Where do we use prime numbers every day?

Answer:

G. I don't exercise and I feel envigorated.

Step-by-step explanation:

In this sentence the I exercise and I feel tired you need to say in order tos ay the negation of this sentence would be:

I do not exercise nor feel tired, since there´s no option that says this, we can choose the one that says I don´t exercise, and envigorated is the opposite than tired, so the correct option would be I don´t exercise and I feel envigorated.

Final answer:

The correct negation of the statement "I exercise and I feel tired" is "I don't exercise or I feel envigorated," which corresponds to option F.

Explanation:

The negation of the compound statement "I exercise and I feel tired" involves negating both parts of the statement and changing the conjunction 'and' to 'or'. This is in line with De Morgan's laws which state that the negation of a conjunction is the disjunction of the negations. Therefore, the negated form of the statement would be "I don't exercise or I don't feel tired." From the given options, the one that best matches this structure is:

F. I don't exercise or I feel envigorated.

Envigorated is understood as the opposite of feeling tired in this context. So, statement F is the correct negation as it correctly captures the negation of both parts of the original statement.

Answer:

150 students take physics.

Step-by-step explanation:

To solve this problem, we must build the Venn's Diagram of this set.

I am going to say that:

-The set A represents the students that take calculus.

-The set B represents the students that take physics

-The set C represents the students that take chemistry.

-The set D represents the students that do not take any of them.

We have that:

[tex]A = a + (A \cap B) + (A \cap C) + (A \cap B \cap C)[/tex]

In which a is the number of students that take only calculus, [tex]A \cap B[/tex] is the number of students that take both calculus and physics, [tex]A \cap C[/tex] is the number of students that take both calculus and chemistry and [tex]A \cap B \cap C[/tex] is the number of students that take calculus, physics and chemistry.

By the same logic, we have:

[tex]B = b + (B \cap C) + (A \cap B) + (A \cap B \cap C)[/tex]

[tex]C = c + (A \cap C) + (B \cap C) + (A \cap B \cap C)[/tex]

This diagram has the following subsets:

[tex]a,b,c,(A \cap B), (A \cap C), (B \cap C), (A \cap B \cap C), D[/tex]

There are 360 people in my school. This means that:

[tex]a + b + c + (A \cap B) + (A \cap C) + (B \cap C) + (A \cap B \cap C) + D = 360[/tex]

The problem states that:

15 take calculus, physics, and chemistry, so:

[tex]A \cap B \cap C = 15[/tex]

15 don't take any of them, so:

[tex]D = 15[/tex]

75 take both calculus and chemistry, so:

[tex]A \cap C = 75[/tex]

75 take both physics and chemistry, so:

[tex]B \cap C = 75[/tex]

30 take both physics and calculus, so:

[tex]A \cap B = 30[/tex]

Solution:

The problem states that 180 take calculus. So

[tex]a + (A \cap B) + (A \cap C) + (A \cap B \cap C) = 180[/tex]

[tex]a + 30 + 75 + 15 = 180[/tex]

[tex]a = 180 - 120[/tex]

[tex]a = 60[/tex]

Twice as many students take chemistry as take physics:

It means that: [tex]C = 2B[/tex]

[tex]B = b + (B \cap C) + (A \cap B) + (A \cap B \cap C)[/tex]

[tex]B = b + 75 + 30 + 15[/tex]

[tex]B = b + 120[/tex]

-------------------------------

[tex]C = c + (A \cap C) + (B \cap C) + (A \cap B \cap C)[/tex]

[tex]C = c + 75 + 75 + 15[/tex]

[tex]C = c + 165[/tex]

----------------------------------

Our interest is the number of student that take physics. We have to find B. For this we need to find b. We can write c as a function o b, and then replacing it in the equations that sums all the subsets.

[tex]C = 2B[/tex]

[tex]c + 165 = 2(b+120)[/tex]

[tex]c = 2b + 240 - 165[/tex]

[tex]c = 2b + 75[/tex]

The equation that sums all the subsets is:

[tex]a + b + c + (A \cap B) + (A \cap C) + (B \cap C) + (A \cap B \cap C) + D = 360[/tex]

[tex]60 + b + 2b + 75 + 30 + 75 + 15 + 15 = 360[/tex]

[tex]3b + 270 = 360[/tex]

[tex]3b = 90[/tex]

[tex]b = \frac{90}{3}[/tex]

[tex]b = 30[/tex]

30 students take only physics.

The number of student that take physics is:

[tex]B = b + (B \cap C) + (A \cap B) + (A \cap B \cap C)[/tex]

[tex]B = b + 75 + 30 + 15[/tex]

[tex]B = 30 + 120[/tex]

[tex]B = 150[/tex]

150 students take physics.

Final answer:

Using a Venn Diagram approach and the information given, we find that 45 students take physics at the school.

Explanation:

To find out how many students take physics at the school, we can use the Venn Diagram principle and the given data. We know that 15 students take calculus, physics, and chemistry together. Additionally, 180 students take calculus, and twice as many students take chemistry as take physics. With 75 students taking both calculus and chemistry, and another 75 taking both physics and chemistry, while only 30 take both physics and calculus, we can establish relationships and solve for the number of students taking each subject.

Let's denote the number of students taking physics as P. Then, the number of students taking chemistry would be 2P.

To avoid double counting, we must subtract those taking all three subjects once for each combination:

With twice as many students in chemistry as in physics, we can write the equation:

180 + 2P + P - (15 + 60 + 15 + 60) + 15 = 360

Solving for P:

Therefore, 45 students take physics at the school.

Answer and Explanation:

To find : Convert the given units ?

Solution :  

a) 2.5 km to mm

[tex]1\ km = 1000000\ mm[/tex]

[tex]2.5\ km = 2.5\times 1000000\ mm[/tex]

[tex]2.5\ km = 2500000\ mm[/tex]

b) 0.05 cm to mm

[tex]1\ cm = 10\ mm[/tex]

[tex]0.05\ cm =0.05\times 10\ mm[/tex]

[tex]0.05\ cm =0.5\ mm[/tex]

c) 200.5 g to kg

[tex]1\ g = 0.001\ kg[/tex]

[tex]200.5\ g =200.5\times 0.001\ kg[/tex]

[tex]200.5\ g =0.2005\ kg[/tex]

d) 0.03 tone into g

[tex]1\ t =1000000\ g[/tex]

[tex]0.03\ t =0.03\times 1000000\ g[/tex]

[tex]0.03\ t =30000\ g[/tex]

e) 3.0412 sec into hour

[tex]1\ sec =\frac{1}{3600}\ hr[/tex]

[tex]3.0412\ sec =3.0412\times \frac{1}{3600}\ hr[/tex]

[tex]3.0412\ sec =0.000844\ hr[/tex]

Answer:

40480 MP3 players expected to be sold in 2002.

Step-by-step explanation:

Sales of a certain MP3 players are approximated by the relationship;

[tex]S(x)=4740x+ 31000[/tex]  (0≤x≤5)

(x = 0 corresponds to the year 2000)

That means 2002 corresponds to x = 2

Now substituting x = 2 in the expression.

[tex]S(x)=4740(2)+ 31000[/tex]

= [tex]9480+31000[/tex]

= 40480.

Hence, 40480 MP3 players expected to be sold in 2002.

Answer:

If the intersection is finite the statement  is true, but if the intersection is infinite the statement is false.

Step-by-step explanation:

From the statement of the problem I am not sure if the intersection is finite or infinite. Then, I will study both cases.

Let us consider first the finite case: [tex]A = \cap_{i=1}^{n}A_i[/tex]. Because the condition A1 ⊇ A2 ⊇ A3 ⊇ A4 ... we can deduce that the set [tex]A_n[/tex] is a subset of each set [tex]A_i[/tex] with [tex] i\leq n[/tex]. Thus,

[tex]\cap_{i=1}^{n}A_i = A_n[/tex].

Therefore, as [tex]A_n[/tex] is infinite, the intersection is infinite.

Now, if we consider the infinite intersection, i.e. [tex]A = \cap_{k=1}^{\infty}A_k[/tex] the reasoning is slightly different. Take the sets

[tex]A_k = (0,1/k)[/tex] (this is, the open interval between 0 and [tex]1/k[/tex].)

Notice that (0,1) ⊇ (0,1/2) ⊇ (0, 1/3) ⊇(0,1/4) ⊇...So, the hypothesis of the problem are fulfilled. But,

[tex]\cap_{k=1}^{\infty}(0,1/k) = \empyset[/tex]

In order to prove the above statement, choose a real number [tex]x[/tex] between 0 and 1. Notice that, no matter how small [tex]x[/tex] is, there is a natural number [tex]K[/tex] such that [tex]1/K<x[/tex]. Then, the number [tex]x[/tex] is not in any interval [tex](0,1/k)[/tex] with [tex]k>K[/tex]. Therefore, [tex]x[/tex] is not in the set [tex]\cap_{k=1}^{\infty}(0,1/k)[\tex].

Answer:

20 L of 22% solution and 8 L of 36% solution

Step-by-step explanation:

Volume of 22% solution + volume of 36% solution = volume of 26% solution

x + y = 28

Acid in 22% solution + acid in 36% solution = acid in 26% solution

0.22x + 0.36y = 0.26(28)

0.22x + 0.36y = 7.28

Solve the system of equations using either elimination or substitution.  I'll use substitution:

x = 28 − y

0.22(28 − y) + 0.36y = 7.28

6.16 − 0.22y + 0.36y = 7.28

0.14y = 1.12

y = 8

x = 28 − y

x = 20

The chemist should use 20 L of 22% solution and 8 L of 36% solution.

Answer:

There should be mixed 20 L of the 22% acid solution with 8L of the 36% acid solution

Step-by-step explanation:

We are mixing two acids.

x = liters of 22% acid solution

y = liters of 36% acid solution

x + y = 28    (total liters)

0.22x +0.36y = 0.26* 28  

Since x+y=28 means y = 28-x

Now we will use substitution to find x

0.22x + 0.36(28-x) = 0.26 * 28

0.22x + 10.08 - 0.36x = 7.28

0.14x = 2.8

x = 20

y = 28 - 20 = 8  

⇒ We use 20 liters of the 22% solution to be mixed with 8 liters of the 36% solution to form  28l of a  26% acid solution.

Answer:

The house with an area 3,600 square feet is more unusual

Step-by-step explanation:

Given:

Number of houses surveyed = 583

Mean price = $176,678

Standard deviation = $61,029

Mean house size = 1,676 square ft

standard deviation = 582 square ft

Now,

the as z score = [tex]\frac{\textup{(X - mean )}}{\textup{standard deviation}}[/tex]

thus,

for selling value of $357,000

z score = [tex]\frac{\textup{(357,000 - 176,678 )}}{\textup{61,029}}[/tex]

or

z score = 2.95

and for house with an area 3,600 square feet

z score =  [tex]\frac{\textup{(3600 - 1676)}}{\textup{582}}[/tex]

or

z score = 3.30

Hence, the house with an area 3,600 square feet is more unusual

Final answer:

To determine the more unusual house in the market, we calculate the z-scores. A house priced at $357,000 has a z-score of 2.95, while a house of 3,600 square feet has a z-score of 3.31. Therefore, the larger house size is more unusual.

Explanation:

To determine which house is more unusual in the given market, we need to calculate the number of standard deviations each value is from the mean, also known as the z-score. The z-score is calculated by taking the difference between the value and the mean, and then dividing by the standard deviation. For the price of the house, the z-score is calculated as follows:

Z = (Value - Mean) / Standard Deviation

For the $357,000 house price:

Z = ($357,000 - $176,678) / $61,029 = 2.95

For the 3,600 square ft house:

Z = (3,600 - 1,676) / 582 = 3.31

The house with an area of 3,600 square ft is 3.31 standard deviations away from the mean, whereas the $357,000 house price is 2.95 standard deviations away from the mean. Hence, the house with an area of 3,600 square ft is more unusual compared to the market's average.

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: (1,0)

Step-by-step explanation: What is the x-intercept of the linear equation y = 4x – 4?

y = 4x - 4

x-intercept ⇒ y = 0

which means that we need to substitute the y by 0.

0 = 4x-4

4x = 4

x = 1

As it is a linear equation, 1st degree, there is only one point.

This way, the linear y = 4x - 4 intercept x on point (1,0)

Answer:

The solution for this system is [tex]x = 5, y = 3[/tex].

Step-by-step explanation:

The Gauss-Jordan elimination method is done by transforming the system's augmented matrix into reduced row-echelon form by means of row operations.

We have the following system:

[tex]5x + 3y = 16[/tex]

[tex]-2x + y = -13[/tex]

This system has the following augmented matrix.

[tex]\left[\begin{array}{ccc}5&3&16\\-2&1&-13\end{array}\right][/tex]

The first step is dividing the first line by 5. So:

[tex]L_{1} = \frac{L_{1}}{5}[/tex]

We now have

[tex]\left[\begin{array}{ccc}1&\frac{3}{5}&\frac{16}{5}\\-2&1&-13\end{array}\right][/tex]

Now i want to reduce the first row, so I do:

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

So we have

[tex]\left[\begin{array}{ccc}1&\frac{3}{5}&\frac{16}{5}\\0&\frac{11}{5}&-\frac{33}{5}\end{array}\right][\tex].

Now, the first step to reduce the second row is:

[tex]L_{2} = \frac{5L_{2}}{11}[/tex]

So we have:

[tex]\left[\begin{array}{ccc}1&\frac{3}{5}&\frac{16}{5}\\0&1&-3\end{array}\right][/tex].

Now, to reduce the second row, we do:

[tex]L_{1} = L_{1} - \frac{3L_{2}}{5}[/tex]

And the augmented matrix is:

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

The solution for this system is [tex]x = 5, y = 3[/tex].

Answer:

This is it:

Step-by-step explanation:

don't click any links

Step-by-step explanation:

Consider the provided statement.

If we lose electricity, then the data will be lost.

We are need to write the negation of the above statement.

First divide the whole statement in two parts

Let us consider p = We lose electricity and q = The data will be lost.

The symbol use for negation is tilde [tex]\sim[/tex]

[tex]\sim(p\rightarrow q)[/tex]

[tex]p\wedge \sim q[/tex]

Represent T for true and F for False.

The required table is shown below:

p           q           [tex](p\wedge \sim q)[/tex]

F            F                F

F            T                F

T            F                T

T            T                F

Hence the required table is shown above.

Answer:

The solution is [tex]f(t)=-\ln \left|\cos \left(t\right)\right|+C[/tex]

Step-by-step explanation:

We know that this ordinary differential equation (ODE) is separable if we can write F(x,y) = f(x)g(y) for some function f(x), g(x).

We can write this ODE in this way

[tex]cos(t) \cdot f'(t)=sin(t)\\f'(t)=\frac{sin(t)}{cos(t)}[/tex]

[tex]\mathrm{If\quad }f^{'} \left(x\right)=g\left(x\right)\mathrm{\quad then\quad }f\left(x\right)=\int g\left(x\right)dx[/tex]

[tex]f(t) =\int\limits{\frac{sin(t)}{cos(t)}} \, dt[/tex]

To solve this integral we need to follow this steps

[tex]\int \frac{\sin \left(t\right)}{\cos \left(t\right)}dt = \\\mathrm{Apply\:u-substitution:}\:u=\cos \left(t\right)\\\int \frac{\sin \left(t\right)}{u}dt \\\mathrm{And \:du=-sin(t)\cdot dt}\\\mathrm{so \>dt=\frac{du}{-sin(t)}}\\\int \frac{\sin \left(t\right)}{u}dt = -\int \frac{1}{u}du[/tex]

[tex]\mathrm{Use\:the\:common\:integral}:\quad \int \frac{1}{u}du=\ln \left(\left|u\right|\right)\\-ln|u|\\\mathrm{Substitute\:back}\:u=\cos \left(t\right)\\-\ln \left|\cos \left(t\right)\right|\\[/tex]

Add the constant of integration

[tex]f(t)=-\ln \left|\cos \left(t\right)\right|+C[/tex]

Answer:

Step-by-step explanation:

A plot of the points connected by straight lines makes it pretty clear they do not all fall on the same line. There is no linear relationship here.

Answer:

Sum = 10,232

Step-by-step explanation:

The given sequence is Arithmetic Progression.

Arithmetic Progression is a sequence in which every two neighbor digits have equal distances.

For finding the sum of given series firstly we find the number of terms in given series.

For finding the nth term, we use formula

aₙ = a + (n - 1) d

where, aₙ = value of nth term

a = First term

n = number of term

d = difference

Now, In given sequence: 3+11+19+27+.....+395+403

a = 3, d = 8, aₙ = 403

∴ 403 = 3 + (n - 1) × 8

⇒ n = 51

Now, the sum of series is determined by formula,

Sₙ = n ÷ 2 [ a + l]

where l = last term

⇒ Sₙ = 51 ÷ 2 [ 3 + 403]

⇒ Sₙ = 51 × 203

⇒ Sₙ = 10,232

Final answer:

The series 3+11+19+27+.....+395+403 is an arithmetic sequence with a common difference of 8. It has 51 terms, and the sum can be found using the formula Sn = n(a1 + an) / 2, which gives us a sum of 10353.

Explanation:

To add the numbers in the series 3+11+19+27+.....+395+403, we need to recognize that this series is an arithmetic sequence, where each term increases by a common difference. In this case, the common difference is 8 (since 11 - 3 = 8, 19 - 11 = 8, and so on).

First, we need to find the number of terms in the series. We know that an arithmetic series can be expressed as ann = a1 + (n - 1)d, where an is the nth term, a1 is the first term, d is the common difference, and n is the number of terms. Plugging in the values we have:

403 = 3 + (n - 1)8
400 = (n - 1)8
n - 1 = 50
n = 51

So there are 51 terms in the series. The sum of an arithmetic series is given by Sn = n(a1 + an) / 2. Plugging in the values we found:

S51 = 51(3 + 403) / 2
S51 = 51(406) / 2
S51 = 10353

Therefore, the sum of the series is 10353.

Answer: There would be $507.39 in the account after 90 days.

Step-by-step explanation:

Since we have given that

Principal = $500

Rate of interest = 6%

Number of days = 90 days

As we know that "Simple interest":

[tex]Interest=\dfrac{P\times R\times T}{100}\\\\Interest=\dfrac{500\times 6\times 90}{100\times 365}\\\\Interest = \$7.39[/tex]

So, Amount = Principal + Interest

Amount = $500 + $ 7.39

Amount = $507.39

Hence, There would be $507.39 in the account after 90 days.

Check the picture below.

let's recall that a straight-line has 180°, and that sum of all interior angles in a triangle is also 180°.

Answer:

Maximum safe height can be reached by ladder = 15.03. ft

Step-by-step explanation:

Given,

Let's assume the maximum safe height of wall = h

angle formed between ladder and ground = 70°

length of ladder = 16 ft

From the given data, it can be seen that ladder will form a right angle triangle structure with the wall

So,from the concept of trigonometry,

[tex]Sin70^o\ =\ \dfrac{\textrm{maximum safe height of wall}}{\textrm{length of ladder}}[/tex]

[tex]=>Sin70^o\ =\ \dfrac{h}{16\ ft}[/tex]

[tex]=>\ h\ =\ 16\times Sin70^o[/tex]

=> h = 16 x 0.9396

=> h = 15.03 ft

So, the maximum safe height that can be reached by the ladder will be 15.03 ft.

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 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.

The given Statement which we have to prove using mathematical induction is

   [tex]5^n\geq 2*2^{n+1}+100[/tex]

for , n≥4.

⇒For, n=4

LHS

[tex]=5^4\\\\5*5*5*5\\\\=625\\\\\text{RHS}=2.2^{4+1}+100\\\\=64+100\\\\=164[/tex]

 LHS >RHS

Hence this statement is true for, n=4.

⇒Suppose this statement is true for, n=k.

 [tex]5^k\geq 2*2^{k+1}+100[/tex]

                      -------------------------------------------(1)

Now, we will prove that , this statement is true for, n=k+1.

[tex]5^{k+1}\geq 2*2^{k+1+1}+100\\\\5^{k+1}\geq 2^{k+3}+100[/tex]

LHS

[tex]5^{k+1}=5^k*5\\\\5^k*5\geq 5 \times(2*2^{k+1}+100)----\text{Using 1}\\\\5^k*5\geq (3+2) \times(2*2^{k+1}+100)\\\\ 5^k*5\geq 3\times (2^{k+2}+100)+2 \times(2*2^{k+1}+100)\\\\5^k*5\geq 3\times(2^{k+2}+100)+(2^{k+3}+200)\\\\5^{k+1}\geq (2^{k+3}+100)+3\times2^{k+2}+400\\\\5^{k+1}\geq (2^{k+3}+100)+\text{Any number}\\\\5^{k+1}\geq (2^{k+3}+100)[/tex]

Hence this Statement is true for , n=k+1, whenever it is true for, n=k.

Hence Proved.

Answer:

  a) y = 4e^(0.06t)

  b) 0.06

  c) 11.55 hours

  d) 6.9 million

Step-by-step explanation:

When the growth rate (millions per hour) is proportional to the number (millions), the relationship is exponential. The growth rate is the constant of proportionality.

a) Formula for y(t):

  y = 4e^(0.06t)

__

b) The growth constant is 0.06, the multiplier of t in the exponential function. It is the constant of proportionality in the given differential equation:

  y' = 0.06y.

__

c) The doubling time is found from ...

  2 = e^(0.06t) . . . the multiplying factor is 2 to double the original number

  ln(2) = 0.06t . . . . taking natural logs

  ln(2)/0.06 = t ≈ 11.55 h . . . . doubling time

__

d) Put t=9 into the formula from part (a). After 9 hours, there will be ...

  y(9) = 4e^(0.06·9) ≈ 6.9 . . . . million bacteria present

Answer:

y = 4e^(0.06t).

Step-by-step explanation:

dy/dt = 0.06y

Solving:

dy = 0.06y dt

dy/y = 0.06dt

Integrating both sides:

ln y = 0.06t + C

y = e^(0.06t + C)

y = Ae^(0.06t)   where A is a constant.

At t = 0 , y = 4 million so

y = 4 = Ae^0 = A

So the formula is

y = 4e^(0.06t).

Answer:

R={(10,2),(10,4),(10,6),(2,2),(2,4),(2,6),(4,2),(4,6)}

Step-by-step explanation:

We are given that

A={10,1,2,3,4,5,61}

B={0,1,2,3,4,5,6,7,81}

We are given that R be the relation from  A to B

R={gcd(a,b)=2,a[tex]\inA,b\inB[/tex]}

Gcd=Greatest common divisor  of a and b.

We have to find the elements in R

(10,2)=2,(10,4)=2,(10,6)=2

(2,2)=2,(2,4)=2,(2,6)=2

(4,2)=2,(4,6)=2

Therefore, R={(10,2),(10,4),(10,6),(2,2),(2,4),(2,6),(4,2),(4,6)}

Answer:

the answer is D

Step-by-step explanation:

Answer:

The answer is [tex]x=a^{-1}cb^{-1}[/tex].

Step-by-step explanation:

First, it is important to recall that the group law is not commutative in general, so we cannot assume it here. In order to solve the exercise we need to remember the axioms of group, specially the existence of the inverse element, i.e., for each element [tex]g\in G[/tex] there exist another element, denoted by [tex]g^{-1}[/tex] such that [tex]gg^{-1}=e[/tex], where [tex]e[/tex] stands for the identity element of G.

So, given the equality [tex] axb=c [/tex] we make a left multiplication by [tex]a^{-1}[/tex] and we obtain:

[tex]a^{-1}axb =a^{-1}c. [/tex]

But, [tex]a^{-1}axb = exb = xb[/tex]. Hence, [tex]xb = a^{-1}c[/tex].

Now, in the equality [tex]xb = a^{-1}c[/tex] we make a right multiplication by [tex]b^{-1}[/tex], and we obtain

[tex] xbb^{-1} = a^{-1}cb^{-1}[/tex].

Recall that [tex]bb^{-1}=e[/tex] and [tex]xe=x[/tex]. Therefore,

[tex]x=a^{-1}cb^{-1}[/tex].

Find the difference per row:

10 seats in the first row

30 seats in the sixth row:

30 -10 = 20 seats difference.

6-1 = 5 rows difference.

20 seats /  5 rows = 4 seats per row.

This means for every additional row, there are 4 more seats per row.

The equation would be:

Sn = S +(n-1)*d

Where d is the difference = 4

S = number of seats from starting row = 10

n = the number of rows wanted

S(11) = 10 + (11-1)*4

S(11) = 10 + 10*4

S(11) = 10 + 40

S(11) = 50

Check:

Row 6 = 30 seats

Row 7 = 30 + 4 = 34 seats

Row 8 = 34 + 4 = 38 seats

Row 9 = 38 + 4 = 42 seats

Row 10 = 42 + 4 = 46 seats

Row 11 = 46 + 4 = 50 seats.

Answer:

a = 2, b = -9, c = 3

Step-by-step explanation:

Replacing x, y values of the points in the equation y = a*x^2 + b*x +c give the following:  

(-1,14)

14 = a*(-1)^2 + b*(-1) + c  

(2,-7)

-7 = a*2^2 + b*2 + c  

(5, 8)  

8 = a*5^2 + b*5 + c  

Rearranging:

a - b + c = 14  

4*a + 2*b + c = -7

25*a + 5*b + c = 8

This is a linear system of equations with 3 equations and 3 unknows. In matrix notation the system is A*x = b whith:

A =

1    -1  1

4    2  1

25  5  1

x =

a  

b

c

b =

14

-7

8

Solving A*x = b gives x = Inv(A)*b, where Inv(A) is the inverse matrix of A. From calculation software (I used Excel) you get:

inv(A) =  

0.055555556 -0.111111111   0.055555556

-0.388888889 0.444444444   -0.055555556

0.555555556 0.555555556 -0.111111111

inv(A)*b

2

-9

3

So, a = 2, b = -9, c = 3

Answer:

$ 679.43

Step-by-step explanation:

Since, the amount formula in compound interest is,

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

Where,

P = initial amount,

r = annual rate of interest,

n = number of compounding periods in a year,

t = number of years,

Here, P = $ 4,000, r = 4% = 0.04, t = 4 years, n = 1,

So, the amount after 4 years would be,

[tex]=4000(1+0.04)^4\approx \$4679.43[/tex]

Hence, the amount of interest,

[tex]I=A-P=4679.43-4000 = \$ 679.43[/tex]

Answer:

There is a 9.6% probability that a person is diagnosed as having cancer.

Step-by-step explanation:

In this problem, we have these following probabilities:

A 5% probability that an adult over 40 has cancer.

This also means that:

There is a 95% probability that an adult over 40 does not have cancer. (Since either the adult has cancer or does not have cancer, and the sum of the probabilities is 100%).

A 78% probability of a person that has cancer being diagnosed,

A 6% probability of a person that does not have cancer being diagnosed.

What is the probability that a person is diagnosed as having cancer?

[tex]P = P_{1} + P_{2}[/tex]

[tex]P_{1}[/tex] is the probability of those who have cancer being diagnosed. So it is 78% of 5%. So

[tex]P_{1} = 0.05*0.78 = 0.039[/tex]

[tex]P_{2}[/tex] is the probability of those who do not have cancer being diagnosed. So it is 6% of 95%. So

[tex]P_{1} = 0.06*0.95 = 0.057[/tex]

So

[tex]P = P_{1} + P_{2} = 0.039 + 0.057 = 0.096[/tex]

There is a 9.6% probability that a person is diagnosed as having cancer.

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

Answer:

The remainder is 3.

Step-by-step explanation:

We have to find out,

[tex]10^z-1(mod 4)=?\text{ where }z\geq 2[/tex]

If z = 2,

[tex]10^{2}-1=100-1=99[/tex]

∵ 99 ( mod 4 ) = 3,

Suppose,

[tex](10^{k}-1)(mod 4)=3\forall \text{ k is an integer greater than 2,}[/tex]

Now,

[tex](10^{k+1}-1) ( mod 4)[/tex]

[tex]= (10^k.10 - 10+9)(mod 4)[/tex]

[tex] = 10(mod 4)\times (10^k-1)(mod 4 ) + 9 ( mod 4)[/tex]

[tex]= (2\times 3)(mod 4) + 1[/tex]

[tex]=2+1[/tex]

[tex]=3[/tex]

Hence, our assumption is correct.

The remainder of [tex]10^z -1[/tex] divided by 4 is 3 where, z ≥ 2.