This sequence has generating function
[tex]F(x)=\displaystyle\sum_{k\ge0}k^3x^k[/tex]
(if we include [tex]k=0[/tex] for a moment)
Recall that for [tex]|x|<1[/tex], we have
[tex]\displaystyle\frac1{1-x}=\sum_{k\ge0}x^k[/tex]
Take the derivative to get
[tex]\displaystyle\frac1{(1-x)^2}=\sum_{k\ge0}kx^{k-1}=\frac1x\sum_{k\ge0}kx^k[/tex]
[tex]\implies\dfrac x{(1-x)^2}=\displaystyle\sum_{k\ge0}kx^k[/tex]
Take the derivative again:
[tex]\displaystyle\frac{(1-x)^2+2x(1-x)}{(1-x)^4}=\sum_{k\ge0}k^2x^{k-1}=\frac1x\sum_{k\ge0}k^2x^k[/tex]
[tex]\implies\displaystyle\frac{x+x^2}{(1-x)^3}=\sum_{k\ge0}k^2x^k[/tex]
Take the derivative one more time:
[tex]\displaystyle\frac{(1+2x)(1-x)^3+3(x+x^2)(1-x)^2}{(1-x)^6}=\sum_{k\ge0}k^3x^{k-1}=\frac1x\sum_{k\ge0}k^3x^k[/tex]
[tex]\implies\displaystyle\frac{x+4x^3+x^3}{(1-x)^4}=\sum_{k\ge0}k^3x^k[/tex]
so we have
[tex]\boxed{F(x)=\dfrac{x+4x^3+x^3}{(1-x)^4}}[/tex]
Linear Algebra Question
The general equation for a circle is
a(x2+y2)+bx+cy+d=0.
There is exactly one circle passing through the points (2,3),(−3,2), and (0,0). Find an equation for this circle.
To find the specific circle passing through points (2,3), (-3,2), and (0,0), we started by substituting these coordinates in the general circle equation. This gave us a system of equations, which when solved, resulted in the coefficients for the circle equation. The equation of the circle we were seeking is (x^2 + y^2)/13 - (3x/32) + (15y/32) = 0.
Explanation:In order to find the equation for a specific circle we need to plugin the coordinates for (x, y) in the general equation a(x^2+ y^2) + bx + cy + d = 0 which represent three points on the circle. Let's start to solve this system using the points (2,3), (−3,2), and (0,0).
First, let's plug in (2,3) in general equation: 13a + 2b + 3c + d = 0Second, we will plugin (−3,2): 13a - 3b + 2c + d = 0Lastly, we will use the point (0,0): 0 = dNow, we have a system of three equations with three variables that we can solve. By subtracting the second from the first to eliminate d and simplifying, we obtain 5b + c = 0. Substituting d = 0 into the first and second equations, we get 13a + 2b + 3c = 0 and 13a - 3b + 2c = 0. Solving this system gives a = 1/13, b = -3/32 and c = 15/32.
Therefore, the equation of the circle is (x^2 + y^2)/13 - (3x/32) + (15y/32) = 0.
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Convert the following to rational numbers: (a) 0.12345, (b) 5.4321.
Answer:
A) [tex]\frac{2469}{20000}[/tex]
B) [tex]\frac{54321}{100000}[/tex]
Step-by-step explanation:
To answer this question first we define what are rational numbers.
A rational number is a number that can be expressed in the form of a fraction [tex]\frac{x}{y}[/tex], where x and y are integers and y ≠ 0.
a) 0.12345
This can be written in fraction form as
[tex]\frac{12345}{100000}[/tex]
Now, we simplify this fraction to lowest form. Simplified form of the above fraction is [tex]\frac{2469}{20000}[/tex]. This simplified form was obtained by dividing both numerator an denominator by 5.
b) This number can be written in fraction form as [tex]\frac{54321}{100000}[/tex]
Now, we simplify this fraction.
The simplified form of the above fraction is [tex]\frac{54321}{100000}[/tex] as the numerator and denominator have no common factor.
Suppose a certain computer virus can enter a system through an email or through a webpage. There is a 40% chance of receiving this virus through the email. There is a 35% chance of receiving it through the webpage. These are not mutually exclusive: the virus enters the system simultaneously by both email and webpage with a probability of 0.17. What is the probability that the virus does not enter the system at all? Enter your answer in decimal form
Answer:
P = 0.42
Step-by-step explanation:
This probability problem can be solved by building a Venn like diagram for each probability.
I say that we have two sets:
-Set A, that is the probability of receiving this virus through the email.
-Set B, that is the probability of receiving it through the webpage.
The most important information in these kind of problems is the intersection. That is, that he virus enters the system simultaneously by both email and webpage with a probability of 0.17. It means that [tex]A \cap B[/tex] = 0.17.
By email only
The problem states that there is a 40 chance of receiving it through the email. It means that we have the following equation:
[tex]A + (A \cap B) = 0.40[/tex]
[tex]A + 0.17 = 0.40[/tex]
[tex]A = 0.23[/tex]
where A is the probability that the system receives the virus just through the email.
The problem states that there is a 40% chance of receiving it through the email. 23% just through email and 17% by both the email and the webpage.
By webpage only
There is a 35% chance of receiving it through the webpage. With this information, we have the following equation:
[tex]B + (A \cap B) = 0.35[/tex]
[tex]B + 0.17 = 0.35[/tex]
[tex]B = 0.18[/tex]
where B is the probability that the system receives the virus just through the webpage.
The problem states that there is a 35% chance of receiving it through the webpage. 18% just through the webpage and 17% by both the email and the webpage.
What is the probability that the virus does not enter the system at all?
So, we have the following probabilities.
- The virus does not enter the system: P
- The virus enters the system just by email: 23% = 0.23
- The virus enters the system just by webpage: 18% = 0.18
- The virus enters the system both by email and by the webpage: 17% = 0.17.
The sum of the probabilities is 100% = 1. So:
P + 0.23 + 0.18 + 0.17 = 1
P = 1 - 0.58
P = 0.42
There is a probability of 42% that the virus does not enter the system at all.
Set P = {x, 4, 6, 10, -2, x, 6, 1, x} In set P, if the value of x is 4 greater than the largest numerical value present, what is the range of the set?
Answer:
16.
Step-by-step explanation:
The largest numerical value present in the set P is 10, then x is 4+10 = 14. Then the set P is
P = {14, 4, 6, 10, -2, 14, 6, 1, 14}.
Now, the range is the difference between the greatest and the smallest element in the set. The greatest number is 14 and the smallest is -2, then the range is
14-(-2) = 14+2 = 16.
Then the range of the set P is 16.
Convert 375 minutes to hours?
Answer:
6.25 hours
Step-by-step explanation:
As we know that,
1 hour = 60 minutes
⇒ [tex]1 \ minute = \frac{1}{60} \ hour[/tex]
⇒ [tex]375 \ minutes =375\times \frac{1}{60} \ hours[/tex]
⇒ 375 mintes = 6.25 hours
Thus, 375 mintes = 6.25 hours
A survey was conducted among 78 patients admitted to a hospital cardiac unit during a two-week period. The data of the survey are shown below. Let B equals the set of patients with high blood pressure. Let C equals the set of patients with high cholesterol levels. Let S equals the set of patients who smoke cigarettes.n(B) equals 36 n(B intersect S) equals 10 n(C) equals 34 n(B intersect C) equals 12 n(S) equals 30 n(B intersect C intersectS) equals 5 n[(B intersect C) union (B intersect S) union (C intersect S)] equals 21
Sets and set operations are ways of organizing, classifying and obtaining information about objects according to the characteristics they possess, as objects generally have several characteristics, the same object can belong to several sets, an example is the subjects of a school , where students (objects) are classified according to the subject they study (set).
The intersection of sets is a new set consisting of those objects that simultaneously possess the characteristics of each intersected set, the intersection of two subjects will be those students who have both subjects enrolled.
The union of sets is a new set consisting of all the objects belonging to the united sets, the union of two subjects will be all students of both courses.
In this case there are three sets B, C and S of which we are given the following information:
Answer
n(BꓵSꓵC)=5
n(BꓵS)=10 – 5 = 5
n(BꓵC)=12 – 5 = 7
n[(BꓵC)ꓴ(BꓵS)ꓴ(CꓵS)]=21 – 5 – 5 – 7 = 4
n(B)=36 – 5 – 5 – 7 = 19
n(S)=30 – 5 – 5 – 4 = 16
n(C)=34 – 5 – 7 – 4 = 18
Estimate the product. Round the first factor to the nearest whole number, round the second factor to the nearest hundred, and then multiply.
5 4/7 * 598
The product is approximately blank
Answer:
The product is approximately 3600.
Step-by-step explanation:
The whole number nearest the first factor is 6.
The hundred nearest the second factor is 600.
The product of these is 6×600 = 3600, your estimated product.
in the following ordinary annuity interest is compounded with each payment and the payment is made at the end of the compounding period. find the accumulated amount of the annuity. 4,500 annually at 6% for 10 years
Answer: $59313.58
Step-by-step explanation:
Formula to find the accumulated amount of the annuity is given by :-
[tex]FV=A(\frac{(1+\frac{r}{m})^{mt})-1}{\frac{r}{m}})[/tex]
, where A is the annuity payment deposit, r is annual interest rate , t is time in years and m is number of periods.
Given : m= $2000 ; m= 1 [∵ its annual] ; t= 10 years ; r= 0.06
Now substitute all these value in the formula , we get
[tex]FV=(4500)(\frac{(1+\frac{0.06}{1})^{1\times10})-1}{\frac{0.06}{1}})[/tex]
⇒ [tex]FV=(4500)(\frac{(1.06)^{10})-1}{0.06})[/tex]
⇒ [tex]FV=(4500)(\frac{0.79084769654}{0.06})[/tex]
⇒ [tex]FV=(4500)(13.1807949423)[/tex]
⇒ [tex]FV=59313.5772407\approx59313.58 \ \ \text{ [Rounded to the nearest cent]}[/tex]
Hence, the accumulated amount of the annuity= $59313.58
0 / 2 pts 5000 kg = _______short tons Round UP to nearest 100th
Answer:
5000kg = 5.5 short tons
Step-by-step explanation:
This can be solved as a rule of three problem.
In a rule of three problem, the first step is identifying the measures and how they are related, if their relationship is direct of inverse.
When the relationship between the measures is direct, as the value of one measure increases, the value of the other measure is going to increase too.
When the relationship between the measures is inverse, as the value of one measure increases, the value of the other measure will decrease.
Unit conversion problems, like this one, is an example of a direct relationship between measures.
Each kg has 0.0011 short tons. How many short tons are there in 5000 kg. So?
1kg = 0.0011 short tons
5000kg - x short tons
[tex]x = 5000*0.0011[/tex]
[tex]x = 5.5[/tex] short tons.
5000kg = 5.5 short tons
To convert 5000 kg to short tons, multiply by the conversion factor and round up.
Explanation:The conversion factor between kilograms and short tons is 0.00110231. To convert 5000 kg to short tons, we multiply by the conversion factor:
5000 kg * 0.00110231 = 5.51155 short tons
Rounding up to the nearest 100th gives us 5.52 short tons.
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dy/dx = y/x , y(1) = −2
Answer with Step-by-step explanation:
The given differential equation is variable separable in nature and hence will be solved accordingly as follows:
[tex]\frac{dy}{dx}=\frac{y}{x}\\\\=\frac{dy}{y}=\frac{dx}{x}\\\\\int \frac{dy}{y}=\int \frac{dx}{x}\\\\ln(y)=ln(x)+ln(c)\\\\ln(y)=ln(cx)\\\\(\because ln(ab)=ln(a)+ln(b))\\\\\therefore y=cx[/tex]
where 'c' is constant of integration whose value shall be obtained using the given condition [tex]y(1)=-2[/tex]
Thus we have
[tex]-2=c\times 1\\\\\therefore c=-2\\[/tex]
Thus solution becomes
[tex]y=-2x[/tex]
Matrices A and B are square matrices of the same size. Prove Tr(c(A + B)) = C (Tr(A) + Tr(B)).
Answer with Step-by-step explanation:
We are given that two matrices A and B are square matrices of the same size.
We have to prove that
Tr(C(A+B)=C(Tr(A)+Tr(B))
Where C is constant
We know that tr A=Sum of diagonal elements of A
Therefore,
Tr(A)=Sum of diagonal elements of A
Tr(B)=Sum of diagonal elements of B
C(Tr(A))=[tex]C\cdot[/tex] Sum of diagonal elements of A
C(Tr(B))=[tex]C\cdot[/tex] Sum of diagonal elements of B
[tex]C(A+B)=C\cdot (A+B)[/tex]
Tr(C(A+B)=Sum of diagonal elements of (C(A+B))
Suppose ,A=[tex]\left[\begin{array}{ccc}1&0\\1&1\end{array}\right][/tex]
B=[tex]\left[\begin{array}{ccc}1&1\\1&1\end{array}\right][/tex]
Tr(A)=1+1=2
Tr(B)=1+1=2
C(Tr(A)+Tr(B))=C(2+2)=4C
A+B=[tex]\left[\begin{array}{ccc}1&0\\1&1\end{array}\right]+\left[\begin{array}{ccc}1&1\\1&1\end{array}\right][/tex]
A+B=[tex]\left[\begin{array}{ccc}2&1\\2&2\end{array}\right][/tex]
C(A+B)=[tex]\left[\begin{array}{ccc}2C&C\\2C&2C\end{array}\right][/tex]
Tr(C(A+B))=2C+2C=4C
Hence, Tr(C(A+B)=C(Tr(A)+Tr(B))
Hence, proved.
mathematical induction of 3k-1 ≥ 4k ( 3k = k power of 3 )
For making mathematical induction, we need:
a base caseAn [tex]n_0[/tex] for which the relation holds true
the induction stepif its true for [tex]n_i[/tex], then, is true for [tex]n_{i+1}[/tex]
base casethe relationship is not true for 1 or 2
[tex]1^3-1 = 0 < 4*1[/tex]
[tex]2^3-1 = 8 -1 = 7 < 4*2 = 8[/tex]
but, is true for 3
[tex]3^3-1 = 27 -1 = 26 > 4*3 = 12[/tex]
induction steplets say that the relationship is true for n, this is
[tex]n^3 -1 \ge 4 n[/tex]
lets add 4 on each side, this is
[tex]n^3 -1 + 4 \ge 4 n + 4[/tex]
[tex]n^3 + 3 \ge 4 (n + 1)[/tex]
now
[tex](n+1)^3 = n^3 +3 n^2 + 3 n + 1[/tex]
[tex](n+1)^3 \ge n^3 + 3 n [/tex]
if [tex]n \ge 1[/tex] then [tex]3 n \ge 3[/tex] , so
[tex](n+1)^3 \ge n^3 + 3 n \ge n^3 + 3 [/tex]
[tex](n+1)^3 \ge n^3 + 3 \ge 4 (n + 1) [/tex]
[tex](n+1)^3 \ge 4 (n + 1) [/tex]
and this is what we were looking for!
So, for any natural equal or greater than 3, the relationship is true.
A frequency polygon having a peak in the center of its distribution is:
symmetric
skewed
kurtosis
a histogram
Which of the following is a true statement regarding the mode?
The mean, median, and mode may be the same value.
The mode is equal to the second quartile, Q2.
A dataset cannot have more than one mode.
The mode is the tallest bar in a histogram.
A measure of spread which magnifies the effect of values far from the mean is the:
median
interquartile range
variance
mode
In Chap4 Practice Exercise Prob #5, the histogram interval width is:
20 units
400
6 classes
Answer:
(a) symmetric
(b) The mean, median, and mode may be the same value.
and, The mode is the tallest bar in a histogram.
(c) variance
Step-by-step explanation:
Data is said to be symmetric if it is not skewed. i.e. if it has peaked at the center of the distribution.
Skewness gives us an idea about the shape of the curve.
Kurtosis enables us to have an idea about the flatness and peakedness of the frequency.
Mode represents the highest repetition of the observation.
The mean, median, and mode can have the same value.
The histogram represents the frequency of observation by the length of the bar this is the same as Mode. Thus, the mode is the tallest bar in a histogram.
Variance is the square of standard deviation and it is defined as, the sum of the square of the distance of an observation from the mean.
So, Variance is the correct option.
The square matrix A is called orthogonal provided
thatAT=A-1. Show that the determinant of such
amatrix must be either +1 or -1.
Answer: The proof is done below.
Step-by-step explanation: Given that the square matrix A is called orthogonal provided that [tex]A^T=A^{-1}.[/tex]
We are to show that the determinant of such a matrix is either +1 or -1.
We will be using the following result :
[tex]|A^{-1}|=\dfrac{1}{|A|}.[/tex]
Given that, for matrix A,
[tex]A^T=A^{-1}.[/tex]
Taking determinant of the matrices on both sides of the above equation, we get
[tex]|A^T|=|A^{-1}|\\\\\Rightarrow |A|=\dfrac{1}{|A|}~~~~~~~~~~~~~~~~~~~~[\textup{since A and its transpose have same determinant}]\\\\\\\Rightarrow |A|^2=1\\\\\Rightarrow |A|=\pm1~~~~~~~~~~~~~~~~~~~~[\textup{taking square root on both sides}][/tex]
Thus, the determinant of matrix A is either +1 or -1.
Hence showed.
determine whether the set is closed for the given operation.
explain your reasoning. -- -the set of integers for division
-the set of positive integers for subtraction.
Closure Property for the set of integers for Subtraction
Take two integers,
A=6
B=9
A-B
=6-9
= -3
Which is also an Integer.Hence Closure Property is satisfied by Integers with respect to Subtraction.
⇒Closure Property for the set of integers for Division
Take two integers,
A=6
B=9
[tex]\rightarrow \frac{A}{B}\\\\=\frac{6}{9}\\\\=\frac{2}{3}-----\text{Not an Integer}[/tex]
Which is not an Integer.Hence Closure Property is not satisfied by Integers with respect to Division.
75 is ___% of 725.50
Answer:
75 is 10.34% of 725.50.
Step-by-step explanation:
Let 75 is x % of 725.50.
It can be written as mathematical equation.
x % of 725.50 = 15
[tex]\frac{x}{100}\times 725.50=75[/tex]
Multiply both sides by 100.
[tex]\frac{x}{100}\times 725.50\times 100=75\times 100[/tex]
[tex]x\times 725.50=7500[/tex]
Divide both sides by 725.50.
[tex]\frac{x\times 725.50}{725.50}=\frac{7500}{725.50}[/tex]
[tex]x=\frac{7500}{725.50}[/tex]
[tex]x=10.3376981392[/tex]
[tex]x\approx 10.34[/tex]
Therefore, 75 is 10.34% of 725.50.
Eli walked 12 feet down the hall of his house to get to the door. He conti in a straight line out the door and across the yard to the mailbox, a distan 32 feet. He came straight back across the yard 14 feet and stopped to pet hil dog. a. Draw a diagram of Eli's walking pattern. b. How far has he walked? c. How far from the house is he now?
Answer:
Eli walked 12 feet down the hall of his house to get to the door. He continued in a straight line out the door and across the yard to the mailbox, a distance of 32 feet.
He came straight back across the yard 14 feet and stopped to pet his dog.
Part A : find the image attached.
Part B : [tex]12+32+14=58[/tex] feet
Part C : [tex]32-14=18[/tex] feet from his house.
Assume that for a particular tire brand, the probability of wearing out before 30,000 miles is 0.25. For someone who buys a set of iour of these tires, what is the probability that all will last at least 30,000 miles? You be a suitable model in this case.
Answer:
The probability that all will last at least 30,000 miles is 0.3164.
Step-by-step explanation:
Consider the provided information.
For a particular tire brand, the probability of wearing out before 30,000 miles is 0.25. Someone who buys a set of four of these tires,
That means the probability of not wearing out is 1-0.25 = 0.75
Now use the binomial distribution formula.
[tex]^nC_r (p^{n-r})(q^r)[/tex]
Substitute p = 0.25, q=0.75, r=4 and n=4
[tex]^4C_4 (0.25)^{4-4}(0.75)^4[/tex]
[tex]1 \times 0.75^4[/tex]
[tex]0.3164[/tex]
The probability that all will last at least 30,000 miles is 0.3164.
If the probability of losing is 7/10, what is the probability of winning?
1/10
7/10
3/10
3/3
Answer:
3/10
Step-by-step explanation:
If the probability of losing is 7/10, then the probability of winning is 3/10. To find out the probability of winning we just have to subtract the probability of losing from the total which is 10:
10/10 = total probability
7/10 = probability of losing
Probability of winning is 3/10.
A company that manufactures small canoes has a fixed cost of $24,000. It costs $100 to prod canoes produced and sold.) a. Write the cost function. C(x) = 24,000 + 100x (Type an expression using x as the variable.) b. Write the revenue function. R(x) = 200x (Type an expression using x as the variable.) c. Determine the break-even point. (Type an ordered pair. Do not use commas in large numbers.)
Answer:
240
Step-by-step explanation:
Given:
The cost function for canoes, c(x) = 24,000 + 100x
Revenue function for canoes, R(x) = 200x
here, x is the number of canoes sold
Fixed cost = $24,000
Cost of production = $100
now at break even point, Revenue = cost
thus,
200x = 24,000 + 100x
or
100x = 24,000
or
x = 240 canoes
Hence, at breakeven, the number of canoes sold is 240
The differential equation xy' = y(in x – Iny) is neither separable, nor linear. By making the substitution y(x) = xv(x), show that the new equation for v(x) equation is separable. N.B. you do not have to actually solve the ODE.
Answer:
We can place everything with v(x) on one side of the equality, everything with x on the other side. This is done on the step-by-step explanation, and shows that the new equation is separable.
Step-by-step explanation:
We have the following differential equation:
[tex]xy' = y(ln x - ln y)[/tex]
We are going to apply the following substitution:
[tex]y = xv(x)[/tex]
The derivative of y is the derivative of a product of two functions, so
[tex]y' = (x)'v(x) + x(v(x))'[/tex]
[tex]y' = v(x) + xv'(x)[/tex]
Replacing in the differential equation, we have
[tex]xy' = y(ln x - ln y)[/tex]
[tex]x(v(x) + xv'(x)) = xv(x)(ln x - ln xv(x))[/tex]
Simplifying by x:
[tex]v(x) + xv'(x) = v(x)(ln x - ln xv(x))[/tex]
[tex]xv'(x) = v(x)(ln x - ln xv(x)) - v(x)[/tex]
[tex]xv'(x) = v(x)((ln x - ln xv(x) - 1)[/tex]
Here, we have to apply the following ln property:
[tex]ln a - ln b = ln \frac{a}{b}[/tex]
So
[tex]xv'(x) = v(x)((ln \frac{x}{xv(x)} - 1)[/tex]
Simplifying by x,we have:
[tex]xv'(x) = v(x)((ln \frac{1}{v(x)} - 1)[/tex]
Now, we can apply the above ln property in the other way:
[tex]xv'(x) = v(x)(ln 1 - ln v(x) -1)[/tex]
But [tex]ln 1 = 0[/tex]
So:
[tex]xv'(x) = v(x)(- ln v(x) -1)[/tex]
We can place everything that has v on one side of the equality, everything that has x on the other side, so:
[tex]\frac{v'(x)}{v(x)(- ln v(x) -1)} = \frac{1}{x}[/tex]
This means that the equation is separable.
Consider the following statement: The square of a prime number is not prime. (a) Write this as an if-then statement, using careful mathematical language and notation. (b) Prove or disprove the statement.
The square of a prime number is not prime.
a) let x ∈ R, If x ∈ {prime numbers}, then [tex]x^{2}[/tex]∉{prime numbers}
there says that if x is a real and x is in the set of the prime numbers, then the square of x isn't in the set of prime numbers.
b) Prove or disprove the statement.
ok, if x is a prime number, then x only can be divided by himself. Now is easy to see that [tex]x^{2}[/tex] = x*x can be divided by himself and x, then x*x is not a prime number, because can be divided by another number different than himself
Final answer:
Rephrased as an if-then statement, 'If p is a prime number, then p² is not prime.' It is proved through contradiction that the square of a prime number cannot be prime, as it would have a factor other than 1 and itself.
Explanation:
The question asks us to consider the statement: "The square of a prime number is not prime." Let's tackle this in two parts:
(a) If-Then Statement
An if-then statement using mathematical language and notation for the given statement might look like this: "If p is a prime number, then p² is not a prime number." Written symbolically, if we let p represent a prime number, then we can express the statement as: "If p is a prime, then p² is not prime," or more formally, "∀p ∈ Primes, (¬ Prime(p²))."
(b) Proof
We can prove this statement by contradiction. Assume the contrary: that there exists a prime number p such that p² is also prime. Since p is prime, p ≥ 2. So p² will be greater than p and have p as a factor, which contradicts the definition of a prime number (a prime number has no positive divisors other than 1 and itself). Hence, the square of a prime number cannot be prime, confirming the original statement.
Suppose that A, B, and C are invertible matrices of the same
size,Show that the product ABC is invertible and that
(ABC)-1= C-1B-1A-1.
Answer:
Step-by-step explanation:
Given that t A, B, and C are invertible matrices of the same
size.
To pr that [tex](ABC)^{-1} = C^{-1}B^{-1}A^{-1}.[/tex]
to pr that [tex]( ABC) (C^{-1}B^{-1}A^{-1})=e[/tex]
LS=[tex]( AB(C (C^{-1})B^{-1}A^{-1})\\=ABIB^{-1}A^{-1})\\=A(BB^{-1})A^{-1})\\=AA^{-1})\\=I[/tex]
Thus proved
To show that the product of invertible matrices A, B, and C (ABC) is invertible and that its inverse is (ABC)-1 = C-1B-1A-1, we use the property that the inverse of a product is the product of inverses in reverse order, and the associativity of matrix multiplication.
Explanation:A student asked to show that if A, B, and C are invertible matrices of the same size, the product ABC is invertible, and that the inverse of this product is (ABC)-1 = C-1B-1A-1.
To prove this, we can use the properties of matrix multiplication and inverses:
First, we confirm that the product of two invertible matrices is itself invertible, and the inverse is given by the product of their inverses in reverse order. This is due to the property that (AB)-1 = B-1A-1.Using associativity, we have the combined product (AB)C. Since AB is invertible, by the first point, we can consider (AB)C as a product of two invertible matrices, (AB) and C.Therefore, ((AB)C)-1 = C-1(AB)-1 = C-1B-1A-1, as required.This completes the proof that the product ABC is invertible and its inverse is C-1B-1A-1
If the earth travels around the sun one time each year and Jupiter travels around the sun one time every 12 years and they met at a point in 2002 when would they meet at the point again?
Answer:
They will meet again in 2014
Step-by-step explanation:
Once every year the Earth will go back through the point where it met with Jupiter. Let's write those years:
Earth = {2002, 2003, 2004, 2005, 2006, 2007, 2008, 2009, 2010, 2011, 2012, 2013, 2014, 2015, 2016...}
Once every 12 years Jupiter will go back through the point where it met with the Earth. Let's write it down:
Jupiter= {2002, 2002+12=2014, 2014+12=2026, 2026+12=2038...}
The Earth and Jupiter will meet again for the first time (after 2002) the year they both go trought that point.
Earth = {2002, 2003, 2004, 2005, 2006, 2007, 2008, 2009, 2010, 2011, 2012, 2013, 2014, 2015, 2016...}
Jupiter= {2002, 2014, 2026, 2038...}
As we can see, 2014 is the first year the'll meet again.
Write a short statement that expresses a possible relationship between the variables. (size of file, time to download the file)
Answer:
So you may know that that the time for downloading a file of A (MB) is B (seconds), where A and B are numbers.
Then suppose if some file size is C, which is 10 times more than A, and you assume that the download speed sees this as downloading 10 times the A (MB) file. Thus the time for downloading this second file will be 10*B (seconds).
As, if [tex]C\ (MB) file=10*A\ (MB) file[/tex]
then [tex]C\ (MB)file =10*B\ (seconds)[/tex]
B(seconds) is the time required to download A (MB) file.
If you roll one die and flip one coin, does the occurrence of a particular outcome on the die affect the probability of a particular outcome on the coin? Why or why not?
Answer:
No, because these events are independent.
Step-by-step explanation:
When we roll one die (a fair die), we will have outcomes in the numbers from 1 to 6.
And when we flip a coin, we will have either a heads or a tails.
Now these two events are independent events as the occurrence of any result in one event is not affecting the result of second event,
Hence, these two are independent events. So, the answer is no, the occurrence of a particular outcome on the die will not affect the probability of a particular outcome on the coin.
Final answer:
The outcomes of a die roll and a coin toss are independent events, meaning the result of the die does not affect the result of the coin toss.
Explanation:
The occurrence of a particular outcome on the die does not affect the probability of a particular outcome on the coin because each event is independent of the other.
For example, if you roll a die and flip a coin, the result of the die does not influence the result of the coin toss as they are separate events.
Additionally, the concept of independence in probability states that the outcomes of the coin toss and die roll do not impact each other.
Sketch the Cartesian product on the x-y plane R^2: Zx Z.
Answer:
[tex]\mathbb{Z}\times \mathbb{Z}=\{(a,b)\lvert a, b\in \mathbb{Z}\}[/tex]
Step-by-step explanation:
In general, the Cartesian product of two sets [tex]A,B[/tex] is a new set defined by
[tex]A\times B=\{(a,b)\lvert a\in A,b\in B\}[/tex]
The pair [tex](a,b)[/tex] is ordered pair because the order is important, that is to say, in general [tex](a,b)\neq (b,a)[/tex].
One of the most important Cartesian products in mathematics is [tex]\mathbb{R}\times \mathbb{R}=\{(x,y) \lvert x,y \in \mathbb{R}\}[/tex] which is precisely the Cartesian Plane xy. The set [tex]\mathbb{Z}\times \mathbb{Z}[/tex] is a subset of [tex]\mathbb{R}\times \mathbb{R}[/tex] which is the set of all the points in the Cartesian plane whose coordinates are integers numbers. So, sketching the set [tex]\mathbb{Z}\times \mathbb{Z}[/tex] we have a picture as the shown below.
Baytown Village Stone Creations is making a custom stone bench. the recommended height for the bench is 18 in. the depth of the stone bench is 3 3/8 in. Each of the two supporting legs is made up of three stacked stones. Two of the stones measure 3 1/2in. and 5 1/4 in. how much must the third stone measure?
Answer:
x = tex]5\frac{\textup{7}}{\textup{8}}\ in[/tex]
Step-by-step explanation:
Given:
Total height of the bench = 18 in
Depth of the stone bench = [tex]3\frac{\textup{3}}{\textup{8}}\ in[/tex] = [tex]\frac{\textup{27}}{\textup{8}}\ in[/tex] = 3.375 in
Measure of stones = [tex]3\frac{\textup{1}}{\textup{2}}\ in[/tex] = [tex]\frac{\textup{7}}{\textup{2}}\ in[/tex] = 3.5 in
measure of another stone = [tex]5\frac{\textup{1}}{\textup{4}}\ in[/tex] = [tex]\frac{\textup{21}}{\textup{4}}\ in[/tex] = 5.25 in
let the height of the third stone be 'x'
Now,
The total height of the bench = depth of bench + Measure of two stones + x
18 = 3.375 + 3.5 + 5.25 + x
or
x = 18 - 12.125
or
x = 5.875 in
or
x = tex]5\frac{\textup{7}}{\textup{8}}\ in[/tex]
Determine,in each of the following cases, whether the described system is or not a group. Explain your answers. Determine what is an Abelian group.
a) G = {set of integers , a ∗ b = a − b}
b) G = {set of matrices of size 2 × 2, A ∗ B = A · B}
c) G = {a0, a1, a2, a3, a4ai ∗aj = aIi+jI , if i+j < 5, ai ∗aj = aIi+j-5I , if i+j ≥ 5}
Answer:
(a) Not a group
(b) Not a group
(c) Abelian group
Step-by-step explanation:
In order for a system <G,*> to be a group, the following must be satisfied
(1) The binary operation is associative, i.e., (a*b)*c = a*(b*c) for all a,b,c in G
(2) There is an identity element, i.e., there is an element e such that a*e = e*a = a for all a in G
(3) For each a in G, there is an inverse, i.e, another element a' in G such that a*a' = a'*a = e (the identity)
If in addition the operation * is commutative (a*b = b*a for every a,b in G), then the group is said to be Abelian
(a)
The system <G,*> is not a group since there are no identity.
To see this, suppose there is an element e such that
a*e = a
then
a-e = a which implies e=0
It is easy to see that 0 cannot be an identity.
For example
2*0 = 2-0 = 2
Whereas
0*2 = 0-2 = -2
So 2*0 is not equal to 0*2
(b)
The system <G,*> is not a group either.
If A is a matrix 2x2 and the determinant of A det(A)=0, then the inverse of A does not exist.
(c)
The table of the operation G is showed in the attachment.
It is evident that this system is isomorphic under the identity map, to the cyclic group
[tex]\mathbb{Z}_{5}[/tex]
the system formed by the subset of Z, {0,1,2,3,4} with the operation of addition module 5, which is an Abelian cyclic group
We conclude that the system <G,*> is Abelian.
Attachment: Table for the operation * in (c)
A ball with mass m kg is thrown upward with initial velocity 28 m/s from the roof of a building 17 m high. Neglect air resistance Use g = 9.8 m/s. Round your answers to one decimal place. (a) Find the maximum height above the ground that the ball reaches. meters (b) Assuming that the ball misses the building on the way down, find the time that it hits the ground. Fend Click If you would like to Show Work for this question: Open Show Work LINK TO TEXT
Answer:
The ball will take 6.3 seconds to reach the maximum height and hit the ground.
Step-by-step explanation:
a). When a ball was thrown upwards with an initial velocity u then maximum height achieved h will be represented by the equation
v² = u² - 2gh
where v = final velocity at the maximum height h
and g = gravitational force
Now we plug in the values in the equation
At maximum height final velocity v = 0
0 = (28)² - 2×(9.8)h
19.6h = (28)²
h = [tex]\frac{(28)^{2}}{19.6}[/tex]
= [tex]\frac{784}{19.6}[/tex]
= 40 meter
B). If the ball misses the building and hits the ground then we have to find the time after which the ball hits the ground that will be
= Time to reach the maximum height + time to hit the ground from the maximum height
Time taken by the ball to reach the maximum height.
Equation to find the time will be v = u - gt
Now we plug in the values in the equation
0 = 28 - 9.8t
t = [tex]\frac{28}{9.8}[/tex]
= 2.86 seconds
Now time taken by the ball to hit the ground from its maximum height.
H = ut + [tex]\frac{1}{2}\times g\times (t)^{2}[/tex]
(17 + 40) = 0 + [tex]\frac{1}{2}\times g\times (t)^{2}[/tex]
57 = 4.9(t)²
t² = [tex]\frac{57}{4.9}[/tex]
t² = 11.63
t = √(11.63)
= 3.41 seconds
Now total time taken by the ball = 2.86 + 3.41
= 6.27 seconds
≈ 6.3 seconds
Therefore, the ball will take 6.3 seconds to reach the maximum height and hit the ground.