How much should you invest now at 7 1/4% simple annual interest in order to earn $190 interest in 4 months?

Non zero vector x perpendicular to u and v : x = [tex][ \frac{-39}{14} x_2 - \frac{26}{14} x_3 , x_2 , x_3, \frac{17}{14}x_2 + \frac{23}{14} x_3 ][/tex]

Given, v= [-2,-8,-7,2] u= [6,7,-2,8]

Let the vector be x = [[tex]x_1 , x_2 , x_3, x_4[/tex]]

Now x is non xero vector perpendicular to vector 'v' and 'u' .

So,

x . v = 0

[tex]-2x_1 - 8x_2 - 7x_3 + 2x_4 = 0[/tex] .........1

x . u = 0

[tex]6x_1 + 7x_2 -2x_3 + 8x_4 = 0[/tex] .........2

Solve 1 and 2 to eliminate [tex]x_4[/tex] .

Multiply 1 with 4 to make the coefficients of [tex]x_4[/tex] same .

[tex]-8x_1 - 32x_2 - 28x_3 + 8x_4 = 0[/tex]

[tex]6x_1 + 7x_2 -2x_3 + 8x_4 = 0[/tex]

Subtract two equations,

[tex]-14x_1 -39x_2 -26x_3 = 0[/tex]

[tex]-14x_1 = 39x_2 + 26x_3[/tex]

[tex]x_1 = \frac{-39}{14} x_2 - \frac{26}{14} x_3[/tex]

From equation 1,

[tex]x_4 = x_1 + 4x_2 + \frac{7}{2} x_3[/tex]

[tex]x_4 = \frac{-39}{14} x_2 - \frac{26}{14} x_3+ 4x_2 + \frac{7}{2} x_3\\\\x_4 = \frac{17}{14}x_2 + \frac{23}{14} x_3[/tex]

Thus x = [tex][ \frac{-39}{14} x_2 - \frac{26}{14} x_3 , x_2 , x_3, \frac{17}{14}x_2 + \frac{23}{14} x_3 ][/tex]

[tex]x_2 = [-39/14 , 1 , 0 , 17/14] + x_3[-26/14, 0 , 1 , 23/14 ][/tex]

[tex]x_1 , x_3[/tex] are arbitrary .

For every value of [tex]x_2 , x_3[/tex] vector x is obtained.

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To find a vector x that is perpendicular to vectors v and u, we can use the cross product.

To find a vector x that is perpendicular to vectors v and u, we can use the cross product. The cross product of two vectors is a vector that is perpendicular to both of them. To find the cross-product, we can use the formula:

x = (v2u3 - v3u2, v3u1 - v1u3, v1u2 - v2u1)

Plugging in the values, we get:

x = (-8(-2) - (-7)(7), (-7)(6) - (-2)(-2), (-2)(8) - (-8)(6)) = (1, 52, -32)

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Answer:  The required solution of the given differential equation is

[tex]x+y\log y=Cy.[/tex]

Step-by-step explanation:  We are given to solve the following ordinary differential equation :

[tex]ydx+(y-x)dy=0~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)[/tex]

We will be using the following formulas for integration and differentiation :

[tex](i)~d\left(\dfrac{x}{y}\right)=\dfrac{ydx-xdy}{y^2},\\\\\\(ii)~\int\dfrac{1}{y}dy=\log y.[/tex]

From equation (i), we have

[tex]ydx+(y-x)dy=0\\\\\Rightarrow ydx+ydy-xdy=0\\\\\\\Rightarrow \dfrac{ydx+ydy-xdy}{y^2}=\dfrac{0}{y^2}~~~~~~~~~~~~~~~~~~~~[\textup{dividing both sides by }y^2]\\\\\\\Rightarrow \dfrac{ydx-xdy}{y^2}+\dfrac{1}{y}dy=0\\\\\\\Rightarrow d\left(\dfrac{x}{y}\right)+d(\log y)=0.[/tex]

Integrating the above equation on both sides, we get

[tex]\int d\left(\dfrac{x}{y}\right)+\int d(\log y)=C~~~~~~~[\textup{where C is the constant of integration}]\\\\\\\Rightarrow \dfrac{x}{y}+\log y=C\\\\\Rightarrow x+y\log y=Cy.[/tex].

Thus, the required solution of the given differential equation is

[tex]x+y\log y=Cy.[/tex].

Answer:  Each muffin tin contains 3 oz of batter.

Step-by-step explanation:  Given that a baker pours  108 oz of batter into 36 muffin tins such that each tin has same amount of batter.

We are to calculate the quantity of batter in each tin.

We will be using the UNITARY method to solve the given problem.

Quantity of batter in 36 muffin tins = 108 oz.

Therefore, the quantity of batter in 1 muffin tin is given by

[tex]Q_t=\dfrac{108}{36}=3~\textup{oz}.[/tex]

Thus, each muffin tin contains 3 oz of batter.

By dividing 108 oz of batter by 36 muffin tins, you find that each tin contains 3 oz of batter. This simple division problem helps distribute the batter evenly. Each tin thus gets exactly 3 oz.

To find out how much batter is in each muffin tin, you need to divide the total amount of batter by the number of muffin tins.

Here are the steps:

So, there are 3 oz of batter in each muffin tin.

I'm going to guess that you meant to include parentheses somewhere, so that the ODE is supposed to be

[tex]y'=\dfrac{y^2x}{y^3+x^3}+\dfrac yx[/tex]

Then substitute [tex]y(x)=xv(x)[/tex] so that [tex]y'(x)=xv'(x)+v(x)[/tex]. Then

[tex]xv'+v=\dfrac{x^3v^2}{x^3v^3+x^3}+v[/tex]

[tex]xv'=\dfrac{v^2}{v^3+1}[/tex]

which is separable as

[tex]\dfrac{v^3+1}{v^2}\,\mathrm dv=\dfrac{\mathrm dx}x[/tex]

Integrate both sides: on the left,

[tex]\displaystyle\int\frac{v^3+1}{v^2}\,\mathrm dv=\int\left(v+\frac1{v^2}\right)\,\mathrm dv=\dfrac12v^2-\dfrac1v[/tex]

The other side is trivial. We end up with

[tex]\dfrac12v^2-\dfrac1v=\ln|x|+C[/tex]

Solve in terms of [tex]y(x)[/tex]:

[tex]\boxed{\dfrac{y^2}{2x^2}-\dfrac xy=\ln|x|+C}[/tex]

Let [tex]A(t)[/tex] denote the amount of salt in the tank at time [tex]t[/tex]. We're told that [tex]A(0)=0[/tex].

For [tex]0\le t\le15[/tex], the salt flows in at a rate of (1/5 lb/gal)*(5 gal/min) = 1 lb/min. When the regulating mechanism fails, 20 lbs of salt is dumped and no more salt flows for [tex]t>15[/tex]. We can capture this in terms of the unit step function [tex]u(t)[/tex] and Dirac delta function [tex]\delta(t)[/tex] as

[tex]\text{rate in}=u(t)-u(t-15)+20\delta(t-15)[/tex]

(in lb/min)

The salt from the mixed solution flows out at a rate of

[tex]\text{rate out}=\left(\dfrac{A(t)\,\mathrm{lb}}{50+(5-5)t\,\mathrm{gal}}\right)\left(5\dfrac{\rm gal}{\rm min}\right)=\dfrac A{10}\dfrac{\rm lb}{\rm min}[/tex]

Then the amount of salt in the tank at time [tex]t[/tex] changes according to

[tex]\dfrac{\mathrm dA}{\mathrm dt}=u(t)-u(t-15)+20\delta(t-15)-\dfrac A{10}[/tex]

Let [tex]\hat A(s)[/tex] denote the Laplace transform of [tex]A(t)[/tex], [tex]\hat A(s)=\mathcal L_s\{A(t)\}[/tex]. Take the transform of both sides to get

[tex]s\hat A(s)-A(0)=\dfrac1s-\dfrac{e^{-15s}}s+20e^{-15s}-\dfrac1{10}\hat A(s)[/tex]

Solve for [tex]\hat A(s)[/tex], then take the inverse of both sides.

[tex]\hat A(s)=\dfrac{\frac{10-10e^{-15s}}{s^2}+\frac{200e^{-15s}}s}{10s+1}[/tex]

[tex]\implies\boxed{A(t)=10-10e^{-t/10}+\left(30e^{3/2-t/10}-10\right)u(t-15)}[/tex]

Answer:  The required solution of the given IVP is

[tex]y(x)=\dfrac{3}{4}e^x+\dfrac{5}{4}e^{-x}.[/tex]

Step-by-step explanation:  We are given to find the solution of the following initial value problem :

[tex]y^{\prime\prime}-y=0,~~~y(0)=2,~~y^\prime(0)=-\dfrac{1}{2}.[/tex]

Let [tex]y=e^{mx}[/tex] be an auxiliary solution of the given differential equation.

Then, we have

[tex]y^\prime=me^{mx},~~~~~y^{\prime\prime}=m^2e^{mx}.[/tex]

Substituting these values in the given differential equation, we have

[tex]m^2e^{mx}-e^{mx}=0\\\\\Rightarrow (m^2-1)e^{mx}=0\\\\\Rightarrow m^2-1=0~~~~~~~~~~~~~~~~~~~~~~~~~~[\textup{since }e^{mx}\neq0]\\\\\Rightarrow m^2=1\\\\\Rightarrow m=\pm1.[/tex]

So, the general solution of the given equation is

[tex]y(x)=Ae^x+Be^{-x},[/tex] where A and B are constants.

This gives, after differentiating with respect to x that

[tex]y^\prime(x)=Ae^x-Be^{-x}.[/tex]

The given conditions implies that

[tex]y(0)=2\\\\\Rightarrow A+B=2~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)[/tex]

and

[tex]y^\prime(0)=-\dfrac{1}{2}\\\\\\\Rightarrow A-B=-\dfrac{1}{2}~~~~~~~~~~~~~~~~~~~~~~~~(ii)[/tex]

Adding equations (i) and (ii), we get

[tex]2A=2-\dfrac{1}{2}\\\\\\\Rightarrow 2A=\dfrac{3}{2}\\\\\\\Rightarrow A=\dfrac{3}{4}.[/tex]

From equation (i), we get

[tex]\dfrac{3}{4}+B=2\\\\\\\Rightarrow B=2-\dfrac{3}{4}\\\\\\\Rightarrow B=\dfrac{5}{4}.[/tex]

Substituting the values of A and B in the general solution, we get

[tex]y(x)=\dfrac{3}{4}e^x+\dfrac{5}{4}e^{-x}.[/tex]

Thus, the required solution of the given IVP is

[tex]y(x)=\dfrac{3}{4}e^x+\dfrac{5}{4}e^{-x}.[/tex]

Explanation:

"Any normal distribution" may have arbitrary mean and standard deviation. The "standard normal distribution" has a mean of zero and a standard deviation of 1.

The probability that has been chosen Bag 1 is 0.2087.

Given that, bag 1 contains 10 blue marbles, while bag 2 contains 15 blue marbles.

Here we have;

Bag 1 contains 10 blue marbles

Bag 2 contains 15 blue marbles

Chosen a bag at random and throw 5 red marbles in it.

[tex]Required Probability = P(\frac{Bag 1}{2 red and 3 blue marbles})[/tex]

= [tex]\frac{P(bag 1)\cap(2 Red \ and \ 3 blue)}{P(2 \ red \ and \ 3 \ blue \ marbles)}[/tex]

= [tex]\frac{\frac{1}{2}\times ^6C_2\times^{10}C_3}{\frac{1}{2}\times^6C_2\times^{10}C_3+\frac{1}{2}\times^6C_2\times^{15}C_3}[/tex]

= 0.2087

Therefore, the probability that has been chosen Bag 1 is 0.2087.

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To find the probability of choosing Bag 1 given there are 2 red marbles among the 5 chosen marbles, we can use Bayes' theorem to calculate the probability.

To solve this problem, we can use Bayes' theorem to find the probability that Bag 1 was chosen given there are exactly 2 red marbles among the 5 chosen marbles. Let's denote Bag 1 as event A and Bag 2 as event B.

The answer to the question is the probability of choosing Bag 1 given there are exactly 2 red marbles among the 5 chosen marbles.

Answer:

JL=2 units

Step-by-step explanation:

we know that

If k is the midpoint in the line JL

then

JL=JK+KL

JK=KL

substitute the given values

2x-5=3x-8

Solve for x

3x-2x=-5+8

x=3

so

JK=2x-5=2(3)-5=1 units

KL=3x-8=3(3)-8=1 units

therefore

JL=JK+KL=1+1=2 units

The answer for the sets corresponding to the given conditions is as follows:

a) A ∩ B = {4, 6, 8, 10, 12}.

b) A ∪ B = {2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13}.

c)  B ∩ C = {}.

d) B ∪ C = {1, 2, 3, 4, 5, 6, 7, 8...}.

e) Set A = {3, 5, 7}   and  Set B = {2,4, 6, 8, 10, 12}

Given:

Set A =  {x ∈ N : 3 ≤ x ≤ 13}

Set B =  {x ∈ N : x is even}

Set C  = {x ∈ N : x is odd}.

Solve each option:

(a) Find A ∩ B (the intersection of sets A and B):

Set A contains natural numbers from 3 to 13 (inclusive): A = {3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13}.

Set B contains even natural numbers: B = {2, 4, 6, 8, 10, 12, ...}.

The intersection of A and B includes even numbers that are between 3 and 13: A ∩ B = {4, 6, 8, 10, 12}.

(b) Find A ∪ B (the union of sets A and B):

Set A contains natural numbers from 3 to 13 (inclusive): A = {3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13}.

Set B contains even natural numbers: B = {2, 4, 6, 8, 10, 12, ...}.

The union of A and B includes all numbers from both sets: A ∪ B = {2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13}.

(c) Find B ∩ C (the intersection of sets B and C):

Set B contains even natural numbers: B = {2, 4, 6, 8, 10, 12, ...}.

Set C contains odd natural numbers: C = {1, 3, 5, 7, 9, 11, ...}.

The intersection of B and C is the empty set, as there are no numbers that are both even and odd.

(d) Find B ∪ C (the union of sets B and C):

Set B contains even natural numbers: B = {2, 4, 6, 8, 10, 12, ...}.

Set C contains odd natural numbers: C = {1, 3, 5, 7, 9, 11, ...}.

The union of B and C includes all natural numbers: B ∪ C = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, ...}.

(e) For the given example:

Set A = {3, 5, 7}

Set B = {2,4, 6, 8, 10, 12}

This example satisfies the conditions A ∩ B = {3, 5} and A ∪ B = {2, 3, 5, 7, 8}

The intersection and Unioun of all the sets is found from A and B.

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The intersection of sets A and B is {4, 6, 8, 10, 12}. The union of sets A and B is {3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13}. The intersection of sets B and C is {}. The union of sets B and C is {3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13}.

(a) To find the intersection of sets A and B, we need to identify the elements that are common to both sets. In set A, we have: {3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13}. In set B, we have: {4, 6, 8, 10, 12}. The elements that are common to both sets are: {4, 6, 8, 10, 12}. Therefore, A ∩ B = {4, 6, 8, 10, 12}.

(b) To find the union of sets A and B, we need to combine all the elements from both sets. In set A, we have: {3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13}. In set B, we have: {4, 6, 8, 10, 12}. Combining these sets gives us: {3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13}. Therefore, A ∪ B = {3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13}.

(c) To find the intersection of sets B and C, we need to identify the elements that are common to both sets. In set B, we have: {4, 6, 8, 10, 12}. In set C, we have: {3, 5, 7, 9, 11, 13}. The elements that are common to both sets are: {}. Therefore, B ∩ C = {}.

(d) To find the union of sets B and C, we need to combine all the elements from both sets. In set B, we have: {4, 6, 8, 10, 12}. In set C, we have: {3, 5, 7, 9, 11, 13}. Combining these sets gives us: {3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13}. Therefore, B ∪ C = {3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13}.

Answer:

<CBA

Step-by-step explanation:

The angle name could be either

<ABC or <CBA

The vertex must be in the middle

Answer:

Yes, the answer is CBA

Step-by-step explanation:

Let [tex]\vec h[/tex] and [tex]\vec\eta[/tex] be two vectors in [tex]H[/tex].

[tex]H[/tex] is a subspace of [tex]V[/tex] if (1) [tex]\vec h+\vec\eta\in H[/tex] and (2) for any scalar [tex]k[/tex], we have [tex]k\vec h\in H[/tex].

(1) True;

[tex]\mathrm{tr}(\vec h+\vec\eta)=\mathrm{tr}(\vec h)+\mathrm{tr}(\vec eta)=0[/tex]

so [tex]\vec h+\vec\eta\in H[/tex].

(2) Also true, since

[tex]\mathrm{tr}(k\vec h)=0k=k[/tex]

Therefore [tex]H[/tex] is a subspace of [tex]V[/tex].

Answer: Yes, H is a subspace of V

Step-by-step explanation:

We know that V is the space of all the 2x2 matrices with real entries.

H is the set of all 2x2 matrices with real entries that have trace equal to 0.

Obviusly the matrices that are in the space H also belong in the space V (because in H you have some selected matrices and in V you have all of them). The thing we need to prove is if H is an actual subspace.

Suppose we have two matrices that belong to H, A and B.

We must see that:

1) if A and B ∈ H, then (A + B)∈H

2) for a scalar number k, k*A ∈ H

lets write this as:

[tex]A = \left[\begin{array}{ccc}a1&a2\\a3&a4\\\end{array}\right] B = \left[\begin{array}{ccc}b1&b2\\b3&b4\\\end{array}\right][/tex]

where a1 + a4 = 0 = b1 + b4

then:

[tex]A + B = \left[\begin{array}{ccc}a1 + b1&a2 + b2\\a3 + b3&a4 + b4\\\end{array}\right][/tex]

the trace is:

a1 + b1 - (a4 + b4) = (a1 - a4) + (b1 - b4) = 0

then the trace is nule, and (A + B) ∈ H

and:

[tex]kA = \left[\begin{array}{ccc}k*a1&k*a2\\k*a3&k*a4\end{array}\right][/tex]

the trace is:

k*a1 - k*a4 = k(a1 - a4) = 0

so kA ∈ H

then H is a subspace of V

Answer:

d

Step-by-step explanation:

13 cant be divided equally nor cubed  because its not an even number u can try to give all thirteen of then

The simpler version of the initial problem is arranging three people in a line. There are three choices for the first spot, two for the second, and one for the third, which results in a total of six possible arrangements. This involves the principle of permutation in combinatorics.

The subject of the given problem can be defined as permutations. If we're looking for a simpler version of it, we should choose a problem which still involves line-up or arrangement of a smaller number of people. Hence, the best option is: 'In how many ways can three people be lined up for a picture?'

To solve this simpler problem, we consider the number of available spots for each person in the line. For the first spot, there are 3 people that could be selected. After the first person is chosen, there are only 2 people left for the second spot. Lastly, there is only 1 person left for the third spot. So, the total number of ways we can line up 3 people for a picture is 3*2*1 = 6 ways.

This is a basic principle called permutation in combinatorics which is a fundamental concept in mathematics that deals with counting, both as a means and an end in obtaining results.

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Answer:

  about 34.9%

Step-by-step explanation:

The probability of not making a marking error is 0.9. The probability of doing that 10 times independently is 0.9^10 ≈ 0.34868 ≈ 34.9%.

Answer:

190578024 ways.

Step-by-step explanation:

We are asked to find the number of ways in which a committee of 5 be chosen from 120 employees to interview prospective applicants.

We will use combinations to solve our given problem.

[tex]_{r}^{n}\textrm{C}=\frac{n!}{(n-r)!r!}[/tex], where,

n = Total number of items,

r = Number of items being chosen at a time.

Upon substituting our given values in above formula, we will get:

[tex]_{5}^{120}\textrm{C}=\frac{120!}{(120-5)!5!}[/tex]

[tex]_{5}^{120}\textrm{C}=\frac{120!}{115!*5!}[/tex]

[tex]_{5}^{120}\textrm{C}=\frac{120*119*118*117*116*115!}{115!*5*4*3*2*1}[/tex]

[tex]_{5}^{120}\textrm{C}=\frac{120*119*118*117*116}{5*4*3*2*1}[/tex]

[tex]_{5}^{120}\textrm{C}=\frac{120*119*118*117*116}{120*1}[/tex]

[tex]_{5}^{120}\textrm{C}=\frac{119*118*117*116}{1}[/tex]

[tex]_{5}^{120}\textrm{C}=\frac{190578024}{1}[/tex]

Therefore, the committee of five can be chosen from 120 employees in 190578024 ways.

Answer:

495

Step-by-step explanation:

To find this you have to find the LCM of the two times which in this case is 33 and 45. The LCM of those two is 495.

The minutes it will be before this happens again is 495.

The unitary method is a method for solving a problem by the first value of a single unit and then finding the value by multiplying the single value.

Assume that an airline’s flights for Miami leave every 33 minutes and flights from Dallas leave every 45 minutes.

To find the LCM of the two times which in this case is 33 and 45.

Factor;

33 = 3 x 11

45 = 5 x 3 x 3

Thus, The LCM of those two is 495.

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Answer:

that would be 4253.28 feet

that would be 1296.4 m

Step-by-step explanation:

I think that's the answer and I hope it helps :)

Answer:

The distance is 9.17 feet.

Step-by-step explanation:

The ramp, vertical distance it is lifted, and the ground form a right triangle, whose hypotenuse the ramp, and whose base and perpendicular are the ground and the lifted distance respectively.

Thus we have a triangle whose hypotenuse [tex]H[/tex] is 10 feet, the perpendicular [tex]P[/tex] is 4 feet, and a base [tex]B[/tex] feet.

The Pythagorean theorem gives:

[tex]H^2=P^2+B^2[/tex]

We substitute the values [tex]H=10[/tex], [tex]P =4[/tex] and solve for B:

[tex]B=\sqrt{H^2-P^2} =\sqrt{10^2-4^2} =9.17.[/tex]

Thus the distance is 9.17 feet.

Answer:

the Answer is ≈ 9.17 feet

Step-by-step explanation:

it is correct on edge  2020

Answer:

$1850.

Step-by-step explanation:

We are asked to find amount of interest earned in one year on an amount of $5000.

We will use simple interest formula to solve our given problem.

[tex]I=Prt[/tex], where,

I = Amount of interest earned,

r = Annual interest rate in decimal form.

t = Time in years.

Let us convert our given interest rate in decimal form.

[tex]37\%=\frac{37}{100}=0.37[/tex]

Upon substituting our given value in simple interest formula, we will get:

[tex]I=\$5000\times 0.37\times 1[/tex]

[tex]I=\$5000\times 0.37[/tex]

[tex]I=\$1850[/tex]

Therefore, an amount of $1850 is earned as interest in one year.

Final answer:

2 to the power of three halves is equivalent to the square root of 2 cubed, which is approximately 2.83.

Explanation:

In mathematics, when we raise a number to a fraction exponent, we are essentially taking the root of that number. In this case, 2 to the power of three halves is equivalent to the square root of 2 cubed.

2 to the power of three halves = [tex]\sqrt{(2^3)}[/tex] = [tex]\sqrt{8}[/tex] = 2.83

[tex]f(t)=\begin{cases}9&\text{for }0\le t<2\\(t-5)^2&\text{for }2\le t<5\\2te^{6t}&\text{for }t\ge5\end{cases}[/tex]

and presumably 0 for [tex]t<0[/tex]. We can express [tex]f(t)[/tex] in terms of the unit step function,

[tex]u(t-c)=\begin{cases}1&\text{for }t\ge c\\0&\text{for }t<c\end{cases}[/tex]

[tex]f(t)=9(u(t)-u(t-2))+(t-5)^2(u(t-2)-u(t-5))+2te^{6t}u(t-5)[/tex]

Quick explanation: [tex]9u(t)=9[/tex] for [tex]t\ge0[/tex], and [tex]9u(t-2)=9[/tex] for [tex]t\ge2[/tex]. So subtracting these will cancel the value of 9 for all [tex]t\ge2[/tex] and leave us with the value of 9 over the interval we want, [tex]0\le t<2[/tex]. The same reasoning applies for the other 3 terms.

Recall the time displacement theorem:

[tex]\mathcal L_s\{f(t-c)u(t-c)\}=e^{-sc}\mathcal L_s\{f(t)\}[/tex]

By this property, we have

[tex]\mathcal L_s\{9u(t)\}=\mathcal L_s\{9\}=\dfrac9s[/tex]

[tex]\mathcal L_s\{9u(t-2)\}=e^{-2s}\mathcal L_s\{9\}=\dfrac{9e^{-2s}}s[/tex]

[tex]\mathcal L_s\{(t-5)^2u(t-2)\}=\mathcal L_s\{((t-2)-3)^2u(t-2)\}[/tex]

[tex]=e^{-2s}\mathcal L_s\{(t-3)^2\}=\left(\dfrac2{s^3}-\dfrac6{s^2}+\dfrac9s\right)e^{-2s}[/tex]

[tex]\mathcal L_s\{(t-5)^2u(t-5)\}=e^{-5s}\mathcal L_s\{t^2\}=\dfrac{2e^{-5s}}{s^3}[/tex]

[tex]\mathcal L_s\{2te^{6t}u(t-5)\}=\mathcal L_s\{2e^{30}(t-5)e^{6(t-5)}+10e^{30}e^{6(t-5)}\}[/tex]

[tex]=2e^{30-5s}\mathcal L_s\{te^{6t}+5e^{6t}\}=2e^{30-5s}\left(\dfrac1{(s-6)^2}+\dfrac5{s-6}\right)[/tex]

Putting everything together, we end up with

[tex]\boxed{\mathcal L_s\{f(t)\}=\dfrac{(2-6s)e^{-2s}-2e^{-5s}}{s^3}+\dfrac9s-\dfrac{2e^{30-5s}(29-5s)}{(s-6)^2}}[/tex]

Answer:

v.w = 22

Step-by-step explanation:

We are given

r = <8, 8, -6>; v = <3, -8, -3>; w = <-4, -2, -6>

and we need to evaluate v.w

Using the formula:  v.w = vxwx+vywy+vzwz

Putting values and solving:

v.w = 3(-4)+(-8)(-2)+(-3)(-6)

v.w = -12+16+18

v.w = 22

So, v.w = 22

Answer with explanation:

(A)

It is given that, A is invertible, That is inverse of matrix exist.

    [tex]|A|=|A^{-1}|\neq 0[/tex]

That is,  [tex]|A|=|A^{-1}|=1[/tex], is incorrect Statement.

False

(B)

If a Matrix has , either any row or column has all entry equal to Zero, then value of Determinant is equal to 0.

Any matrix with a row of all zeros has a determinant of 1 ,is incorrect Statement.

False

(C)

The Meaning of Singular matrix is that , then Determinant of Singular Matrix is equal to Zero.

For, a n×n , matrix, whether n is Odd or even

  [tex]A^{T}= -A\\\\|A^{T}|=|-A|=(-1)^n|A|[/tex]

So, the statement, If A is a skew symmetric matrix,  [tex]A^{T}= -A[/tex],and A has size n x n then A must be singular if n is odd ,is incorrect Statement.

False

Answer:

The length of a side of the square is 22 inches.

Step-by-step explanation:

Let each side of square be = s

Let each side of triangle be = s - 8

Perimeter of square, p₁ = 4s

Perimeter of triangle = p₂ = 3s

                                          = 3(s-8)

                                          = 3s - 24

Therefore, according to the question

p₁ - p₂ = 46

4s - (3s - 24) = 46

4s - 3s + 24 = 46

s = 46 - 24

s = 22

The length of a side of the square is 22 inches.

Answer:

  105/252 = 0.41666...

Step-by-step explanation:

There are (7C4)(3C1) = (35)(3) = 105 ways to choose exactly 4 boys. There are 10C5 = 252 ways to choose 5 youths, so the probability that a randomly chosen team will consist of exactly 4 boys is ...

  105/252

_____

nCk = n!/(k!(n-k!))

Answer:

There is a 41.67% probability that exactly 4 boys are picked in this team of 5.

Step-by-step explanation:

The order is not important, so we use the combinations formula.

[tex]C_{n,x}[/tex] is the number of different combinatios of x objects from a set of n elements, given by the following formula.

[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]

Number of desired outcomes.

Four boys and one girl: So

[tex]C_{7,4}*C_{3,1} = \frac{7!}{4!(7-4)!}*\frac{3!}{1!(3-1)!} = 35*3 = 105[/tex]

Number of total outcomes:

Combination of five from a set of 10.

So

[tex]C_{10,5} = \frac{10!}{5!(10-5)!} = 252[/tex]

What is the probability that exactly 4 boys are picked in this team of 5?

[tex]P = \frac{105}{252} = 0.4167[/tex]

There is a 41.67% probability that exactly 4 boys are picked in this team of 5.

Answer:

Step-by-step explanation:

Given : The readings on thermometers are normally distributed with

Mean : [tex]\mu=\ 0[/tex]

Standard deviation : [tex]\sigma= 1[/tex]

The formula to calculate the z-score :-

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

For x = -1.52

[tex]z=\dfrac{-1.52-0}{1}=-1.52[/tex]

For x = -0.81

[tex]z=\dfrac{-0.81-0}{1}=-0.81[/tex]

The p-value = [tex]P(-1.52<z<-0.81)=P(z<-0.81)-P(z<-1.52)[/tex]

[tex]0.2089701-0.0642555=0.1447146\approx0.1447[/tex]

Hence, the probability that a randomly selected thermometer reads between negative 1.52 and negative 0.81 = 0.1447

To find the probability, standardize the values using z-scores and find the area under the normal curve between the z-scores.

To find the probability that a randomly selected thermometer reads between -1.52 and -0.81, we need to find the area under the normal curve between these two values. First, we need to standardize the values by finding the z-scores for these values using the formula z = (x - μ) / σ, where x is the given value, μ is the mean, and σ is the standard deviation. After finding the z-scores, we can then use the normal distribution table or a calculator to find the area between these z-scores.

The z-score for -1.52 is z = (-1.52 - 0) / 1.00 = -1.52 and the z-score for -0.81 is z = (-0.81 - 0) / 1.00 = -0.81. Using a normal distribution table or a calculator, we can find the area to the left of -1.52 and the area to the left of -0.81. The probability that a randomly selected thermometer reads between -1.52 and -0.81 is the difference between these two areas: P(-1.52 < X < -0.81) = P(X < -0.81) - P(X < -1.52).

Using the normal distribution table or a calculator, we can find that P(X < -0.81) is approximately 0.2123 and P(X < -1.52) is approximately 0.0655. Therefore, the probability that a randomly selected thermometer reads between -1.52 and -0.81 is approximately 0.2123 - 0.0655 = 0.1468, or 14.68%. The sketch of the region would be a shaded area under the standard normal curve between -1.52 and -0.81.

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Step-by-step explanation:

Assuming that the 350 taking history and the 300 taking math each includes the 270 taking both history and math, then:

N(H or M) = N(H) + N(M) − N(H and M)

N = 350 + 300 − 270

N = 380

There are 380 first-year students taking history or mathematics.

Answer:

1577.20 ft³

Step-by-step explanation:

Cube of length = 12 ft = a

Hole diameter which is cutout = 4 ft = d

Hole radius which is cutout = 4/2 =2 ft = r

Volume of the cube = a³

⇒Volume of the cube = a×a×a

⇒Volume of the cube = 12×12×12

⇒Volume of the cube = 1728 ft³

The hole cut out will be in the shape of a cylinder

Volume of cylinder = πr²h

⇒Volume of cylinder = π×2²×12

⇒Volume of cylinder = 150.79 ft³

Now volume of the solid figure with hole cut out is

Volume of the cube - Volume of cylinder

=1728 - 150.79

=1577.20 ft³

∴ Volume of solid figure not including hole cutout is 1577.20 ft³

Answer:

26

Step-by-step explanation:

Converting 11010 to base 10.

1*24=16

1*23=8

0*22=0

1*21=2

0*20=0

Adding all to get Ans=26_10

Step2 converting 26_10 to 10

The equation calculation formula for 26_10 number to 10 is like this below.

10|26  

10|2|6

10|2|2

Ans:26_10

Assuming the given number is in base 2, we have

[tex]11010_2=2^4+2^3+2^1=16+8+2=26_{10}[/tex]

Answer:

The required answer is [tex]x+C=\frac{1}{2}\ln|\frac{2+x-y}{y-x}|[/tex].

Step-by-step explanation:

The given differential equation is

[tex]\frac{dy}{dx}=(x-y+1)^2[/tex]

Substitute u=x-y+1 in the above equation.

[tex]\frac{du}{dx}=1-\frac{dy}{dx}[/tex]

[tex]\frac{dy}{dx}=1-\frac{du}{dx}[/tex]

[tex]1-\frac{du}{dx}=u^2[/tex]

[tex]1-u^2=\frac{du}{dx}[/tex]

Using variable separable method, we get

[tex]dx=\frac{du}{1-u^2}[/tex]

Integrate both the sides.

[tex]\int dx=\int \frac{du}{1-u^2}[/tex]

[tex]x+C=\frac{1}{2}\ln|\frac{1+u}{1-u}|[/tex]      [tex][\because \int \frac{dx}{a^2-x^2}=\frac{1}{2a}\n|\frac{a+x}{a-x}|+C][/tex]

Substitute u=x-y+1 in the above equation.

[tex]x+C=\frac{1}{2}\ln|\frac{1+x-y+1}{1-(x-y+1)}|[/tex]

[tex]x+C=\frac{1}{2}\ln|\frac{2+x-y}{y-x}|[/tex]

Therefore the required answer is [tex]x+C=\frac{1}{2}\ln|\frac{2+x-y}{y-x}|[/tex].