Which of the following is a solution to the second order differential equation LaTeX: y''=-4y y ″ = − 4 y ? To answer this question, attempt to verify each of the following proposed solutions. a. LaTeX: y=\sin2t y = sin ⁡ 2 t b. LaTeX: y=-\frac{2}{3}t^3 y = − 2 3 t 3 c. LaTeX: y=\cos2t y = cos ⁡ 2 t d. LaTeX: y=e^{2t} y = e 2 t e. LaTeX: y=\frac{y''}{-4}

Answer:

0.592

Step-by-step explanation:

The volume of the solid, rounded to three decimal places is 18.257.

The volume of the solid obtained by rotating the region around the y-axis can be found using the method of discs or washers. Since the region is bounded by the x-axis, the y-axis, the line y=2, and the curve [tex]y=e^x[/tex], we will integrate with respect to y.

The volume V of the solid of revolution is given by the integral:

[tex]\[ V = \pi \int_{a}^{b} [R(y)]^2 dy - \pi \int_{a}^{b} [r(y)]^2 dy \][/tex]

where [tex]\( R(y) \)[/tex] is the outer radius and [tex]\( r(y) \)[/tex] is the inner radius of the discs or washers.

In this case, the outer radius [tex]\( R(y) \)[/tex] is given by the line y=2, which is a horizontal line, so the outer radius is constant and equal to 2. The inner radius [tex]\( r(y) \)[/tex] is given by the curve [tex]y=e^x[/tex]. To express x in terms of y, we take the natural logarithm of both sides to get [tex]\( x = \ln(y) \)[/tex].

Now we can set up our integrals:

[tex]\[ V = \pi \int_{0}^{2} [2]^2 dy - \pi \int_{0}^{2} [\ln(y)]^2 dy \][/tex]

[tex]\[ V = \pi \int_{0}^{2} 4 dy - \pi \int_{0}^{2} [\ln(y)]^2 dy \][/tex]

The first integral is straight forward:

[tex]\[ \pi \int_{0}^{2} 4 dy = \pi \left[ 4y \right]_{0}^{2} = \pi [4(2) - 4(0)] = 8\pi \][/tex]

The second integral requires integration by parts. Let [tex]\( u = [\ln(y)]^2 \)[/tex]and [tex]\( dv = dy \)[/tex], then [tex]\( du = \frac{2\ln(y)}{y} dy \)[/tex] and [tex]\( v = y \)[/tex]. Applying integration by parts gives:

[tex]\[ \pi \int_{0}^{2} [\ln(y)]^2 dy = \pi \left[ y[\ln(y)]^2 - \int_{0}^{2} 2\ln(y) dy \right] \][/tex]

Now, we need to integrate [tex]\( 2\ln(y) \)[/tex] by parts again, with [tex]\( u = \ln(y) \)[/tex] and [tex]\( dv = dy \)[/tex], then [tex]\( du = \frac{1}{y} dy \)[/tex] and [tex]\( v = y \)[/tex]. This gives:

[tex]\[ \int_{0}^{2} 2\ln(y) dy = \left[ 2y\ln(y) - \int_{0}^{2} 2 dy \right] \][/tex]

[tex]\[ \int_{0}^{2} 2\ln(y) dy = \left[ 2y\ln(y) - 2y \right]_{0}^{2} \][/tex]

[tex]\[ \int_{0}^{2} 2\ln(y) dy = 2(2)\ln(2) - 2(2) - (0) \][/tex]

[tex]\[ \int_{0}^{2} 2\ln(y) dy = 4\ln(2) - 4 \][/tex]

Putting it all together:

[tex]\[ \pi \int_{0}^{2} [\ln(y)]^2 dy = \pi \left[ y[\ln(y)]^2 - (4\ln(2) - 4) \right]_{0}^{2} \][/tex]

[tex]\[ \pi \int_{0}^{2} [\ln(y)]^2 dy = \pi \left[ 2[\ln(2)]^2 - (4\ln(2) - 4) \right] - \pi \left[ \lim_{y \to 0} y[\ln(y)]^2 - (4\ln(2) - 4) \right] \][/tex]

The limit as y approaches 0 of [tex]\( y[\ln(y)]^2 \)[/tex] is 0, so we have:

[tex]\[ \pi \int_{0}^{2} [\ln(y)]^2 dy = \pi \left[ 2[\ln(2)]^2 - 4\ln(2) + 4 \right] \][/tex]

[tex]\[ \pi \int_{0}^{2} [\ln(y)]^2 dy = 2\pi[\ln(2)]^2 - 4\pi\ln(2) + 4\pi \][/tex]

Now, subtract this from the first integral:

[tex]\[ V = 8\pi - (2\pi[\ln(2)]^2 - 4\pi\ln(2) + 4\pi) \][/tex]

[tex]\[ V = 8\pi - 2\pi[\ln(2)]^2 + 4\pi\ln(2) - 4\pi \][/tex]

[tex]\[ V = 4\pi + 4\pi\ln(2) - 2\pi[\ln(2)]^2 \][/tex]

[tex]\[ V = 4\pi(1 + \ln(2) - \frac{1}{2}[\ln(2)]^2) \][/tex]

Rounded to three decimal places, the volume is:

[tex]\[ V \approx 4\pi(1 + \ln(2) - \frac{1}{2}[\ln(2)]^2) \approx 4\pi(1 + 0.693 - \frac{1}{2}(0.693)^2) \][/tex]

[tex]\[ V \approx 4\pi(1 + 0.693 - 0.240) \][/tex]

[tex]\[ V \approx 4\pi(1.453) \][/tex]

[tex]\[ V \approx 5.812\pi \][/tex]

[tex]\[ V \approx 18.257 \][/tex]

Therefore, the volume of the solid, rounded to three decimal places, is:

[tex]\[ \boxed{18.257} \][/tex].

The complete question is:

Find the volume of the solid obtained by rotating the region bounded by the x-axis, the y-axis, the line y=2, and [tex]y=e^x[/tex] about the y-axis. Round your answer to three decimal places.

Answer:

x₃=1.599997=1.6

Step-by-step explanation:

f(x)=cos x, x₁=1.6

[tex]f'(x)=\frac{d}{dx}cos\ x\\=-sin\ x[/tex]

First iteration

At x₁=1.6

f(x₁)=f(1.6)

=cos(1.6)

=-0.0292

f'(x₁)=f'(1.6)

=-sin(1.6)

=-0.9996

[tex]\frac{f(x_1)}{f'(x_1)}=\frac{-0.0292}{-0.9996}\\=0.0292\\x_2=x_1-\frac{f(x_1)}{f'(x_1)}=1.6-(0.0292)\\\therefore x_2=1.5708[/tex]

[tex]f(x_2)=f(1.5708)\\=cos 1.5708\\=0.000003\\f'(x_2)=-sin1.5708\\=-1\\x_3=x_2-\frac{f(x_2)}{f'(x_2)}=1.6-\frac{0.000003}{-1}\\\therefore x_3=1.599997=1.6[/tex]

Answer:

(2/13)²

Step-by-step explanation:

-3/13 x -8/78 ( - x - is + )

+ 3 x 8 / 13 x 78 = 24/1014 = 4/169  =(2/13)²

P - existing home sales will decrease

Q - new housing starts will increase

R - unemployment will decrease

S - the Federal Reserve decreases long-term interest rates

T - foreign trade deficits decrease

U - foreign trade deficits increase

V - manufacturing rates increase

[tex][(P \vee Q) \wedge R] \iff (S \wedge T) \vee (U \wedge V)[/tex]

P - Germany will vote to limit the number of immigrants it admits

Q - will reject Angela Merkel’s international policies

R - neighboring EU countries agree to a multi-national work visa program

S - the World Bank revalues the Euro relative to the US dollar

T - the US-Russia brokered peace treaty is signed by Syria.

[tex]\neg (R \wedge (S \vee T))\implies (P \implies R)\\[/tex]

Substitute [tex]v(x)=-2x+y(x)[/tex], so that [tex]\dfrac{\mathrm dv}{\mathrm dx}=-2+\dfrac{\mathrm dy}{\mathrm dx}[/tex]. Then the ODE is equivalent to

[tex]\dfrac{\mathrm dv}{\mathrm dx}+2=v^2-7[/tex]

which is separable as

[tex]\dfrac{\mathrm dv}{v^2-9}=\mathrm dx[/tex]

Split the left side into partial fractions,

[tex]\dfrac1{v^2-9}=\dfrac16\left(\dfrac1{v-3}-\dfrac1{v+3}\right)[/tex]

so that integrating both sides is trivial and we get

[tex]\dfrac{\ln|v-3|-\ln|v+3|}6=x+C[/tex]

[tex]\ln\left|\dfrac{v-3}{v+3}\right|=6x+C[/tex]

[tex]\dfrac{v-3}{v+3}=Ce^{6x}[/tex]

[tex]\dfrac{v+3-6}{v+3}=1-\dfrac6{v+3}=Ce^{6x}[/tex]

[tex]\dfrac6{v+3}=1-Ce^{6x}[/tex]

[tex]v=\dfrac6{1-Ce^{6x}}-3[/tex]

[tex]-2x+y=\dfrac6{1-Ce^{6x}}-3[/tex]

[tex]y=2x+\dfrac6{1-Ce^{6x}}-3[/tex]

Given the initial condition [tex]y(0)=0[/tex], we find

[tex]0=\dfrac6{1-C}-3\implies C=-1[/tex]

so that the ODE has the particular solution,

[tex]\boxed{y=2x+\dfrac6{1+e^{6x}}-3}[/tex]

A second order linear, non - homogeneous ODE has a form of [tex]ay''+by'+cy=g(x)[/tex]

The general solution to, [tex]a(x)y''+b(x)y'+c(x)y=0[/tex]

Can be written as,

[tex]y=y_h+y_p[/tex]

Where [tex]y_h[/tex] is a solution to the homogeneous ODE and [tex]y_p[/tex] the particular solution, function that satisfies the non - homogeneous equation.

We can solve [tex]y_h[/tex] by rewriting the equation,

[tex]ay''+by'+cy=0\Longrightarrow(e^{xy})''+4e^{xy}=0[/tex]

Which simplifies to,

[tex]e^{xy}(y^2+4)=0[/tex]

From here we get two solutions,

[tex]y_{h1}=2i, y_{h2}=-2i[/tex]

So the form here refines,

[tex]y_h=c_1\cos(2x)+c_2\sin(2x)[/tex]

The same thing we do with [tex]y_p[/tex] to get form of,

[tex]y_p=-2x\cos(2x)[/tex]

From here the final form emerges,

[tex]y=\boxed{c_1\cos(2x)+c_2\sin(2x)-2x\cos(2x)}[/tex]

Hope this helps.

r3t40

Answer:

The maximum volume of the open box is 24.26 cm³

Step-by-step explanation:

The volume of the box is given as [tex]V=g(x)[/tex], where [tex]g(x)=x(6-2x)(8-2x)[/tex] and [tex]0\le x\le3[/tex].

Expand the function to obtain:

[tex]g(x)=4x^3-28x^2+48x[/tex]

Differentiate  wrt  x to obtain:

[tex]g'(x)=12x^2-56x+48[/tex]

To find the point where the maximum value occurs, we solve

[tex]g'(x)=0[/tex]

[tex]\implies 12x^2-56x+48=0[/tex]

[tex]\implies x=1.13,x=3.54[/tex]

Discard x=3.54 because it is not within the given domain.

Apply the second derivative test to confirm the maximum critical point.

[tex]g''(x)=24x-56[/tex], [tex]g''(1.13)=24(1.13)-56=-28.88\:<\:0[/tex]

This means the maximum volume occurs at [tex]x=1.13[/tex].

Substitute [tex]x=1.13[/tex] into [tex]g(x)=x(6-2x)(8-2x)[/tex] to get the maximum volume.

[tex]g(1.13)=1.13(6-2\times1.13)(8-2\times1.13)=24.26[/tex]

The maximum volume of the open box is 24.26 cm³

See attachment for graph.

Answer: 0.9554

Step-by-step explanation:

Given : The proportion of  the registered voters in a country are Republican = P=0.56

The number of voters = 447

The test statistic for proportion :-

[tex]z=\dfrac{p-P}{\sqrt{\dfrac{P(1-P)}{n}}}[/tex]

For p= 0.60

[tex]z=\dfrac{0.60-0.56}{\sqrt{\dfrac{0.56(1-0.56)}{447}}}\approx1.70[/tex]

Now, the probability that the sample proportion of Republicans will be less than 60% (by using the standard normal distribution table):-

[tex]P(x<0.60)=P(z<1.70)=0.9554345\approx0.9554[/tex]

Hence, the probability that the sample proportion of Republicans will be less than 60% = 0.9554

Answer:

  1/16

Step-by-step explanation:

Substituting the given value of x into the equation, we get ...

  [tex]y = 4\times 2^x=4\times 2^{-6}=\dfrac{4}{2^6}=\dfrac{4}{64}=\dfrac{1}{16}[/tex]

Answer:

[tex]2\frac{11}{12}[/tex] cups of flour are nedeed

Step-by-step explanation:

we know that

3 dozen peanut butter cookies calls for 1 and 1/4 cups of flour

Convert mixed number to an improper fraction

[tex]1\frac{1}{4}\ cups=\frac{1*4+1}{4}=\frac{5}{4}\ cups[/tex]

using proportion

Find out how much flour would you need to make 7 dozen cookies

Let

x ----> the number of cups of flour

[tex]\frac{3}{(5/4)}\frac{dozen}{cups}=\frac{7}{x}\frac{dozen}{cups} \\ \\x=7*(5/4)/3\\ \\x=\frac{35}{12}\ cups[/tex]

Convert to mixed number

[tex]\frac{35}{12}\ cups=\frac{24}{12}+\frac{11}{12}=2\frac{11}{12}\ cups[/tex]

Answer:

2.92 or 2 and 23/25 cups are required for 7 dozen cookies

Step-by-step explanation:

Determine the required number of flour for 3 dozen cookies. Use it to find flour required for 7 dozen cookies.

Dozen cookies               Flour

        3                             1 + 1/4

        7                                x

Cross multiply to find the value of x

3x = 7(1+1/4)

3x = 7(5/4)

12x = 35

x = 2.92 cups or 2 and 23/25 cups

Therefore, 2.92 cups or 2 and 23/25 cups of flour are required for 7 dozen peanut butter cookies.

!!

The answer is:

[tex](f+g)(2)=20[/tex]

To solve the problem, we need to add the given functions, and then, evaluate the resultant function with the given value of "x" which is equal to 2.

We need to remember that:

[tex](f+-g)(x)=f(x)+-g(x)[/tex]

So, we are given the functions:

[tex]f(x)=3x^2-2\\g(x)=4x+2\\[/tex]

Then, adding the functions , we have:

[tex](f+g)(x)=f(x)+g(x)=(3x^2-2)+(4x+2)[/tex]

[tex](f+g)(x)=3x^2-2+4x+2=3x^2+4x-2+2=3x^2+4x[/tex]

Therefore, we have that:

[tex](f+g)(x)=3x^2+4x[/tex]

Now, evaluating the function, we have:

[tex](f+g)(2)=3(2)^2+4(2)=3*4+4*2=12+8=20[/tex]

Hence, we have that the answer is:

[tex](f+g)(2)=20[/tex]

Have a nice day!

Answer:

Method A.

Step-by-step explanation:

For solving this question we need to find out the z-scores for both methods,

Since, the z-score formula is,

[tex]z=\frac{x-\mu}{\sigma}[/tex]

Where, [tex]\mu[/tex] is mean,

[tex]\sigma[/tex] is standard deviation,

Given,

For method A,

[tex]\mu = 33[/tex]

[tex]\sigma=8[/tex]

Thus, the z score for 34 is,

[tex]z_1=\frac{34-33}{8}=0.125[/tex]

While, for method B,

[tex]\mu = 37[/tex]

[tex]\sigma = 4[/tex]

Thus, the z score for 34 is,

[tex]z_2=\frac{34-37}{4}=-0.75[/tex],

Since, [tex]z_1 > z_2[/tex]

Hence, method A is preferred if the procedure must be completed within 34 minutes.

Comparison of two normal distribution can be done via intermediary standard normal distribution. The procedure to be preferred for  getting the procedure completed within 34 minutes is: Method A

If we've got a normal distribution, then we can convert it to standard normal distribution and its values will give us the z score.

If we have

[tex]X \sim N(\mu, \sigma)[/tex]

(X is following normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex] )

then it can be converted to standard normal distribution as

[tex]Z = \dfrac{X - \mu}{\sigma}, \\\\Z \sim N(0,1)[/tex]

(Know the fact that in continuous distribution, probability of a single point is 0, so we can write

[tex]P(Z \leq z) = P(Z < z) )[/tex]

Also, know that if we look for Z = z in z tables, the p value we get is

[tex]P(Z \leq z) = \rm p \: value[/tex]

Firstly, we need to figure out what the problem is asking, and a method which we can apply. Two normal distributions have to be compared. We can convert them to standard normal distribution for comparison. Then, we will get the p-value for 34 minutes(converted to standard normal variate's value) which will tell about the probability of obtaining time as 34 minutes(or under it)(this can be obtained with p value) in method A or B. The more the probability is there, the more chances for that method would be for getting completed within 34 minutes (compared to other method).

Let X = time taken for completion of surgical procedure by method A,

Then, by given data, we have: [tex]X \sim N(\mu = 33, \sigma = 8)[/tex]

The probability that X will fall within value 34 is [tex]P(X \leq 34)[/tex]

Converting this whole thing to standard normal distribution, we get the needed probability as:

[tex]P(X \leq 34) = P(Z = \dfrac{X - \mu}{\sigma} \leq \dfrac{34 - 33}{8} ) = P(Z \leq 0.15)[/tex]

From the z-tables, the p value for Z = 0.15 is 0.5596

Thus, we get:

[tex]P(X \leq 34) = P(Z \leq 0.15 ) \approx 0.5596[/tex]

Let Y = time taken for completion of surgical procedure by method B,

Then, by given data, we have: [tex]Y \sim N(\mu = 37, \sigma = 4)[/tex]

The probability that X will fall within value 34 is [tex]P(Y \leq 34)[/tex]

Converting this whole thing to standard normal distribution, we get the needed probability as:

[tex]P(Y \leq 34) = P(Z = \dfrac{Y - \mu}{\sigma} \leq \dfrac{34 - 37}{4} ) = P(Z \leq -0.25)[/tex]

From the z-tables, the p value for Z = -0.25 is 0.4013

Thus, we get:

[tex]P(Y \leq 34) = P(Z \leq -0.25 ) \approx 0.4013[/tex]

Thus, we see that:

P(method A will make surgical procedure last within 34 minutes) = 0.5596   > P(method B will make surgical procedure last within 34 minutes) =  0.4013

Thus, method A should be preferred, as there is higher chances for method A to get the surgery completed within 34 minutes than method B.

Learn more about standard normal distribution here:

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Answer:

Slope is 0.094,

It represents the average rate of change.

Step-by-step explanation:

Since, the slope is the ratio of difference in y-coordinates and the difference in x-coordinates,

Also, in a order pair, first element shows the x-coordinate and second element shows the y-coordinate.

Here, the data points are (1950, 0.75) and (1997, 5.15),

Thus, the slope is,

[tex]m=\frac{5.15-0.75}{1997-1950}[/tex]

[tex]=\frac{4.4}{47}[/tex]

[tex]=0.0936170212766[/tex]

[tex]\approx 0.094[/tex]

Also, Slope represents the average rate of change.

Answer:

Step-by-step explanation:

Great question, it is always good to ask away and get rid of any doubts that you may be having.

This process is called Rule-Making. It is basically when the Federal Government makes regulations on certain topics. These Regulations help advance and set strict boundaries on projects so that people know what they can and cannot do, as well as protect others from being scammed or robbed by these projects. This is all done by the administrative process known as Rule Making.

I hope this answered your question. If you have any more questions feel free to ask away at Brainly.

Answer:

The system of linear equations has infinitely many solutions. x=t, y=2-t and z=t-8.

Step-by-step explanation:

The given educations are

[tex]x+2y+z=-4[/tex]

[tex]-2x-3y-z=2[/tex]

[tex]2x+4y+2z=-8[/tex]

Using the Gauss-Jordan elimination method, we get

[tex]\begin{bmatrix}1 & 2 & 1\\ -2 & -3 & -1\\ 2 & 4 & 2\end{bmatrix}\begin{bmatrix}x\\ y\\ z\end{bmatrix}=\begin{bmatrix}-4\\ 2\\ -8\end{bmatrix}[/tex]

[tex]R_3\rightarrow R_3-2R_1[/tex]

[tex]\begin{bmatrix}1 & 2 &1\\ -2 & -3 & -1\\ 0 & 0 & 0\end{bmatrix}\begin{bmatrix}x\\ y\\ z\end{bmatrix}=\begin{bmatrix}-4\\ 2\\ 0\end{bmatrix}[/tex]

Since elements of bottom row are 0, therefore the system of equations have infinitely many solutions.

[tex]0x+0y+0z=0\Rightarrow 0=0[/tex]

[tex]R_2\rightarrow R_2+2R_1[/tex]

[tex]\begin{bmatrix}1 & 2 &1\\ 0 & 1 & 1\\ 0 & 0 & 0\end{bmatrix}\begin{bmatrix}x\\ y\\ z\end{bmatrix}=\begin{bmatrix}-4\\ -6\\ 0\end{bmatrix}[/tex]

[tex]R_1\rightarrow R_1-R_2[/tex]

[tex]\begin{bmatrix}1 & 1 &0\\ 0 & 1 & 1\\ 0 & 0 & 0\end{bmatrix}\begin{bmatrix}x\\ y\\ z\end{bmatrix}=\begin{bmatrix}2\\ -6\\ 0\end{bmatrix}[/tex]

[tex]x+y=2[/tex]

[tex]y+z=-6[/tex]

Let x=t

[tex]t+y=2\rightarrow y=2-t[/tex]

The value of y is 2-t.

[tex](2-t)+z=-6[/tex]

[tex]z=-6-2+t[/tex]

[tex]z=t-8[/tex]

The value of z is t-8.

Therefore the he system of linear equations has infinitely many solutions. x=t, y=2-t and z=t-8.

To solve the system of linear equations using the Gauss-Jordan elimination method, perform row operations on the augmented matrix to obtain the reduced row-echelon form. The solution is x = -2, y = 1, and z = 0.

To solve the system of linear equations using the Gauss-Jordan elimination method, we need to perform row operations to transform the augmented matrix into row-echelon form and then into reduced row-echelon form. Let's start by representing the system of equations as an augmented matrix:

[1 2 1 -4; -2 -3 -1 2; 2 4 2 -8]

Performing row operations, you can transform the augmented matrix into reduced row-echelon form, obtaining:
[1 0 0 -2; 0 1 0 1; 0 0 1 0]

The solution to the system is x = -2, y = 1, and z = 0. Therefore, (x, y, z) = (-2, 1, 0).

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Answer:

Step-by-step explanation:

Given that there are 3 events as

rolling 3 dice, are the events A: sum divisible by 3 and B: sum divisible 5

The sample space will have

(1,1,1)...(6,6,6)

Sum will start with 3 and end with 18

Sum divisible by 3 are 3,6,9..18

Sum divisible by 5 are 5,10,....15

The common element is 15

Hence these two events are not mutually exclusive.

Answer:

The correct option is C.

Step-by-step explanation:

Consider the provided information.

Four conditions of a binomial experiment:

There should be fixed number of trials

Each trial is independent with respect to the others

The maximum possible outcomes are two

The probability of each outcome remains constant.

Now, observe the provided options:

Option A and B are not possible as they doesn't satisfy the conditions of binomial experiment which is there must be fixed number of trials.

Now observe the option C which state that there must be fixed number of trials, it satisfy the condition of a binomial experiment.

Therefore, the correct option is C.

The number of trials in a binomial experiment must be fixed, with two possible outcomes and independent and identical trial conditions.

The number of trials in a binomial experiment must be fixed as one of the key characteristics of a binomial experiment is that there is a fixed number of trials (n). This characteristic distinguishes it from other types of experiments.

In a binomial experiment, there are only two possible outcomes (success and failure) for each trial, and these probabilities do not change from trial to trial. The trials are independent and identical in conditions.

For example, tossing a fair coin multiple times or conducting a series of independent Bernoulli trials involve a fixed number of trials, making them fit the criteria for a binomial experiment.

4, 7 and 9 are mutually coprime, so you can use the Chinese remainder theorem.

Start with

[tex]x=7\cdot9+4\cdot2\cdot9+4\cdot7\cdot5[/tex]

Taken mod 4, the last two terms vanish and we're left with

[tex]x\equiv63\equiv64-1\equiv-1\equiv3\pmod4[/tex]

We have [tex]3^2\equiv9\equiv1\pmod4[/tex], so we can multiply the first term by 3 to guarantee that we end up with 1 mod 4.

[tex]x=7\cdot9\cdot3+4\cdot2\cdot9+4\cdot7\cdot5[/tex]

Taken mod 7, the first and last terms vanish and we're left with

[tex]x\equiv72\equiv2\pmod7[/tex]

which is what we want, so no adjustments needed here.

[tex]x=7\cdot9\cdot3+4\cdot2\cdot9+4\cdot7\cdot5[/tex]

Taken mod 9, the first two terms vanish and we're left with

[tex]x\equiv140\equiv5\pmod9[/tex]

so we don't need to make any adjustments here, and we end up with [tex]x=401[/tex].

By the Chinese remainder theorem, we find that any [tex]x[/tex] such that

[tex]x\equiv401\pmod{4\cdot7\cdot9}\implies x\equiv149\pmod{252}[/tex]

is a solution to this system, i.e. [tex]x=149+252n[/tex] for any integer [tex]n[/tex], the smallest and positive of which is 149.

The problem is about finding a number x that satisfies a system of modular arithmetic equations. It can be solved using the Chinese Remainder Theorem which is part of number theory in mathematics. More information is needed to solve this specific system.

The problem at hand is to find a number x which satisfies the conditions x ≡ 1 (mod 4), x ≡ 2 (mod 7), and x ≡ 5 (mod 9). This falls under the mathematical concept of modular arithmetic.

Modular arithmetic is a system of arithmetic for integers, where numbers wrap around once reaching a certain value—the modulus.

The expressions x ≡ 1 (mod 4), x ≡ 2 (mod 7), and x ≡ 5 (mod 9) mean that when x is divided by 4, the remainder is 1; when x is divided by 7, the remainder is 2; and when x is divided by 9, the remainder is 5 respectively.

This is a type of problem known as a system of linear congruences, which can be solved by applying the Chinese Remainder Theorem. However, the information provided is insufficient to provide a specific numerical solution to the system of congruences. It is recommended that the student consults the section of their classroom material that discusses the Chinese Remainder Theorem and its applications.

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Answer:

15 subsets of cardinality 4 contain at least one odd number.

Step-by-step explanation:

Here the given set,

S={1,2,3,4,5,6},

Since, a set having cardinality 4 having 4 elements,

The number of odd digits = 3 ( 1, 3, 5 )

And, the number of even digits = 3 ( 2, 4, 6 )

Thus, the total possible arrangement of a set having 4 elements out of which atleast one odd number = [tex]^3C_1\times ^3C_3+^3C_2\times ^3C_2+^3C_3\times ^3C_1[/tex]

By using [tex]^nC_r=\frac{n!}{r!(n-r)!}[/tex],

[tex]=3\times 1+3\times 3+1\times 3[/tex]

[tex]=3+9+3[/tex]

[tex]=15[/tex]

Hence, 15 subsets of cardinality 4 contain at least one odd number.

Answer:

Right angles are congruent

Step-by-step explanation:

One right angle can be transformed to another using rigid transformations such as translation ,rotation and reflection.This is basically the definition of congruence because the idea is to transform one object to another.

Answer:

future payment is $77056.92

total interest is paid after 45 year is  $23056.42

Step-by-step explanation:

Given data

payment (P) = $100

No of installment (n) = 12

rate of interest ( r ) = 1.5 %  i.e. = 0.015

time period (t) = 45 years

to find out

future payment and interest after 45 year

solution

we know future payment formula i.e. given below

future payment = payment × [tex](1+\frac{r}{n})^{nt} - 1) / (r/n)[/tex]

now put all these value in equation

future payment = $ 100  × [tex](1+\frac{0.015}{12})^{12*45} - 1) / (0.015/12)[/tex]  

future payment = $ 77056.92

payment paid in 45 year @ $100 total money is paid is 45 × 12 × $100 i.e. = $54000

total interest = future payment  - money paid

total interest = $77056.42 - $54000

total interest = $23056.42

We can use reduction of order. Given that [tex]y_1(x)=x[/tex] is a known solution, we look for a solution of the form [tex]y_2(x)=v(x)y_1(x)[/tex]. It has derivatives [tex]{y_2}'=v'y_1+v{y_1}'[/tex] and [tex]{y_2}''=v''y_1+2v'{y_1}'+v{y_1}''[/tex]. Substituting these into the ODE gives

[tex]x(xv''+2v')-x(xv'+v)+xv=0[/tex]

[tex]x^2v''+(2x-x^2)v'=0[/tex]

Let [tex]w(x)=v'(x)[/tex] so that [tex]w'(x)=v''(x)[/tex] and we get an ODE linear in [tex]w[/tex]:

[tex]x^2w'+(2x-x^2)w=0[/tex]

Divide both sides by [tex]e^x[/tex]:

[tex]x^2e^{-x}w'+(2x-x^2)e^{-x}w=0/tex]

Since [tex](x^2e^{-x})=(2x-x^2)e^{-x}[/tex], we can condense the left side as the derivative of a product:

[tex](x^2e^{-x}w)'=0[/tex]

Integrate both sides and solve for [tex]w(x)[/tex]:

[tex]x^2e^{-x}w=C\implies w=\dfrac{Ce^x}{x^2}[/tex]

Integrate both sides again to solve for [tex]v(x)[/tex]. Unfortunately, there is no closed form for the integral of the right side, but we can leave the result in the form of a definite integral:

[tex]v=\displaystyle C_2+C_1\int_{x_0}^x\frac{e^t}{t^2}\,\mathrm dt[/tex]

where [tex]x_0[/tex] is any point on an interval over which a solution to the ODE exists.

Finally, multiply by [tex]y_1(x)[/tex] to solve for [tex]y_2(x)[/tex]:

[tex]y_2=\displaystyle C_2x+C_1x\int_{x_0}^x\frac{e^t}{t^2}\,\mathrm dt[/tex]

[tex]y_1(x)[/tex] already accounts for the [tex]C_2x[/tex] term above, so the second independent solution is

[tex]y_2=x\displaystyle\int_{x_0}^x\frac{e^t}{t^2}\,\mathrm dt[/tex]

Answer:

276 feet

Step-by-step explanation:

The best way to do this is to complete the square, which puts the quadratic in vertex form. The vertex of a negative parabola, which is what this is, is the highest point of the function...the max value.  The k coordinate of the vertex will tell us that highest value.  To complete the square, we will first set the quadratic equal to 0, then move the constant over to the other side:

[tex]-16t^2+112t=-80[/tex]

The rule for completing the square is that the leading coefficient must be a positive 1.  Ours is a negative 16, so we have to factor out -16:

[tex]-16(t^2-7t)=-80[/tex]

Now the next thing is to take half the linear term, square it, and then add it to both sides.  Our linear term is 7, half of that is 7/2.  Squaring 7/2 gives you 49/4.  So we add 49/9 into the parenthesis on the left.  However, we can't forget that there is a -16 out front there that refuses to be ignored.  We have to add then (-16)(49/4) onto the right:

[tex]-16(t^2-7t+\frac{49}{4})=-80-196[/tex]

The purpose of this is to create a perfect square binomial that serves as the h value of the vertex (h, k).  Stating that perfect square on the left and doing the addition on the right:

[tex]-16(t-\frac{7}{2})^2=-276[/tex]

Now we finalize by moving the constant back over and setting it back equal to y:[tex]y=-16(t-\frac{7}{2})^2+276[/tex]

The vertex is [tex](\frac{7}{2},276)[/tex]

That translates to "at 3.5 seconds the particle is at its max height of 276 feet".

The correct answer is b)276 feet.

The standard form of a quadratic equation is[tex]H(t) = at^2 + bt + c.[/tex]For the given equation, a = -16, b = 112, and c = 80. The vertex (h, k) of the parabola can be found using the formula h = -b/(2a).

Let's calculate h:

h = -b/(2a) = -112 / (2 * -16) = -112 / -32 = 3.

Now we substitute h back into the original equation to find k, the maximum height:

k = H(h) = -16(3.5)^2 + 112(3.5) + 80

Calculating k:

k = -16(12.25) + 392 + 80

k = -196 + 392 + 80

k = 196 + 80

k = 276

So the maximum height is 276 feet, which corresponds to option 2.

Answer:

a. 37 rooms are need to break even.

b. 73 rooms are required to make twice as much profit as you spent.

Step-by-step explanation:

Revenue of the company from each additional room is $1200.

Total construction cost = $44,000.

At break even condition, total revenue is equal to total cost. In other words, the profit of the firm is zero at break even.

Let x be the number of rooms that are need to break even.

Total revenue of x room is

[tex]TR=1200x[/tex]

At break even,

Total revenue = Total cost

[tex]1200x=44000[/tex]

Divide both sides by 1200.

[tex]x=\frac{44000}{1200}=36.6667\approx 37[/tex]

Therefore, 37 rooms are need to break even.

Let y be the number of rooms to make twice as much profit as you spent.

Total revenue of y room is

[tex]TR=1200y[/tex]

Total revenue = 2 × Total cost

[tex]1200x=2\times 44000[/tex]

[tex]1200x=88000[/tex]

Divide both sides by 1200.

[tex]y=\frac{88000}{1200}=73.33\approx 73[/tex]

Therefore 73 rooms are required to make twice as much profit as you spent.

Answer:

The system of linear equations has infinitely many solutions

Step-by-step explanation:

Let's modified the equations and find the answer.

Using the first equation:

[tex]2x-y=5[/tex] we can multiply by 2 in both sides, obtaining:

[tex]2*(2x-y)=2*5[/tex] which can by simplified as:

[tex]4x-2y=10[/tex] which is equal to:

[tex]4x=2y+10[/tex]

Considering the second equation:

[tex]=4x+ky=2[/tex]

Taking into account that from the first equation we know that: [tex]4x=2y+10[/tex], we can express the second equation as:

[tex]2y+10+ky=2[/tex], which can be simplified as:

[tex](2+k)y=2-10[/tex]

[tex](2+k)y=-8[/tex]

[tex]y=-8/(2+k)[/tex]

Because (-8) is being divided by (2+k), then (2+k) can't be equal to 0, so:

[tex]2+k=0[/tex] if [tex]k=-2[/tex]

This means that k can be any number different than -2, and for each of these solutions, there is a different solution for y, allowing also, different solutions for x.

For example, if k=0 then

[tex]y=-8/(2+0)[/tex] which give us y=-4, and, because:

[tex]4x=2y+10[/tex] if y=-4 then [tex]x=(-8+10)/4=0.5[/tex]

Now let's try with k=-1, then:

[tex]y=-8/(2-1)[/tex] which give us y=-8, and, because:

[tex]4x=2y+10[/tex] if y=-8 then [tex]x=(-16+10)/4=-1.5[/tex].

Then, the system of linear equations has infinitely many solutions

Answer:

Laplace transformation of [tex]L(t^\frac{3}{2}-e^{-10t})[/tex]=[tex]\frac{3\sqrt{\pi} }{4 s^\frac{5}{2} }\ -\frac{1}{s+10}[/tex]

Step-by-step explanation:

Laplace transformation of [tex]L(t^\frac{3}{2}-e^{-10t}  )[/tex]

[tex]L(t^\frac{3}{2})=\int_{0 }^{\infty}t^\frac{3}{2}e^{-st}dt\\substitute \ u =st\\L(t^\frac{3}{2})=\int_{0 }^{\infty}\frac{u}{s} ^\frac{3}{2}e^{-u}\frac{du}{s}=\frac{1}{s^{\frac{5}{2}}}\int_{0 }^{\infty}{u} ^\frac{3}{2}e^{-u}{du}[/tex]

the integral is now in gamma function form

[tex]\frac{1}{s^{\frac{5}{2}}}\int_{0 }^{\infty}{u} ^\frac{3}{2}e^{-u}{du}=\frac{1}{s^{\frac{5}{2}}}\Gamma(\frac{5}{2})=\frac{1}{s^{\frac{5}{2}}}\times\frac{3}{2}\times\frac{1}{2} }\Gamma (\frac{1}{2} )=\frac{3\sqrt{\pi} }{4 s^\frac{5}{2} }[/tex]

now laplace of [tex]L(e^{-10t})[/tex]

[tex]L(e^{-10t})=\frac{1}{s+10}[/tex]

hence

Laplace transformation of [tex]L(t^\frac{3}{2}-e^{-10t})[/tex]=[tex]\frac{3\sqrt{\pi} }{4 s^\frac{5}{2} }\ -\frac{1}{s+10}[/tex]

The Laplace transform of the function [tex]t^3/2 - e^-10t[/tex] is (3/2√π)/[tex]s^5/2 - 1/(s + 10)[/tex].

To find the Laplace transform of the function[tex]t^3/2 - e-10t[/tex], we use the Laplace transform properties and tables.

Set up a ratio:

You drove 72 minutes and 100 km = 72/100

You want the number of minutes (x) to drive 150 km = x/150

Set the ratios to equal each other and solve for x:

72/100 = x/150

Cross multiply:

(72 * 150) = 100 * x)

Simplify:

10,800/100x

Divide both sides by 100:

x = 10800/100 = 108

This means it would take 108 minutes to drive 150 km.

Now subtract the time you have already driven to fin how much more you need:

180 - 72 = 36 more minutes.

Answer:

36 min

Step-by-step explanation:

It takes 72 min to drive 100 km. 50 km are left to drive.

Half of the driving above is: It takes 36 min to drive 50 km.

Answer: 0.3446

Step-by-step explanation:

Given  : Mean : [tex]\mu = 90[/tex]

Standard deviation : [tex]\sigma = 10[/tex]

Also, the reading speed of second grade students in a large city is approximately​ normal.

Then , the formula to calculate the z-score is given by :_

[tex]z=\dfrac{x-\mu}{\sigma}[/tex]

For x =  94

[tex]z=\dfrac{94-90}{10}=0.4[/tex]

The p-value = [tex]P(z>0.4)=1-P(z<0.4)=1-0.6554217[/tex]

[tex]\\\\=0.3445783\approx0.3446[/tex]

Hence, the  probability a randomly selected student in the city will read more than 94 words per​ minute =0.3446

Answer:

None

Step-by-step explanation:

The slopes are the same. The y intercepts are 2 and - 6. These lines are parallel which means they never intersect. No intersection means no solution. Just to show you what this means, I graphed this on Desmos for you.

Red: y = 3x - 6

Blue: y = 3x + 2

These two never meet.

Answer:

Step-by-step explanation:

Explanation:

Graph both the equations and read the coordinates of the point of intersection as shown in the graph below:

Answer:

Step-by-step explanation:

We are given that a and b are rational numbers where [tex]b\neq0[/tex] and x is irrational number .

We have to prove a+bx is irrational number by contradiction.

Supposition:let  a+bx is a rational number then it can be written in [tex]\frac{p}{q}[/tex] form

[tex]a+bx=\frac{p}{q}[/tex] where [tex]q\neq0[/tex] where p and q are integers.

Proof:[tex]a+bx=\frac{p}{q}[/tex]

After dividing p and q by common factor except 1 then we get

[tex]a+bx=\frac{r}{s}[/tex]

r and s are coprime therefore, there is no common factor of r and s except 1.

[tex]a+bx=\frac{r}{s}[/tex] where r and s are integers.

[tex]bx=\frac{r}{s}-a[/tex]

[tex]x=\frac{\frac{r}{s}-a}{b}[/tex]

When we subtract one rational from other rational number then we get again a rational number and we divide one rational by other rational number then we get quotient number which is also rational.

Therefore, the number on the right hand of equal to is rational number but x is a irrational number .A rational number is not equal to an irrational number .Therefore, it is contradict by taking a+bx is a rational number .Hence, a+bx is an irrational number.

Conclusion: a+bx is an irrational number.