanation:

This function passes the vertical line test.

The vertical line test test is done by drawing a vertical line anywhere on the graph. If the line goes through two points or more then it isn't a function.

The solution to the given differential equation is yp=−14xcos(2x)

The characteristic equation for this differential equation is:

P(s)=s2+4

The roots of the characteristic equation are:

s=±2i

Therefore, the homogeneous solution is:

yh=c1sin(2x)+c2cos(2x)

Notice that the forcing function has the same angular frequency as the homogeneous solution. In this case, we have resonance. The particular solution will have the form:

yp=Axsin(2x)+Bxcos(2x)

If you take the second derivative of the equation above for  yp , and then substitute that result,  y′′p , along with equation for  yp  above, into the left-hand side of the original differential equation, and then simultaneously solve for the values of A and B that make the left-hand side of the differential equation equal to the forcing function on the right-hand side,  sin(2x) , you will find:

A=0

B=−14

Therefore,

yp=−14xcos(2x)

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