Show that the following argument with hypotheses on lines 1–2 and conclusion on line c is
valid, by supplementing steps using the rules of inference and/or logical equivalences. Clearly label which rule you used in each step.

1. p → q
2. ∼ (q ∨ r)
c. ∼ p

Answer:

  The digit 6 in February is worth 1/100 of its value in January.

Step-by-step explanation:

The value of a given digit in a number can be found by setting the other digits to zero.

In February, the digit 6 has a value of 00600 = 600.

In January, the digit 6 has a value of 060000 = 60,000.

The ratio of the value in February to the value in January is ...

  600/60000 = 1/100

The digit 6 in February is one hundredth of the digit 6 in January.

Answer:

5 + 2          

Step-by-step explanation:

We have to rewrite the given statement in addition form.

The integers have  property of:

Negative(-)  Negative(-) = Positive(+)

Positive(+)  Positive(+)  = Positive(+)

Positive(+)  Negative(-) = Negative(-)

Negative(-) Positive(+)  = Negative(-)

The given statement is:

5-(-2)

Since we have two negative together, it is converted into a positive.

Thus, the given statement can be written in positive form as

5 + 2

Answer: 0.0475

Step-by-step explanation:

Given : The amount of money spent on red balloon in a certain college town when the football team is in town is a normal random variable with

[tex]\mu=\$50000[/tex] and [tex]\sigma=\$3000[/tex]

Let x be the random variable that represents the  amount of money spent on red balloon.

Using formula [tex]z=\dfrac{x-\mu}{\sigma}[/tex], the z-score corresponding to x= 45000 will be :_

[tex]z=\dfrac{45000-50000}{3000}\approx-1.67[/tex]

Now, by using the standard normal distribution table for z, we have

P value : [tex]P(z<-1.67)=1-P(z<1.67)=1-0.9525=0.0475[/tex]

∴The proportion of home football game days in this town is less than $45000 worth of red balloons sold = 0.0475

Answer:

The rabbit should be fed:

[tex]6 + 2z[/tex] grams of food A

[tex]12 - 3z[/tex] grams of food B

[tex]z[/tex] grams of food C

For [tex]z \leq 4[/tex].

Step-by-step explanation:

This can be solved by a system of equations.

I am going to say that x is the number of grams of food A, y is the number of grams of food B and z is the number of grams of Food C.

The problem states that:

A researcher wants to provide a rabbit exactly 162 units of ​protein:

There are 5 units of protein in each gram of food A, 11 units of protein in each gram of food B and 23 units of protenin in each gram of food C. So

[tex]5x + 11y + 23z = 162[/tex]

A researcher wants to provide a rabbit exactly 72 units of carbohydrates:

There are 2 units of carbohydrates in each gram of food A, 5 units of carbohydrates in each gram of food B and 11 units of carbohydrates in each gram of food C. So:

[tex]2x + 5y + 11z = 72[/tex]

A researcher wants to provide a rabbit exactly 30 units of Vitamin A:

There is 1 unit of Vitamin A in each gram of food A, 2 units of Vitamin A in each gram of food B and 4 units of Vitamin A in each gram of food C. So:

[tex]x + 2y + 4z = 30[/tex].

We have to solve the following system of equations:

[tex]5x + 11y + 23z = 162[/tex]

[tex]2x + 5y + 11z = 72[/tex]

[tex]x + 2y + 4z = 30[/tex].

I think that the easier way to solve this is reducing the augmented matrix of this system.

This system has the following augmented matrix:

[tex]\left[\begin{array}{cccc}5&11&23&162\\2&5&11&72\\1&2&4&30\end{array}\right][/tex]

To help reduce this matrix, i am going to swap the first line with the third

[tex]L_{1} <-> L_{3}[/tex]

Now we have the following matrix:

[tex]\left[\begin{array}{cccc}1&2&4&30\\2&5&11&72\\5&11&23&162\end{array}\right][/tex]

Now i am going to do these following operations, to reduce the first row:

[tex]L_{2} = L_{2} - 2L_{1}[/tex]

[tex]L_{3} = L_{3} - 5L_{1}[/tex]

Now we have

[tex]\left[\begin{array}{cccc}1&2&4&30\\0&1&3&12\\0&1&3&12\end{array}\right][/tex]

Now, to reduce the second row, i do:

[tex]L_{3} = L_{3} - L_{2}[/tex]

The matrix is:

[tex]\left[\begin{array}{cccc}1&2&4&30\\0&1&3&12\\0&0&0&0\end{array}\right][/tex]

This means that z is a free variable, so we are going to write y and x as functions of z.

From the second line, we have

[tex]y + 3z = 12[/tex]

[tex]y = 12 - 3z[/tex]

From the first line, we have

[tex]x + 2y + 4z = 30[/tex]

[tex]x + 2(12 - 3z) + 4z = 30[/tex]

[tex]x + 24 - 6z + 4z = 30[/tex]

[tex]x = 6 + 2z[/tex]

Our solution is: [tex]x = 6 + 2z, y = 12 - 3z, z = z[/tex].

However, we can not give a negative number of grams of a food. So

[tex]y \geq 0[/tex]

[tex]12 - 3z \geq 0[/tex]

[tex]-3z \geq -12 *(-1)[/tex]

[tex]3z \leq 12[/tex]

[tex]z \leq 4[/tex]

The rabbit should be fed:

[tex]6 + 2z[/tex] grams of food A

[tex]12 - 3z[/tex] grams of food B

[tex]z[/tex] grams of food C

For [tex]z \leq 4[/tex].

Step-by-step explanation:

Consider the provided information.

We have given that H and k are the subgroups of orders 5 and 8, respectively.

We need to prove that H∩K = {e}.

As we know "Order of element divides order of group"

Here, the order of each element of H must divide 5 and every group has 1 identity element of order 1.

1 and 5 are the possible order of 5 order subgroup.

For subgroup order 8: The possible orders are 1, 2, 4 and 8.

Now we want to find the intersection of these two subgroups.

Clearly both subgroup H and k has only identity element in common.

Thus, H∩K = {e}.

Answer:

[tex]f(x) = 4sin(\frac{\pi}{2}x) - 3[/tex], the third one

Step-by-step explanation:

Explaining the sine function:

The sine function is defined by:

[tex]S = Asin(p(x - x_{0})) + V[/tex]

In which A is the amplitude, [tex]p = \frac{2\pi}{N}[/tex] is the period, [tex]x_{0}[/tex] is the horizontal shift and V is the vertical shift.

So, in your problem:

The amplitude is 4, so A = 4.

The period is [tex]\frac{\pi}{2}[/tex], so [tex]p = \frac{\pi}{2}[/tex].

There is no horizontal shift, so [tex]x_{0} = 0[/tex].

The vertical shift is −3, so V = -3.

The sine function that represents these following conditions is

[tex]f(x) = 4sin(\frac{\pi}{2}x) - 3[/tex], the third one

Answer: In fraction : [tex]\dfrac{373}{500}[/tex]

In Decimal :  0.746

Step-by-step explanation:

Given : A ream of paper contains 500 sheets of paper.

Norm has 373 sheets of paper left from a ream.

Then, the fraction of ream Norm has will be :-

[tex]\dfrac{\text{Number of sheets left from ream }}{\text{Total sheets in a ream }}\\\\=\dfrac{373}{500}[/tex]

To convert in decimal we divide 373 by 500, we get 0.746.

Hence, The portion of a ream Norm has as = [tex]\dfrac{373}{500}[/tex] or 0.746

Answer:

Maximum height of the arrow is 203 feets

Step-by-step explanation:

It is given that,

The height of the arrow as a function of time t is given by :

[tex]h(t)=-16t^2+64t+11[/tex]..........(1)

t is in seconds

We need to find the maximum height of the arrow. For maximum height differentiating equation (1) wrt t as :

[tex]\dfrac{dh(t)}{dt}=0[/tex]

[tex]\dfrac{d(-16t^2+64t+11)}{dt}=0[/tex]

[tex]-32t+64=0[/tex]

t = 2 seconds

Put the value of t in equation (1) as :

[tex]h(t)=-16(2)^2+64(2)+11[/tex]

h(t) = 203 feet

So, the maximum height reached by the arrow is 203 feet. Hence, this is the required solution.

Answer:

equation of plane, 5x+3y-z-36=0

Distance of point (2,2,2) from plane = 4.05 units

Step-by-step explanation:

Given,

Plane passing through the point = (8, -2, 0)

Let's say, [tex]x_1\ =\ 8[/tex]

               [tex]y_1\ =\ -2[/tex]

                [tex]z_1\ =\ 0[/tex]

Plane perpendicular to the vector, a= 5i + 3j- k

Since, the vector is perpendicular to the plane, hence the equation of plane can be given by

[tex](5i + 3j- k).((x-x_1)i+(y-y_1)j+(z- z_1)k)=\ 0[/tex]

[tex]=>(5i + 3j- k).((x-8)i+(y+2)j+(z-0)k)=\ 0[/tex]

[tex]=>\ 5(x-8)+3(y+2)-z=0[/tex]

[tex]=>\ 5x\ -\ 40\ +\ 3y\ +\ 6\ -\ z\ =\ 0[/tex]

[tex]=>\ 5x\ +\ 3y\ -\ z\ -\ 36\ =\ 0[/tex]

Hence, the equation of plane can be given by, 5x+3y-z-36=0

Now, we have to calculate the distance of the point O(2,2,2) from the plane 5x+3y-z-36=0

Let's say,

a= 5, b= 3, c= -1, d=-36

[tex]x_0=2,\ y_0=2,\ z_0=2[/tex]

So, distance of a point from the plane can be given by,

[tex]d=\dfrac{ax_0+by_0+cz_0+d}{\sqrt{a^2+b^2+c^2}}[/tex]

 [tex]=\dfrac{\left |5\times 2+3\times 2+(-1)\times 2-36\right |}{\sqrt{5^2+3^2+(-1)^2}}[/tex]

 [tex]=\dfrac{24}{\sqrt{35}}[/tex]

  = 4.05 units

So, the distance of the point O(2,2,2) from the given plane will be 4.05 units.

Answer:

Net Present Value = - $99,360

Step-by-step explanation:

As provided,

Cash outlay = $65,000 each year for 10 years

Since the first outlay will be immediately, the cumulative discounting factor for cash outlay will be @ 16% = 1 for year 0 + 4.606 for 9 years = 5.606

Therefore, cumulative present value of total cash outlay = $65,000 [tex]\times[/tex] 5.606 = $364,390

Cash inflows beginning in year 11 = $170,000 for another continuous 20 years.

these cash flow will occur in the beginning of year 11 and end of year 10

Discounting factor will be [tex]\frac{1}{(1+0.16)^1^0}[/tex] = 0.2267

For, consecutive 20 years = 1.559

Therefore, value of inflows = $170,000 [tex]\times[/tex] 1.559 = $265,030

Net Present Value = Present Value of Cash Inflows - Present Value of Cash Outflows = $265,030 - $364,390 = - $99,360

Answer:

1mL should be administered to provide a 25-mg dose of aminophylline.

Step-by-step explanation:

The problem states that one 20-mL ampul contains 0.5 g of aminophylline, and asks how many milliliters should be administered to provide a 25-mg dose of aminophylline.

The first step is converting 0.5g to mg.

Each g has 1000mg, so:

1g - 1000mg

0.5g - xmg

x = 1000*0.5

x = 500mg.

Now we have that one 20-mL ampul contains 500mg of aminophylline. How many milliliters should be administered to provide a 25-mg dose of aminophylline?

As the dose increases, so does the quantity of aminophylline. It means that we have a direct rule of a three, there is a cross multiplication. So:

20mL - 500mg

x mL - 25mg

500x = 500

[tex]x = \frac{500}{500}[/tex]

x = 1 mL

1mL should be administered to provide a 25-mg dose of aminophylline.

Final answer:

1 mL of the aminophylline solution should be administered to provide a 25-mg dose. This calculation was made by first converting the total amount of aminophylline to milligrams and then establishing the amount per milliliter to solve for the necessary volume.

Explanation:

To determine how many milliliters of aminophylline should be administered to provide a 25-mg dose, we need to use the following information:

First, we need to convert 0.5 g of aminophylline to milligrams:

0.5 g × 1000 mg/g = 500 mg

Now that we have the total amount of aminophylline in milligrams, we can determine the amount per milliliter:

500 mg / 20 mL = 25 mg/mL

To find the volume that contains 25 mg, we set up a proportion:

(25 mg of aminophylline) / (X mL) = (25 mg/mL)

By solving for X, we find:

X = 1 mL

So, 1 mL of the aminophylline solution should be administered to provide a 25-mg dose.

Answer:

The equation to find the number of gallons purchased is:

[tex]C(n) = 3.05n[/tex]

You purchased 12 gallons of gasoline

Step-by-step explanation:

This problem can be modeled by the following first order function

[tex]C(n) = P_{G}n[/tex]

Where C(n) is the cost in function to the number of gallons, P is the price of the gallon and n is the number of gallons

The problem states that the price of gasoline is $3.05 per gallon, so P = 3.05

The equation to find the number of gallons purchased is:

[tex]C(n) = 3.05n[/tex]

If you pay $36.60, you have C = 36.60, and want to find n, so:

[tex]36.60 = 3.05n[/tex]

[tex]n = \frac{36.60}{3.05}[/tex]

[tex]n = 12[/tex]

You purchased 12 gallons of gasoline

Answer:

No, whole numbers are numbers that are whole, which would mean that they are not a decimal or a fraction. This is because a fraction or a decimal is a portion of a number, which would mean that the decimal or fraction is not complete/ not whole!

Answer:

  x ≈ 181/128 ≈ 1.41406

Step-by-step explanation:

The attached table shows the iterations. At each step, the interval containing the root is bisected and the function value at the midpoint of the interval is found. The sign of it relative to the signs of the function values at the ends of the interval tell which half interval contains the root. The process is repeated until the interval width is less than 10^-2.

Interval: [0, 2], signs [+, -], midpoint: 1; sign at midpoint: +

             [1, 2]                                      3/2                           -

             [1, 3/2]                                   5/4                           +

...

the rest is in the attachment. The listed table values are the successive interval midpoints.

The final midpoint is 181/128 ≈ 1.41406. This is within 0.0002 of the actual root.

_____

The actual solution in the interval [0, 2] is √2 ≈ 1.41421.

To find a solution utilizing the Bisection Method, one needs to establish the function and verify it satisfies the bisection condition. The process is iteratively repeated by adjusting the interval to the midpoint until the error tolerance is reached or the function value of the midpoint is within the desired accuracy.

The subject of your question concerns utilizing the Bisection method to solve a certain polynomial equation from a given interval [0, 2] with an accuracy of within 10^-2. The Bisection Method is a root-finding method in numerical analysis to solve for roots in given intervals.

This is an example of how the Bisection Method would be applied in solving for roots of polynomial equations. Do take note that this method only provides approximate solutions and it can be a lengthy process for equations with multiple roots.

https://brainly.com/question/32563551

#SPJ2

Answer: 0.1038

Step-by-step explanation:

We assume that oil in each container is filled will normal distribution.

Given : Population mean : [tex]\mu=12[/tex]

Standard deviation: [tex]\sigma=0.25[/tex]

Sample size : [tex]n=40[/tex]

Let x be the random variable that denotes the amount of oil filled in container.

z-score : [tex]z=\dfrac{x-\mu}{\dfrac{\sigma}{\sqrt{n}}}[/tex]

For x= 12.05

[tex]z=\dfrac{12.05-12}{\dfrac{0.25}{\sqrt{40}}}=1.26491106407\approx1.26[/tex]

Now by using the standard normal table for z, we have the probability that the sample mean,X is greater  then 12.05:-

[tex]P(z>1.26)=1-0.8961653=0.1038347\approx0.1038[/tex]

Hence, the probability that the sample mean,X is greater  then 12.05 = 0.1038

Answer: 0.9575465

Step-by-step explanation:

Let the random variable X is normally distributed with mean [tex]\mu=50[/tex] and standard deviation[tex]\sigma=7[/tex] .

Using the formula , [tex]z=\dfrac{x-\mu}{\sigma}[/tex] , we have the z-value for x= 34

[tex]z=\dfrac{34-50}{7}\approx-2.29[/tex]

For x= 63

[tex]z=\dfrac{63-50}{7}\approx1.86[/tex]

P-value : P(34<x<63)=P(-2.29<z<1.86)

[tex]=P(z<1.86)-P(z<-2.29)\\\\=0.9685572-(1-P(z<2.29))\\\\1=0.9685572-(1-0.9889893)\\\\=0.9575465[/tex]

Hence, the required probability = 0.9575465

Answer:

$5265.71

Step-by-step explanation:

We have been given that you deposit $3000 into a money-market savings account which pays 4.8% compounded quarterly.

We will use future value formula to solve our given problem.

[tex]FV=C_0\times (1+r)^n[/tex], where,

[tex]C_0=\text{Initial amount}[/tex],

r = Rate of return in decimal form,

n = Number of periods.

[tex]4.8\%=\frac{4.8}{100}=0.048[/tex]

[tex]n=3\times 4=12[/tex]

[tex]FV=\$3,000\times (1+0.048)^{12}[/tex]

[tex]FV=\$3,000\times (1.048)^{12}[/tex]

[tex]FV=\$3,000\times 1.7552354909370114[/tex]

[tex]FV=\$5265.7064\approx \$5265.71[/tex]

Therefore, there will be $5265.71 in your account at the end of those 3 years.

Step-by-step explanation:

We will prove by mathematical induction that, for every natural n,  

[tex](x-y)(x^{n}+x^{n-1}y+...+xy^{n-1}+y^{n})=x^{n+1}-y^{n+1}[/tex]

We will prove our base case (when n=1) to be true:

Base case:

[tex](x-y)(x^{n}+x^{n-1}y+...+xy^{n-1}+y^{n})=(x-y)(x^{1}+y^{1})=x^2-y^2=x^{1+1}-y^{1+1}[/tex]

Inductive hypothesis:  

Given a natural n,  

[tex]x^{n+1}-y^{n+1}=(x-y)(x^{n}+x^{n-1}y+...+xy^{n-1}+y^{n})[/tex]

Now, we will assume the inductive hypothesis and then use this assumption, involving n, to prove the statement for n + 1.

Inductive step:

Observe that, for y=0 the conclusion is clear. Then we will assume that [tex]y\neq 0.[/tex]

[tex](x-y)(x^{n+1}+x^{n}y+...+xy^{n}+y^{n+1})=(x-y)y(\frac{x^{n+1}}{y}+x^{n}+...+xy^{n-1}+y^{n})=(x-y)y(\frac{x^{n+1}}{y})+(x-y)y(x^{n}+...+xy^{n-1}+y^{n})=(x-y)y(\frac{x^{n+1}}{y})+y(x^{n+1}-y^{n+1})=(x-y)x^{n+1}+y(x^{n+1}-y^{n+1})=x^{n+2}-yx^{n+1}+yx^{n+1}-y^{n+2}=x^{n+2}-y^{n+2}\\[/tex]

With this we have proved our statement to be true for n+1.    

In conlusion, for every natural n,

[tex](x-y)(x^{n}+x^{n-1}y+...+xy^{n-1}+y^{n})=x^{n+1}-y^{n+1}[/tex]

Answer:  21

Step-by-step explanation:

Given : A bag containing five red marbles, two green ones, one transparent one, four yellow ones, and two orange ones .

Total marbles other than red or green = 1+4+2=7

Now, the number of combinations to select five marbles from the set of 7 will be :-

[tex]7C_5=\dfrac{7!}{5!(7-5)!}=\dfrac{7\times6\times5!}{5!\times2!}=21[/tex]

Hence, the number of  possible sets of five marbles are there in which none of them are red or green =21

Answer:

[tex]y(t)=-\frac{1}{t+C}[/tex]

Step-by-step explanation:

We are given that

[tex]\frac{dy}{dt}=y^2[/tex]

We have to find the value of y(t).

[tex]\frac{dy}{dt}=y^2[/tex]

[tex]\frac{dy}{y^2}=dt[/tex]

Integrating on both sides

[tex]\int y^{-2}dy=\int dt[/tex]

We know that [tex]\int x^n dx=\frac{x^{n+1}}{n+1}+C[/tex]

Using the formula

[tex]\frac{y^{-1}}{-1}=t+C[/tex]

[tex]-\frac{1}{y}=t+C[/tex]

[tex]a^{-1}=\frac{1}{a}[/tex]

Taking the reciprocal on both side then , we get

[tex]-y=\frac{1}{t+C}[/tex]

[tex]y(t)=-\frac{1}{t+C}[/tex]

Answer: I'm sure its 2/21

Step-by-step explanation: you just need to multiply cross sides then divide by any number that works on both of them.

I hope that I answer your question.

Answer:

Step-by-step explanation:

Assume a solution

Assume that 60 is the length. The width is then 1/4 less, or 60 -60/4 = 45.

The diagonal of this rectangle is found using the Pythagorean theorem:

  d = √(60² +45²) = √5625 = 75

Make the adjustment

This is a factor of 75/40 larger than the actual diagonal, so the actual dimensions must be 40/75 = 8/15 times those we assumed.

  length = (8/15)×60 = 32

  width = (8/15)×45 = 24

The length and width of the rectangle are 32 and 24, respectively.

_____

Comment on this solution method

This method is suitable for problems where variables are linearly related. If we were concerned with the area, for example, instead of the diagonal, we would have to adjust the linear dimensions by the square root of the ratio of desired area to "false" area.

Final answer:

This detailed answer explains how to use the Babylonian method of false position to find the length and width of a rectangle.

Explanation:

The Babylonian method of false position involves making an initial assumption about the solution and then iteratively honing in on the correct answer.

Step-by-step:

Let's start with an assumption: Length = 60. Then calculate the width based on the given conditions.

Adjust your assumption based on the calculated width until you reach the correct solution.

By using this method, you can find the length and width of the rectangle described in the problem.

Answer: 1503

Step-by-step explanation:

Given : Significance level : [tex]\alpha:1-0.98=0.02[/tex]

Critical value : [tex]z_{\alpha/2}=2.326[/tex]

Margin of error : [tex]E=\pm0.03[/tex]

We know that the formula to find the sample size (the prior true proportion is not available) is given by :-

[tex]n=0.25(\dfrac{z_{\alpha/2}}{E})^2[/tex]

i.e. [tex]n=0.25(\dfrac{2.326}{0.03})^2=1502.85444444\approx1503[/tex]

Hence, the minimum sample size should be 1503.

Answer:

A simple random sample.

Step-by-step explanation:

A simple random sample is an statistical sample in which each member of a group has the same probability of being chosen. Since the researcher doesn't really have specific characteristics added to the sample other than being from public or private universities, this would be a simple random sample.

Answer: 200 mm

Step-by-step explanation:

The perimeter of rectangle is given by :-

[tex]P=2(l+w)[/tex], where l is length and w is width of the rectangle.

Given : Two sides of a rectangle are 4 cm in length. The other two sides are 6 cm in length.

The perimeter of the rectangle will be :_

[tex]P=2(4+6)=2(10)=20\ cm[/tex]

We know that 1 cm = 10 mm

Therefore,  perimeter of the rectangle = [tex]20\times10=200\ mm[/tex]

Answer: There are 4.8 grams in 100 mL of Drug A

Step-by-step explanation:

In order to determinate how many grams are in 100 mL of Drug A you can multiply and divide the expression of the concentration by 10 (To obtain 100 mL in the denominator)

Notice that the value remains unaltered.

(475 mg/10 mL )(10/10) = 4750 mg/ 100 mL

But the question is how many grams are in 100 mL, so you have to convert the value from mg to g.

The prefix m (mili) is equivalent to 0,001 so you can use 0.001 instead of the prefix

4750(0.001) g/ 100 mL

4.75 g/ 100 mL

The rounded result is 4.8 g/ 100 mL

Answer:

The basis for the null space of A is [tex]{\left[\begin{array}{c}-1&-1&1&0\end{array}\right],\left[\begin{array}{c}-1&1&0&1\end{array}\right]}[/tex]

Step-by-step explanation:

[tex]\left[\begin{array}{cccc}1&0&1&1\\3&2&5&1\\0&4&4&-4\end{array}\right][/tex]

[tex]\left[\begin{array}{cccc}1&0&1&1\\0&2&2&-2\\0&4&4&-4\end{array}\right][/tex]

[tex]\left[\begin{array}{cccc}1&0&1&1\\0&2&2&-2\\0&0&0&0\end{array}\right][/tex]

[tex]\left[\begin{array}{cccc}1&0&1&1\\0&1&1&-1\\0&0&0&0\end{array}\right][/tex]

    2. Convert the matrix equation back to an equivalent system and solve the matrix equation

[tex]1x_{1} +x_{3} +1x_{4}=0\\ 1x_{2} +x_{3} -1x_{4}=0\\0=0[/tex]

[tex]\left[\begin{array}{cccc}1&0&1&1\\0&1&1&-1\\0&0&0&0\end{array}\right] \left[\begin{array}{c}x_{1} &x_{2} &x_{3}&x_{4} \end{array}\right]=\left[\begin{array}{c}0&0&0\end{array}\right][/tex]

If we take [tex]x_{3}=t, x_{4}=s[/tex] then [tex]x_{1}=-s-t,x_{2}=s-t,x_{3}=t,x_{4}=s[/tex]

Therefore,

[tex]\boldsymbol{x}=\left[\begin{array}{c}-s-t&s-t&t&s\end{array}\right]=\left[\begin{array}{c}-1&-1&1&0\end{array}\right]t+\left[\begin{array}{c}-1&1&0&1\end{array}\right]s\\\boldsymbol{x}=\left[\begin{array}{c}-1&-1&1&0\end{array}\right]x_{3} +\left[\begin{array}{c}-1&1&0&1\end{array}\right]x_{4}[/tex]

The null space has a basis formed by the set {[tex]{\left[\begin{array}{c}-1&-1&1&0\end{array}\right],\left[\begin{array}{c}-1&1&0&1\end{array}\right]}[/tex]}

Answer:

4 rows

Step-by-step explanation:

It is given that two statements two be connected with AND, the statements may be either true or false.

The output of the AND will be true if both the statements will be true otherwise false.

We can construct the table as follows:

statement 1            statement 2         output

false                        false                        false

false                         true                         false

true                          false                        false

true                          true                          true

Hence, the number of rows the truth will have = 4

Answer:

[tex]\frac{25}{8}[/tex]

Step-by-step explanation:

A fraction is a part of a whole .

A proper fraction is a fraction whose numerator is less than denominator .

For example [tex]\frac{2}{3}\,,\,\frac{3}{4}[/tex]

An improper fraction is a fraction whose numerator is greater than denominator .

For example [tex]\frac{5}{4}\,,\,\frac{8}{7}[/tex]

A mixed fraction is made up of whole number and a proper fraction .

Given : [tex]4\frac{1}{6}\,,1\frac{1}{3}[/tex]

Solution :

[tex]4\frac{1}{6}=\frac{4\times 6+1}{6}=\frac{25}{6}\\1\frac{1}{3}=\frac{3\times 1+1}{3}=\frac{4}{3}[/tex]

We need to divide [tex]4\frac{1}{6}[/tex] by [tex]1\frac{1}{3}[/tex] .

[tex]4\frac{1}{6}\div 1\frac{1}{3}=\frac{25}{6}\div\frac{4}{3} \\\Rightarrow 4\frac{1}{6}\div 1\frac{1}{3}= \frac{25}{6}\times \frac{3}{4}=\frac{25}{8}[/tex]

Answer:

  see below for the working

Step-by-step explanation:

Dividing by a number is the same as multiplying by the inverse of that number.

[tex]\displaystyle\frac{\left(\frac{1}{4}\right)}{\left(-\frac{2}{3}\right)}=-\frac{1}{4}\cdot\frac{3}{2}=-\frac{3}{4\cdot 2}=-\frac{3}{8}[/tex]