AB=
BC=
Help me please!! Asap thanks so much

(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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Answer:

2/9 - 2/15

Solution

LCM = 45

45 ÷9 ×2 = 10

45÷ 15 ×2 = 6

10 - 6 = 4

ANSWER = 4/45

Answer:

l

Step-by-step explanation:

take the lcm of 9 and 15. it will be 45 . than continue

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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[tex]\sum^{\infty}_{n=1} (a/b)^n=5 \\ \\ =\frac{a/b}{1-\frac{a}{b}}=5 \\ \\ \frac{a}{b-a} =5 \\ \\ \frac{a}{b}=\frac{5}{6}[/tex]

So, we need to find

[tex]\sum^{\infty}_{n=1} n(5/6)^n

[/tex]

Let this sum be S.

Then,

[tex]S=(5/6)+2(5/6)^2 +3(5/6)^3+\cdots \\ \\ \frac{5}{6}S=(5/6)^2 + 2(5/6)^3+\cdots \\ \\ \implies \frac{1}{6}S=(5/6)+(5/6)^2+(5/6)^3+\cdots=5 \\ \\ \implies S=\boxed{30}[/tex]

Answer: line ZX is congruent to line VX (option 4)

Step-by-step explanation:

We already know <X is congruent to <X, we also know that line YX is congruent to line XW. Now all we need is one more line adjacent to X which is going to be ZX ad VX

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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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:

x=2

Step-by-step explanation:

2^x *2^4=2^3x

using law of indices

2^x+4=2^3x

x+4=3x

4=3x-x

4=2x

2=x

Step-by-step explanation:

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: 45

Step-by-step explanation: Given the way the formula is formatted, the first term is 1. The common difference can be found by subtracting a number from the number that follows (ex. 3-2 or 4-3), therefore it's 1. The desired term is what you're trying to find so 44-1=43. When you put it all together, the formula should be 2+1(44-1) which equals 45 when you follow the rules of PEMDAS.

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.

[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: 243

Step-by-step explanation:

If you divide the 18 inch by two to get the r, you will get 9. So you should multiple the 3 and the 9 square. So it should be 3x81=243

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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In order to solve this IVP using Laplace transforms, we must first write f(t) in terms of the Heaviside function.

f(t)=0*(u(t)-u(t-Pi))+1*(u(t-Pi)-u(t-2Pi))+0*(u(t-2Pi))

f(t)=u(t-π)-u(t-2π)

So, the rewritten IVP is

y'' +y = u(t-π)-u(t-2π)y(0)=0, y'(0)=1

Taking the Laplace transform of both sides of the equation, we get:

s2L{y}-sy(0)-y'(0)+L{y}=(1/s)*e-πs-(1/s)*e-2πs

s2L{y}-1+L{y}=(1/s)*e-πs-(1/s)*e-2πs

(s2+1)L{y}=1+(1/s)*e-πs-(1/s)*e-2πs

L{y}=1/(s2+1)+(1/s(s2+1))e-πs-(1/s(s2+1))*e-2πs

Now, we must take the inverse transform of both sides to solve for y.

The first inverse transform is easy enough. By definition, it is sin(t).

The second two inverse transforms will be a little tougher, we will have to use partial fraction decomposition to break them down into terms that are easier to compute.

A/s+(Bs+C)/(s^2+1)=1/(s(s^2+1))

A(s^2+1)+(Bs+C)(s)=1

As^2+A+Bs^2+Cs=1

Rewriting this system in matrix form, we get:

1  1  0    A     0

0  0  1  * B  = 0

1  0  0    C     1

Using row-reduction we find that A=1, B=-1, and C=0. So, our reduced inverse transforms are:

L-1{(e-πs)(1/s-s/(s2+1))}

and

L-1{(e-2πs)(1/s-s/(s2+1))}

Using the first and second shifting properties, these inverse transforms can be computed as.

L-1{(e-πs)(1/s-s/(s2+1))}=u(t-π)-cos(t-π)u(t-π)

L-1{(e-2πs)(1/s-s/(s2+1))}=u(t-2π)-cos(t-2π)u(t-2π)

Combining all of our inverses transforms, we get the solution the IVP as:

y=sin(t)+u(t-π)-cos(t-π)u(t-π)+u(t-2π)-cos(t-2π)u(t-2π)

In mathematics, the Laplace transform, named after its discoverer Pierre Simon Laplace (/ləˈplɑːs/), transforms a function of real variables (usually in the time domain) into a function of complex variables (in the time domain). is the integral transform that Complex frequency domain, also called S-area or S-plane).

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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:

680

Step-by-step explanation:

x = total number

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

.6x = 408

x = 408/.6 = 680

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 area of the shaded part is 30. 89 cm²

We have the shape to be a rectangle

The area of the shaded part should be;

Area of rectangle - 2 ( area of a semi circle)

The formula for area of a rectangle

Area = length × width

Area = 12 × 12

Area = 144 cm²

Area of a semicircle = 1/2 πr²

Area = 1/ 2 × 3. 142 × 6²

Area = 56. 56 cm²

Area of shaded part = area of rectangle - 2( area of semicircle)

Area of shaded part = 144 - 2(56. 56)

Area of shaded part = 144 - 113. 11

Area of shaded part = 30. 89 cm²

Thus, the area of the shaded part is 30. 89 cm²

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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.

The distance the player is from the right post is = 5.1 meters

The Pythagorean theorem states that the square of the length of the hypotenuse of a right triangle equals the sum of the squares of the lengths of the other two sides.

Formula for the Pythagorean theorem =

a²= b²+c²

From the diagram given,

The hypotenuse (a) = x²

b= 1.82²

c = 4.8²

x² = 1.82² + 4.8²

x²= 3.3124 + 23.04

x²= 26.3524

a= √26.3524

a= 5.1 meters.

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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 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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Answer:

20 cookies

Step-by-step explanation:

8 in 1.5 minutes

so we want to find how many in 3.75 minutes since 3 + 45/60 = 3.75

so then its 1.5*2 = 3 so 8*2 = 16 to get that 16 cookies in 3 minutes

then we still have .75 left so then divide 8/2 to get 4 cookies in 0.75 minutes

16+4 = 20

you can also just find how many in 0.25 minutes (15 seconds) you get 6/8

multiply that by 3.75/0.25 = 15 you get 15*(8/6) = 20

Answer:

B

Step-by-step explanation:

if an angle is congruent, it is the same measurement