A new building that costs $1,000,000 has a useful life of 25 years and a scrap value of $600,000. Using straight-line depreciation, find the equation for the value V in terms of t, where t is in years. (Make sure you use t and not x in your answer.)
V(t) =



Find the value after 1 year, after 2 years, and after 20 years.

Value after 1 year $
Value after 2 years $
Value after 20 years $

Answer:

Thousandths: 7

Ten-thousandths: 4

Tenths: 3

Step-by-step explanation:

The given number is 59.3274659

The place values of a numner is given in the followwing format:

Tens  Ones.Tenths  Hundredths  Thousandths Tenthousandths  and so on

We can fit our number

Tens  Ones.Tenths  Hundredths  Thousandths Tenthousandths HT M

  5          9 .     3                 2                     7                       4        

Thousandths: 7

Ten-thousandths: 4

Tenths: 3

Answer:

A bi-conditional statement represents the idea of "if and only if." Its symbol is a [tex]\leftrightarrow[/tex].

Step-by-step explanation:

We have been given an incomplete sentence. We are supposed to fill in the given blank.

A _______ represents the idea of "if and only if." Its symbol is a [tex]\leftrightarrow[/tex].

We know that "if and only if" stands for bi-conditional statement, which represents either both statements are true or both are false.

The symbol [tex]\leftrightarrow[/tex] represent a bi-conditional statement.

Therefore, our complete statement would be: A bi-conditional statement represents the idea of "if and only if." Its symbol is a [tex]\leftrightarrow[/tex].

A bi-conditional statement in mathematics signifies 'if and only if' situation. Represented by the symbol ↔, it means both conditions in the statement must be true concurrently.

In mathematics, a bi-conditional statement represents the concept of 'if and only if'. Its symbol is ↔ which is a double arrow or a left right arrow. This statement essentially means that both the conditions present need to be true for the entire statement to be true. For example, in the statement 'x is even if and only if x is divisible by 2', both parts (x being even and x being divisible by 2) need to be true together.

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Let [tex]q[/tex] be the unknown amount of the chemical originally in the pond, so [tex]Q(0)=q[/tex].

a. The incoming water introduces the chemical at a rate of

[tex]Q'_{\rm in}=\left(0.1\dfrac{\rm kg}{\rm L}\right)\left(4\dfrac{\rm L}{\rm hr}\right)=\dfrac25\dfrac{\rm kg}{\rm hr}[/tex]

and the mixture flows out at a rate of

[tex]Q'_{\rm out}=\left(\dfrac Q{2760}\dfrac{\rm kg}{\rm L}\right)\left(4\dfrac{\rm L}{\rm hr}\right)=\dfrac Q{690}\dfrac{\rm kg}{\rm hr}[/tex]

so that the net rate of change (in kg/hr) of the chemical in the pond is given by the differential equation,

[tex]\boxed{Q'=\dfrac25-\dfrac Q{690}}[/tex]

b. The ODE is linear; multiplying both sides by [tex]e^{t/690}[/tex] gives

[tex]e^{t/690}Q'+\dfrac{e^{t/690}}{690}Q=\dfrac{2e^{t/690}}5[/tex]

Condense the left side into the derivative of a product:

[tex]\left(e^{t/690}Q\right)'=\dfrac{2e^{t/690}}5[/tex]

Integrate both sides to get

[tex]e^{t/690}Q=276e^{t/690}+C[/tex]

and solve for [tex]Q[/tex] to get

[tex]Q=276+Ce^{-t/690}[/tex]

The pond starts with [tex]q[/tex] kg of the chemical, so when [tex]t=0[/tex] we have

[tex]q=276+C\implies C=q-276[/tex]

so that the amount of chemical in the water at time [tex]t[/tex] is

[tex]Q(t)=276+(q-276)e^{-t/690}[/tex]

As [tex]t\to\infty[/tex], the exponential term will converge to 0, leaving a fixed amount of 276 kg of the chemical in the pond.

The differential equation for the amount of chemical in the pond is dQ/dt = 0.01 kg/L * 4 L/h. After a long time, the amount of chemical in the pond is Q = 0.04 kg/h * t.

(a)

To write a differential equation for the amount of chemical in the pond at any time, we need to consider the rate of change of the chemical in the pond. The chemical flows into the pond at a rate of 0.01 kg/L and flows out at the same rate, so the rate of change of the chemical Q(t) in the pond is 0.01 kg/L multiplied by the rate of change of the volume of water in the pond, which is 4 L/h. Therefore, the differential equation for the amount of chemical in the pond is:

dQ/dt = 0.01 kg/L * 4 L/h

(b)

To determine how much chemical will be in the pond after a long time, we can solve the differential equation.

We can rewrite the differential equation as:

dQ = 0.01 kg/L * 4 L/h * dt

Integrating both sides:

∫dQ = ∫0.01 kg/L * 4 L/h dt

Q = 0.04 kg/h * t + C

Where C is a constant of integration. Given that the amount of chemical in the pond is initially 0 (since the pond starts with only pure water), we can substitute Q = 0 and solve for C:

0 = 0.04 kg/h * 0 + C

C = 0

Therefore, the amount of chemical in the pond after a long time is given by:

Q = 0.04 kg/h * t

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Answer:

5

Step-by-step explanation:

We have to see the pattern, we start with -8, the second number is -2, the distance between -8 and -2 is 6 units. Now from -2 to 2 the distance is 4 units. From 2 to 4 distance is 2 units. Then we can conclude that with each step the distance is divided by 2, then the next number is 5 because the distance between 4 and 5 is 1 unit.

Answer: Fraction of all eligible voters voted for Shelly is [tex]\dfrac{3}{20}[/tex] .

Morgan received the most votes .

Step-by-step explanation:

Given : The fraction of all eligible voters cast votes =[tex]\dfrac{1}{2}[/tex]

The fraction of votes received by Shelly = [tex]\dfrac{3}{10}[/tex]        (1)

Now, the fraction of all eligible voters voted for Shelly is given by :_

[tex]\dfrac{3}{10}\times\dfrac{1}{2}=\dfrac{3}{20}[/tex]

Thus, the fraction of all eligible voters voted for Shelly is [tex]\dfrac{3}{20}[/tex] .

The fraction of vote received by Morgan= [tex]\dfrac{5}{8}[/tex]          (2)

To compare the fractions given in (1) and (2), we need to find least common multiple of 10 and 8 .

LCM (10, 8)=40

Now, make denominator 40 (to make equivalent fraction)  in (1), (2) we get

Fraction of all eligible voters voted for Shelly = [tex]\dfrac{3\times4}{10\times4}=\dfrac{12}{40}[/tex]

Fraction of all eligible voters voted for Morgan =[tex]\dfrac{5\times5}{8\times5}=\dfrac{25}{40}[/tex]

Since, [tex]\dfrac{25}{40}>\dfrac{12}{40}[/tex]  [By comparing numerators]

Therefore, Morgan received the most votes .

Answer: The approximate absentee rate that day would be 8.09%.

Step-by-step explanation:

Since we have given that

Number of students who were absent = 36

Total number of  students = 445

We need to find the approximate absentee rate that day :

Rate of absentee of that day would be

[tex]\dfrac{\text{Number of absentee}}{\text{Total number of students}}\times 100\\\\=\dfrac{36}{445}\times 100\\\\=8.09\%[/tex]

Hence, the approximate absentee rate that day would be 8.09%.

Assume a standard deck of 52 cards with 4 suits of 13 cards each.

a. There are [tex]\dbinom{13}8\dbinom{39}2[/tex] ways of being dealt a hand consisting of 8 hearts and 2 non-hearts, so the probability of being dealt such a hand is

[tex]\dfrac{\dbinom{13}8\dbinom{39}2}{\dbinom{52}{10}}\approx0.0000602823[/tex]

b. This time, the non-hearts specifically belong to the suit of diamonds, for which there are [tex]\dbinom{13}2[/tex] ways of getting drawn, so the probability is

[tex]\dfrac{\dbinom{13}8\dbinom{13}2}{\dbinom{52}{10}}\approx0.0000063455[/tex]

a. The probability P of being dealt exactly 8 hearts is:

[tex]\[ P = \frac{k}{n} = \frac{\binom{13}{8} \times \binom{39}{2}}{\binom{52}{10}} \][/tex]

b. The probability  P'  of being dealt exactly 8 hearts given the first two cards were diamonds is:

[tex]\[ P' = \frac{k'}{n'} = \frac{\binom{8}{6} \times \binom{42}{2}}{\binom{50}{8}} \][/tex]

To solve this problem using combinatorics, we can calculate the probability by considering the total number of possible outcomes and the number of favorable outcomes.

Let's denote:

-  n  as the total number of ways to deal 10 cards from a standard deck (52 cards).

-  k  as the number of ways to deal exactly 8 hearts and 2 non-hearts from the remaining 44 cards in the deck.

a. To calculate the probability a player is dealt exactly 8 hearts:

Total number of ways to choose 8 hearts out of 13 hearts:

[tex]\[ \binom{13}{8} \][/tex]

Total number of ways to choose 2 non-hearts out of 39 non-hearts:

[tex]\[ \binom{39}{2} \][/tex]

Therefore, the number of favorable outcomes is:

[tex]\[ k = \binom{13}{8} \times \binom{39}{2} \][/tex]

The total number of ways to deal 10 cards from a deck of 52 cards is:

[tex]\[ n = \binom{52}{10} \][/tex]

So, the probability P of being dealt exactly 8 hearts is:

[tex]\[ P = \frac{k}{n} = \frac{\binom{13}{8} \times \binom{39}{2}}{\binom{52}{10}} \][/tex]

b. To calculate the probability a player is dealt exactly 8 hearts if the first two cards dealt were diamonds:

If the first two cards are diamonds, then there are 50 cards remaining, out of which 8 are hearts and 42 are non-hearts.

Total number of ways to choose 6 more hearts out of the remaining 8 hearts:

[tex]\[ \binom{8}{6} \][/tex]

Total number of ways to choose 2 non-hearts out of the remaining 42 non-hearts:

[tex]\[ \binom{42}{2} \][/tex]

Therefore, the number of favorable outcomes is:

[tex]\[ k' = \binom{8}{6} \times \binom{42}{2} \][/tex]

The total number of ways to deal 8 cards from the remaining 50 cards is:

[tex]\[ n' = \binom{50}{8} \][/tex]

So, the probability  P'  of being dealt exactly 8 hearts given the first two cards were diamonds is:

[tex]\[ P' = \frac{k'}{n'} = \frac{\binom{8}{6} \times \binom{42}{2}}{\binom{50}{8}} \][/tex]

Answer:

No

.3 = .30

.32 > .30

Step-by-step explanation:

Answer:

a. y = 7500 - 500x

b. $ 4,500

Step-by-step explanation:

a. Given,

The original value of the phone = $ 75,00

Let the linear function that shows the value of booth after x years,

y = b - ax,

For x = 0 years, y = 7500

⇒ 7500 = b - a(0) ⇒ b = 7500

For x = 15, y = 0,

0 = b - 15x ⇒ 0 = 7500 - 15a ⇒ 15a = 7500 ⇒ a = 500

Hence, the function that shows the value of car after x years,

y = 7500 - 500x

b. if x = 6,

The value of the booth after 6 years would be,

y = 7500 - 500(6) = 7500 - 3000 = $ 4,500

Answer:

the equation of the least-squares line for the data is: [tex]\hat Y=9.3+1.3x[/tex]

Step-by-step explanation:

In a simple linear regression model, such as, [tex]\hat Y=b_0+b_1x[/tex], the coefficients bo and b1 are estimated through the method of least squares by the use of the equations:

[tex]b_1=\frac{S{xy}}{S_x^2}\\\\b_0=\bar{y}+b_1 \bar{x}[/tex]

For the data provided you have to:

[tex]S_{xy}=\frac{\sum {(x_i-\bar x)(y_i-\bar y)}}{n-1}=3.25\\\\S_x^2=\frac{\sum {(x_i-\bar x)^2}}{n-1}=2.5\\\\\bar y=5.4[/tex], thus:

[tex]b_1=\frac{3.25}{2,5}=1.3\\\\b_0=5.4+1.3(3.0)=9.3[/tex]

the equation of the least-squares line for the data is:

[tex]Y=9.3+1.3x[/tex]

Answer:

The draw in the file is a realization of a graph of order 4 and size zero.

In the book of Douglas West, Introduction to Graph Theory the name of this graph is 'Trivial graph'

Step-by-step explanation:

Remember that the order of a graph is the number of vertices and the size of the graph is the number of edges of the graph.

Answer:

The possible combinations are (4 & 6) ,(8 &3), (2 &12), (24 &1)

Step-by-step explanation:

The area of a flower bed is 24 square feet.

Now, the factors of 24 are 2 x 2 x 2 x 3.

Hence, if the other sides were whole number dimensions, then the possible combinations will be (4 & 6) ,(8 &3), (2 &12), (24 &1)

Answer:

A= 0,2

B= 0,2

C= 0,4

D=0,2

Step-by-step explanation:

We know that only one team can win, so the sum of each probability of wining  is one

P(A)+P(B)+P(C)+P(D)=1

then we Know that the probability of Team A and B are the same, so

P(A)=P(B)

And that the  the probability that either team A or team C wins the tournament is 0.6, so P(A)+Pc)= 0,6, then P(C)= 0.6-P(A)

Also, we know that team C is twice as likely to win the tournament as team D, so P(C)= 2 P(D) so P(D) = P(C)/2= (0.6-P(A))/2

Now if we use the first formula:

P(A)+P(B)+P(C)+P(D)=1

P(A)+P(A)+0.6-P(A)+(0.6-P(A))/2=1

0,5 P(A)+0.9=1

0,5 P(A)= 0,1

P(A)= 0,2

P(B)= 0,2

P(C)=0,4

P(D)=0,2

Teams A and B each have a 0.2 probability of winning, team C has a 0.4 probability, and team D has a 0.2 probability of winning the tournament. The given conditions were used to calculate these probabilities step-by-step.

To find the probabilities of each team (A, B, C, and D) winning the tournament, let's denote the probability for each team as follows: P(A), P(B), P(C), and P(D). According to the problem, we are given these conditions:

Let's express everything in terms of P(D):

From the total probability, we know that:

Substituting the given conditions:

From condition 3:

We now have two equations:

First, solve for x in terms of P(D) from equation 2:

Substitute this into equation 1:

Now substitute P(D) back to find x:

Therefore, P(A) = P(B) = 0.2.

Summarizing the probabilities:

Answer:

0.00125 Km

Step-by-step explanation:

Data provided in the question:

Speed of the horse = 9 Km/h

Duration for which the Lysera's eyes were shut = 0.50 seconds

now,

1 hour = 3600 seconds

or

1 second = [tex]\frac{\textup{1}}{\textup{3600}}\ textup{hours}[/tex]

Thus,

0.50 seconds =  [tex]\frac{\textup{0.50}}{\textup{3600}}\ textup{hours}[/tex]

Also,

Distance = speed × Time

on substituting the values, we get

Distance = 9 × [tex]\frac{\textup{0.50}}{\textup{3600}}[/tex]

or

Distance = 0.00125 Km

All of the statements are true.

If the limit of a function f(x) at x = a is exist .

            [tex]\lim_{x \to a+} f(x)= \lim_{x \to a-} f(x)=f(a)[/tex]

Given that,

          [tex]\lim_{x \to 2-} f(x)= \lim_{x \to 2+} f(x)=7[/tex]

But f(2) is not defined.

It means that function f(x) has jump discontinuity at x = 2

A removable discontinuity is a point on the graph that is undefined or does not fit the rest of the graph.

So that, Function f(x) has removable discontinuity at x = 2

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The false statement among the given options is that redefining f(2) to 7 will make the new function continuous at x = 2.

The false statement among the options given is (c) If we re-define f so that f(2) = 7 then the new function will be continuous at x = 2.

The given information states that both the left and right limits as x approaches 2 are equal to 7, which suggests that the limit as x approaches 2 exists and is equal to 7. This means that option (a) lim x→2 f(x)=7 is true.

We know that f(2) is not defined in the original function, meaning there is a hole in the graph at x = 2. Therefore, option (d) f has a removable discontinuity at x = 2 is also true.

However, if we redefine f(2) = 7, the new function will still have a jump discontinuity at x = 2 since there will be a discontinuity between the values of f(2) before and after the redefinition. Therefore, option (c) is false.

So, the correct answer is (c) If we redefine f so that f(2) = 7 then the new function will be continuous at x = 2.

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Answer:

Distributive property

Subtraction property of equality

Division property of equality

Step-by-step explanation:

Given

-6(x-4)=42

-6*x+-4*-6=42-------------------distributive property

-6x+24-24=42-24-----------------subtraction property of equality

-6x=18

-6x/-6=18/-6--------------------------division property of equality

x= -3

Step-by-step explanation:

As the hint says, for any function [tex]f:\Omega\to\mathcal{P}(\Omega)[/tex], we can think of the set [tex] X=\{ \omega\in\Omega : \omega \notin f(\omega)\}[/tex] (which is the set of all those elements of [tex]\Omega[/tex] which don't belong to their image). So [tex]X[/tex] is made of elements of [tex]\Omega[/tex], and so it belongs to [tex]\mathcal{P}(\Omega)[/tex].

Now, this set [tex]X[/tex] is NOT the image of any element in [tex] \Omega[/tex], since if there was some [tex]a\in\Omega[/tex] such that [tex]f(a)=X[/tex], then the following would happen:

If [tex]a\in X=f(a)[/tex], then by definition of the set [tex]X[/tex], [tex]a\notin f(a)[/tex], so we're getting that [tex]a\in f(a)[/tex] and also [tex] a\notin f(a)[/tex], which is a contradiction.

On the other hand, if [tex]a\notin f(a)[/tex], then by definition of the set [tex]X[/tex], we would get that [tex]a\in X=f(a)[/tex], so we're getting that [tex]a\in f(a)[/tex] and also [tex] a\notin f(a)[/tex], which is a contradiction again.

So in any case, the assumption that this set [tex]X[/tex] is the image of some element in [tex]\Omega[/tex] leads us to a contradiction, therefore this set [tex]X[/tex] is NOT the image of any element in [tex]\Omega[/tex], and so there cannot be a bijection from [tex]\Omega[/tex] to [tex]\mathcal{P}(\Omega)[/tex], and so the two sets cannot have the same cardinality.

Answer:

It is sufficient to prove that  [tex] p_1\implies p_2, p_2\implies p_3, p_3\implies p_1[/tex]

Step-by-step explanation:

The propositions [tex] p_1,p_2,p_3[/tex] being equivalent means they should always have the same truth value. If one of them is true, then all of them must be true. And if one of them is false, then all of them must be false.

Suppose we've proven that [tex] p_1\implies p_2, p_2\implies p_3, p_3\implies p_1[/tex] (call these first, second and third implications).

If [tex]p_1[/tex] was true, then by the first implication that we proved, it would follow that [tex]p_2[/tex] is also true. And then by the second implication that we prove it would follow then that [tex]p_3 [/tex] is also true. Therefore the three of them would be true. Notice the reasoning would have been the same if we had started assuming that the one that was true was either [tex]p_2~or~p_3[/tex]. So one of them being true makes all of them be true.

On the other hand, if [tex]p_1[/tex] was false, then by the third implication that we proved, it would follow that [tex]p_3[/tex] has to be false (otherwise [tex]p_1[/tex] would have to be true, which would be a contradiction). And then, since [tex]p_3[/tex] is false, by the second implication that we proved it would follow that [tex] p_2[/tex] is false (otherwise [tex] p_3[/tex] would have to be true, which would be a contradiction). Therefore the three of them would be false. Notice the reasoning would have been the same if we had started assuming that the one that was false was either [tex]p_2~or~p_3[/tex]. So one of them being false makes all of them be false.

So, the three propositions always have the same truth value, and so they're all equivalent.

Answer:

The solutions are: [tex]x=4,\:x=-4,\:x=3i,\:x=-3i[/tex]

Step-by-step explanation:

Consider the provided equation.

[tex]x^4-7x^2-144=0[/tex]

Substitute [tex]u=x^2\mathrm{\:and\:}u^2=x^4[/tex]

[tex]u^2-7u-144=0[/tex]

[tex]u^2-16u+9u-144=0[/tex]

[tex](u-16)(u+9)=0[/tex]

[tex]u=16,\:u=-9[/tex]

Substitute back [tex]\:u=x^2[/tex] and solve for x.

[tex]x^2=16\\x=\sqrt{16}\\ \quad x=4,\:x=-4[/tex]

Or

[tex]x^2=-9\\x=\sqrt{-9}\\ \quad x=3i,\:x=-3i[/tex]

Hence, the solutions are: [tex]x=4,\:x=-4,\:x=3i,\:x=-3i[/tex]

Check:

Substitute x=4 in provided equation.

[tex]4^4-7(4)^2-144=0[/tex]

[tex]256-112-144=0[/tex]

[tex]0=0[/tex]

Which is true.

Substitute x=-4 in provided equation.

[tex](-4)^4-7(-4)^2-144=0[/tex]

[tex]256-112-144=0[/tex]

[tex]0=0[/tex]

Which is true.

Substitute x=3i in provided equation.

[tex](3i)^4-7(3i)^2-144=0[/tex]

[tex]81+63-144=0[/tex]

[tex]0=0[/tex]

Which is true.

Substitute x=-3i in provided equation.

[tex](-3i)^4-7(-3i)^2-144=0[/tex]

[tex]81+63-144=0[/tex]

[tex]0=0[/tex]

Which is true.

Approximately 347,200 books can be shelved in the college library with 3500 m² of floor space.

Given:

Floor space of the library = 3500 m²

To estimate how many books can be shelved in a college library with 3500 m² of floor space, we can use the following assumptions:

To calculate the number of books that can be shelved, we need to find the volume of the shelving space and divide it by the volume of each book.

To find the volume of the shelving space, we need to subtract the volume of the corridors from the total volume of the library. The total volume of the library is:

[tex]V_{library} = (63\ yards) \times (32 \ yards) \times (6\ yards)[/tex]

Converting yards to meters, we get:

[tex]V_{library} = (63 \times 0.9144 \ meters) \times (32 \times 0.9144 \ meters) \times (6 \times 0.9144\ meters)[/tex]

Simplifying the equation, we get:

[tex]V_{library} \approx 1407\ m^3[/tex]

The volume of the corridors can be calculated as follows:

[tex]V_{corridors} = (8\ shelves) \times (0.5\ m + 1.5\ m) \times (1.5\ m) \times (63\ m + 32\ m)[/tex]

Simplifying the equation, we get:

[tex]V_{corridors} = 756\ m^3[/tex]

Therefore, the volume of the shelving space is:

[tex]V_{shelving} = V_{library} - V_{corridors} \\V_{shelving} \approx 651 \ m^3[/tex]

To find the volume of each book, we can multiply the depth, width, and height of each book:

[tex]V_{book} = (0.25\ m) \times (0.05\ m) \times (0.15\ m)[/tex]

Simplifying the equation, we get:

[tex]V_{book} = 0.001875 \ m^3[/tex]

Finally, we can divide the volume of the shelving space by the volume of each book to find the number of books that can be shelved:

[tex]Number\ of\ books = \frac{V_{shelving}} {V_{book}} \\Number\ of\ books \approx 347,200[/tex]

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The number of books can be shelved in a college library with [tex]3500 \ m^2[/tex] of floor space is 700,000 books.

Assume that the room is square-shaped.

The size of an average book is [tex]0.05\times0.25 \ m[/tex]. Thus, the thickness of the shelf is 0.5 m.

Area of the room, [tex]A=3500 \ m^2[/tex]

Width of the corridor space, [tex]W_c=1.5 \ m[/tex]

As the room is square-shaped, its width is as follows:

[tex]W_{room}=\sqrt{3500}[/tex]

[tex]= 59.16 m[/tex]

The area of a square with side a is [tex]a^2[/tex].

The total number of rows (r) is

[tex]r=\frac{W_{room}}{W_c+shelfsize}[/tex]

[tex]= \frac{59.16}{1.5+0.5}[/tex]

[tex]= 29.28 m[/tex]

The total number of shelves

[tex]= r\times 8(height of the shelves)\times2(facing both sides)[/tex]

[tex]= 473.28[/tex]

As the room is square-shaped, the length of each shelf is [tex]59.16 \ m[/tex].

The number of books on each shelf (n) is as follows:

[tex]n=\frac{Length \ of \ each \ row}{Thickness \ of \ each \ book}[/tex]

= [tex]\frac{59.16}{0.04}[/tex]

[tex]= 1479[/tex]

The total number of books is the sum of books on all shelves. Thus, the total number of books (N) is as follows:

[tex]N=n\times (Total \ number \ of \ shelves)[/tex]

[tex]= 1479\times473.28[/tex]

[tex]= 699981.12[/tex]

[tex]\approx 700000[/tex]

Thus, there are approximately 700,000 books.

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Answer:

The general solution: [tex]C_{1}e^{-4x} + xC_{2}e^{-4x}[/tex]

Step-by-step explanation:

Differential equation: y'' + 8y' + 16y = 0

We have to find the general solution of the above differential equation.

The auxiliary equation for the above equation can be writtwn as:

m² + 8m +16 = 0

We solve the above equation for m.

(m+4)² = 0

[tex]m_{1}[/tex] = -4, [tex]m_{2}[/tex] = -4

Thus we have repeated roots for the auxiliary equation.

Thus, the general solution will be given by:

y = [tex]C_{1}e^{m_{1}x} + xC_{2}e^{m_{2}x}[/tex]

y = [tex]C_{1}e^{-4x} + xC_{2}e^{-4x}[/tex]

Final answer:

To find the general solution of y'' + 8y' + 16y = 0, determine the characteristic roots, and utilize them to form the solution expression with constants. The primary topic is solving second-order homogeneous differential equations.

Explanation:

To find a general solution of the differential equation y'' + 8y' + 16y = 0, we first look for the characteristic equation by assuming y = e^(rt). Substituting this into the equation gives us the characteristic equation r^2 + 8r + 16 = 0. Solving this quadratic equation gives us the roots r = -4, -4. Therefore, the general solution is y(x) = c1e^(-4x) + c2xe^(-4x), where c1 and c2 are constants.

Answer:

Option A. y= -1/5x + 5

Step-by-step explanation:

step 1

Find the slope of the given line

we have

y=5x-2

The slope m is

m=5

step 2

Find the slope of a line perpendicular to the given line

Remember that

If two lines are perpendicular, then their slopes are opposite reciprocal (the product of their slopes is -1)

m1*m2=-1

we have

m1=5

(5)*m2=-1

m2=-1/5

therefore

The equation of a line perpendicular to the given line could be

y= -1/5x + 5

Answer:

The system is consistent with infinitely many solutions.

Step-by-step explanation:

Consider the provided information.

It is given that the 4 × 6 coefficient matrix for a system of linear equations has 4 pivot columns.

4 pivot columns means that there is a  pivot in every row but not in every column as we know that the number of columns are greater than 4.

In a coefficient matrix the columns without pivot elements correspond to free variables.

If a system of linear equations has no solution then it is known as inconsistent otherwise its called consistent.

Infinite solution: The system of equation is consistent but  at least one of the variables is free.

As each row has  pivot that means that means system of linear equations has solution. Also at least one of the variable is free that means it has infinitely many solutions.

Thus, the system is consistent with infinitely many solutions as there is a pivot in every row but not every column.

Final answer:

A 4 × 6 coefficient matrix with 4 pivot columns indicates an underdetermined system of linear equations, leading to an infinite number of solutions within a two-dimensional solution space.

Explanation:

When a 4 × 6 coefficient matrix for a system of linear equations has 4 pivot columns, it indicates the presence of 4 independent equations for solving the variables. However, since there are 6 columns in total and 4 of them are pivot columns, it implies that there are 6 variables in the system. The presence of 4 pivot columns means that 4 of these variables can be solved in terms of the remaining 2 variables, assuming the system has consistent equations. Therefore, the system does not have a unique solution; instead, it has an infinite number of solutions that form a two-dimensional solution space, because the system is underdetermined (more variables than independent equations).

The concept of pivot columns is crucial in linear algebra for understanding the solvability of linear equation systems. A pivot column in a matrix corresponds to an independent equation in the system, which directly affects the nature of the solutions. When dealing with a system of linear equations, it is essential to determine the number of pivot columns to understand the dimensions of the solution space and whether the system is over-determined, underdetermined, or exactly determined.

Answer:

To accomplish this goal, how much should adam invest now $12800.7525 in a CD that pays 1.24% interest compounded monthly.

Step-by-step explanation:

Amount = 16000

Time = 18 years

Interest = 1.24% interest compounded monthly

So, Formula : [tex]A=P(1+\frac{r}{n})^{nt}[/tex]

Formula : [tex]16000=P(1+\frac{1.24}{100 \times 12})^{12 \times 18}[/tex]

[tex]16000=P(1.24992651224)[/tex]

[tex]\frac{16000}{1.24992651224}=P[/tex]

[tex]12800.7525=P[/tex]

Hence To accomplish this goal, how much should adam invest now $12800.7525 in a CD that pays 1.24% interest compounded monthly.

Answer:

v and or.

Step-by-step explanation:

A ∨conveys the notion of "or". We use the symbol logical ∨ to represent a disjunction.

The "or" in logical lenguage indicates that one thing can happen but another thing also can happen, that is what we call a disjunction. The symbol to represent it is ∨.

The "and" in logical lenguage indicates that a situation is true only if all the statements are true. In other words we see all the statements as needs, not as options, that is called a conjunction. The symbol to represent it is ^.

Answer:

231

Step-by-step explanation:

Let's compute the probability of this event and then multiply it by 100.

As each deck has 4 suits and each suit has 3 faces, there are 12 faces in a deck.

As a deck has 52 cards, the probability of drawing a face is

12/52 = 0.23076

This means that if you choose a card 1000 times from a deck of poker cards, you probably will select 231  times a face (rounding to the nearest integer).

Robby Billity is expected to draw approximately 500 red cards and 231 face cards out of 1000 draws, based on the probability of each type of card in a standard deck.

To determine how many times we would expect Robby Billity to draw a specific type of card over 1000 draws, we first need to calculate the probability of drawing each type of card from a standard deck. Since there are 52 cards in the deck and Robby replaces the card each time before drawing again, each draw is independent with the same probability.

Calculating Probabilities:

Red Cards: There are 26 red cards (hearts and diamonds) out of 52. The probability of drawing a red card is 26/52, which simplifies to 1/2 or 0.5.

Face Cards: Each suit has 3 face cards (Jack, Queen, King), totaling 4 x 3 = 12 face cards. Therefore, the probability of drawing a face card is 12/52, which simplifies to 3/13 or approximately 0.231.

Expected Number of Draws:

Therefore, Robby Billity is expected to draw approximately 500 red cards and 231 face cards out of 1000 draws.

Step-by-step explanation:

a) Give two pairs which are in the relation [tex]\equiv \mod 4[/tex] and two pairs that are not.

As stated before, a pair [tex](x,y)\in \mathbb{Z}\times\mathbb{Z}[/tex] is equal mod m (written [tex]x\equiv y\mod m[/tex]) if [tex]m\mid (x-y)[/tex]. Then:

b) Show the [tex]\equiv \mod m[/tex] is an equivalence relation.

An equivalence relation is a binary relation that is reflexive, symmetric and  transitive.

By definition [tex]\equiv \mod m[/tex] is a binary relation. Observe that:

[tex]m\mid [(y-x)+(z-y)] \implies m\mid (z-x) \implies x\equiv z \mod m[/tex].

In conclusion, [tex]\equiv \mod m[/tex] defines an equivalence relation.

Answer:

The augmented matrix for each set of linear equations is:

a) [tex]x_1-2x_2=0\\3x_1+4x_2=-1\\2x_1-x_2=3[/tex]

Augmented matrix:

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

b) [tex]x_1+x_3=1\\-x_1+2x_2-x_3=3[/tex]

Augmented matrix:

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

c) [tex]x_1+x_3=1\\2x_2-x_3+x_5=2\\2x_3+x_4=3[/tex]

Augmented matrix:

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

d) [tex]x_1=1\\x_2=2[/tex]

Augmented matrix:

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

Step-by-step explanation:

In order to find the augmented matrix, you have to take the numeric values of each variable and make a matrix with them. For example, in the linear system a) you can make a matrix out of the numeric values accompanying x_1 and x_2, this matrix will be:  

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

Then you have to make a vector with the constants in the linear equations, for the case of system a) the vector will be:  

[tex]\left[\begin{array}{c}0&-1&3\end{array}\right][/tex]

To construct the augmented matrix, you append those matrices together and create a new one:

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

Answer: The negations are:

1. The train is not late and my watch is not fast.

2. Dogs don't bark or cats don't meow.

Step-by-step explanation:

Hi!

De Morgan's laws for two propositions P and Q are:

1. ¬(P ∨ Q) = (¬P) ∧ (¬Q)

2. ¬(P ∧ Q) = (¬P) ∨ (¬Q)

where the symbols are,

¬ = not

∨ = or

∧ = and

1. In this case the proposition is P ∨  Q, with

P = "the train is late"

Q = "my watch is fast"

Then by law 1:  ¬(P ∨ Q) =  "The train is not late and my watch is not fast"

2. In this case the proposition is P  ∧ Q, with

P = "dogs bark"

Q = "cats meow"

Then by law 2: ¬(P ∧ Q) = "Dogs don't bark or cats don't meow"

Answer:

2

Step-by-step explanation:

Answer:

$2.24

Step-by-step explanation:

Given:

Threshold limit =  2.0 × 10⁻¹¹ grams per liter of air

Current price of 50.0 g vanillin = $112

Volume of aircraft hanger = 5.0 × 10⁷ m³

Now,

1 m³ = 1000 L

thus,

5.0 × 10⁷ m³ = 5.0 × 10⁷ × 1000 = 5 × 10¹⁰ L

therefore,

The mass of vanillin required = 2 × 10⁻¹¹ × 5 × 10¹⁰ = 1 g

Now,

50 grams of vanillin costs = $112

thus,

1 gram of vanillin will cost = [tex]\frac{\textup{112}}{\textup{50}}[/tex] = $2.24