Write a short statement that expresses a possible relationship between the variables. (size of file, time to download the file)

Answer:

{ b, d, f }

Step-by-step explanation:

In the roster form we write the elements of a set by separating commas and enclose them within {} bracket.

We have give,

[tex]U = \{a, b, c, d, e, f, g\}[/tex],

[tex]A = \{a, c, e, g\}[/tex],

[tex]B = \{a, b, c, d\}[/tex],

[tex]\because A' = U - A[/tex]

[tex]\implies A' = \{a, b, c, d, e, f, g\} - \{a, c, e, g\}[/tex]

= { b, d, f }

Final answer:

A' is the complement of set A, consisting of elements in the universal set U that are not in A. Given U = {a, b, c, d, e, f, g} and A = {a, c, e, g}, A' is found to be {b, d, f}.

Explanation:

The complement of a set A, denoted by A', consists of all the elements in the universal set U that are not in A. Given the universal set U = {a, b, c, d, e, f, g} and set A = {a, c, e, g}, to find A' we need to identify all elements in U that are not in A.

To find A', we compare each element in U with the elements in A:

If the element is in A, we do not include it in A'.

If the element is not in A, we include it in A'.

After comparing, we find that A' consists of the elements {b, d, f} since these are the elements in U that are not present in set A.

Therefore, the roster form of A' with a single space after each comma is:

{b, d, f}

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Answer: The relation is a function.

Step-by-step explanation: In this situation, we are given a relation in the form of a graph and we are asked if it represents a function. In this situation, we would you something called the vertical line test. In other words, if we can draw a vertical line that passes through more than one point on the graph, then the relation is not a function. Notice that in this problem, it's impossible to draw a vertical line that passes through more than one point on the graph so the relation is a function.

Therefore, this relation must be a function.

Answer:

2 mL

Step-by-step explanation:

Given:

Volume of water for injection = 23 mL

Resulting concentration = 200,000 units per mL

Amount of drug in the vial = 5,000,000 units

Now,

Let the final volume of the solution be 'x' mL

Now, concentration = [tex]\frac{\textup{units of the powder}}{\textup{Total volume of the soltuion}}[/tex]

thus,

200,000 = [tex]\frac{\textup{5,000,000}}{\textup{x}}[/tex]

or

x = 25 mL

also,

Total volume 'x' = volume of water + volume of powder

or

25 mL = 23 + volume of powder

or

Volume of powder = 2 mL

Answer:  300

Step-by-step explanation:

Given : Hugh and Janet are dining out at a cafe that has on its menu 6 entrees, 5 salads and 10 desserts.

They decide thy will each order a different entree, salad, and dessert, and share the portions.

Then, the number of different meals are possible :-

[tex]6\times5\times10=300[/tex]

Hence, the number of different meals are possible =300

Answer: There were 469,532 American prison for drug related offenses in 2015.

Step-by-step explanation:

Since we have given that

Number of Americans in prison for drug related offenses in 1980 = 40,900

Rate of increment in Americans in prison from 1980 to 2015 = 1048%

So, Number of Americans who were in prison for drug related offenses in 2015 is given by

[tex]\dfrac{100+1048}{100}\times 40900\\\\\dfrac{1148}{100}\times 40900\\\\=1148\times 409\\\\=469,532[/tex]

Hence, there were 469,532 American prison for drug related offenses in 2015.

Answer:

Proved

Step-by-step explanation:

To prove that if n is a perfect square, then n+1 can never be a perfect square

Let n be a perfect square

[tex]n=x^2[/tex]

Let [tex]n+1 = y^2[/tex]

Subtract to get

[tex]1 = y^2-x^2 =(y+x)(y-x)[/tex]

Solution is y+x=y-x=1

This gives x=0

So only 0 and 1 are consecutive integers which are perfect squares

No other integer satisfies y+x=y-x=1

Answer:

Y = [tex]e^{t}[/tex] +  [tex]te^{t}[/tex] + [tex]t^{2} e^{t}[/tex] + t - 18

Step-by-step explanation:

y''' − 3y'' + 3y' − y = ex − x + 21

Homogeneous solution:

First  we propose a solution:

Yh = [tex]e^{r*t}[/tex]

Y'h = [tex]r*e^{r*t}[/tex]

Y''h = [tex]r^{2}*e^{r*t}[/tex]

Y'''h = [tex]r^{3}*e^{r*t}[/tex]

Now we solve the following equation:

Y'''h - 3*Y''h + 3*Y'h - Yh = 0

[tex]r^{3}*e^{r*t}[/tex] - 3*[tex]r^{2}*e^{r*t}[/tex] + 3*[tex]r*e^{r*t}[/tex] - [tex]e^{r*t}[/tex] = 0

[tex]r^{3} - 3r^{2} + 3r - 1 = 0[/tex]

To solve the equation we must propose a solution to the  polynomial :

r = 1

To find the other r we divide the polynomial by (r-1) as you can see  

attached:

solving the equation:

(r-1)([tex]r^{2} - 2r + 1[/tex]) = 0

[tex]r^{2} - 2r + 1[/tex] = 0

r = 1

So we have 3 solution [tex]r_{1} = r_{2} =r_{3}[/tex] = 1

replacing in the main solution

Yh =  [tex]e^{t}[/tex] +  [tex]te^{t}[/tex] + [tex]t^{2} e^{t}[/tex]

The t and [tex]t^{2}[/tex] is used because we must have 3 solution  linearly independent

Particular solution:

We must propose a Yp solution:

Yp = [tex]c_{1} (t^{3} + t^{2} + t + c_{4} )e^{t} + c_{2} t + c_{3}[/tex]

Y'p = [tex]c_{1}(t^{3} + t^{2} + t + c_{4} )e^{t} + c_{1}( 3t^{2} + 2t + 1 )e^{t} + c_{2}[/tex]

Y''p = [tex]c_{1}(t^{3} + t^{2} + t + c_{4} )e^{t} + c_{1}(6t + 2)e^{t}[/tex]

Y'''p = [tex]c_{1}(t^{3} + t^{2} + t + c_{4} )e^{t} + 6c_{1}e^{t}[/tex]

Y'''p - 3*Y''p + 3*Y'p - Yp = [tex]e^{t} - t + 21[/tex]

[tex]6c_{1}e^{t} - 18c_{1} te^{t} - 6c_{1} e^{t} + 6c_{1} te^{t} + 9c_{1} t^{2} e^{t} + 3c_{1}e^{t} + 3c_{2} - c_{2} t -  c_{3}[/tex] = [tex]e^{t} - t + 21[/tex]

equalizing coefficients of the same function:

- 12c_{1} = 0

9c_{1} = 0

3c_{1} = 0

c_{1} = 0

3c_{2} - c_{3} = 21 => c_{5} = [tex]\frac{1}{3}[/tex]

-c_{2} = -1

c_{2} = 1

c_{3} = -18

Then we have:

Y = [tex]e^{t}[/tex] +  [tex]te^{t}[/tex] + [tex]t^{2} e^{t}[/tex] + t - 18

Answer:

[tex]y(x) = e^{3x} [Acos x+Bsin x][/tex]

Step-by-step explanation:

Given is a differential equation

[tex]y" - 6y' +10y=0.[/tex]

We have characteristic equation as

[tex]m^2-6m+10 =0[/tex]

The above quadratic cannot be factorised hence use formula

[tex]m=\frac{6+/-\sqrt{36-40} }{2} \\=3+i, 3-i[/tex]

Hence general solution would be

[tex]y(x) = e^{3x} [Acos x+Bsin x][/tex]

Answer:

3 : 8

Step-by-step explanation:

Let x quantity of 10% ethanol is mixed with y quantity of 65% ethanol to obtain 50% ethanol mixture,

Thus, the total quantity of resultant mixture = x + y

Also, ethanol in 10% ethanol mixture + ethanol in 65% ethanol mixture = ethanol in resultant mixture,

⇒ 10% of x + 65% of y = 50% of (x+y)

[tex]\implies \frac{10x}{100}+\frac{65y}{100}=\frac{50(x+y)}{100}[/tex]

⇒ 10x + 65y = 50(x+y)

⇒ 10x + 65y = 50x+50y

⇒ 10x - 50x = 50y - 65y

⇒ -40x = -15y

[tex]\implies \frac{x}{y}=\frac{15}{40}=\frac{3}{8}[/tex]

Answer:

The dosage has 1.2ml

Step-by-step explanation:

This problem 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.

First step: The first step is the conversion from mcg to mg.

The bottle has 60 mcg. How much is this in ml? 1 mcg has 0.001mg. So:

1 mcg - 0.001mg

60mcg - x mg

x = 60*0.001

x = 0.06mg

Final step: The bottle of elixir contains 0.05 mg per 1 ml. Calculate the dosage in ml.

The dosage has 0.06 mg, so:

0.05mg - 1 ml

0.06mg - xml

0.05x = 0.06

[tex]x = \frac{0.06}{0.05}[/tex]

x = 1.2ml

The dosage has 1.2ml

Answer:

The solution is: [tex](x_{1}, x_{2}, x_{3}) = (1,0,-4)[/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]2x_{1} - x_{2} + 3x_{3} = -10[/tex]

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

[tex]x_{1} - 5x_{2} + 2x_{3} = -7[/tex]

This system has the following augmented matrix:

[tex]\left[\begin{array}{ccc}2&-1&3|-10\\1&-2&1|-3\\1&-5&2| -7\end{array}\right][/tex]

To make the reductions easier, i am going to swap the first two lines. So

[tex]L1 <-> L2[/tex]

Now the matrix is:

[tex]\left[\begin{array}{ccc}1&-2&1|-3\\2&-1&3|-10\\1&-5&2| -7\end{array}\right][/tex]

Now we reduce the first row, doing the following operations

[tex]L2 = L2 - 2L1[/tex]

[tex]L3 = L3 - L1[/tex]

So, the matrix is:

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

Now we divide L2 by 3

[tex]L2 = \frac{L2}{3}[/tex]

So we have

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

Now we have:

[tex]L3 = 3L2 + L3[/tex]

So, now we have our row reduced matrix:

[tex]\left[\begin{array}{ccc}1&-2&1|-3\\0&1&\frac{1}{3}|\frac{-4}{3}\\0&0&2| -8\end{array}\right][/tex]

We start from the bottom line, where we have:

[tex]2x_{3} = -8[/tex]

[tex]x_{3} = \frac{-8}{2}[/tex]

[tex]x_{3} = -4[/tex]

At second line:

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

[tex]x_{2} - \frac{4}{3} = -\frac{4}{3}[/tex]

[tex]x_{2} = 0[/tex]

At the first line

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

[tex]x_{1} - 4 = -3[/tex]

[tex]x_{1} = 1[/tex]

The solution is: [tex](x_{1}, x_{2}, x_{3}) = (1,0,-4)[/tex]

Answer:

number of 1 bedroom units are 96

number of 1 bedroom units are 64

number of 1 bedroom units are 32

Step-by-step explanation:

Let the number of 1 bedroom units be 'a'

number of 2 bedroom units be 'b'

and,

number of 3 bedroom units be 'c'

now,

according to the question

a + b + c = 192 ................. (1)

also,

b + c = a ..............(2)

and,

a = 3c ...................(3)

now,

substituting value of 'a' from 3  into 2, we get

b + c = 3c

or

b = 2c ...................(4)

also,

from 3 and 1

3c + b + c = 192

or

4c + b = 192  ................(5)

now from 4 and 5,

4c + 2c = 192

or

6c = 192

or

c = 32 units

now, substituting c in equation 4, we get

b = 2 × 32 = 64 units

and, substituting c in equation (3), we get

a = 3 × 32 = 96 units

Therefore,

number of 1 bedroom units are 96

number of 1 bedroom units are 64

number of 1 bedroom units are 32

Answer:

The equation of the line is:

[tex]y = x - 3[/tex]

Step-by-step explanation:

The general equation of a straight line is given by:

[tex]y = ax + b[/tex]

Being given two points, we can replace x and y, solve the system and find the values for a and b.

Solution:

The line goes through the point [tex](2,-1)[/tex]. It means that when [tex]x = 2, y = -1[/tex]. Replacing in the equation:

[tex]y = ax + b[/tex]

[tex]-1 = 2a + b[/tex]

[tex]2a + b = -1[/tex]

The line also goes through the point [tex](5,2)[/tex]. It means that when [tex]x = 5 y = 2[/tex]. Replacing in the equation:

[tex]y = ax + b[/tex]

[tex]2 = 5a + b[/tex]

[tex]5a + b = 2[/tex]

Now we have to solve the following system of equations:

[tex]1) 2a + b = -1[/tex]

[tex]2) 5a + b = 2[/tex]

From 1), we have:

[tex]b = -1 - 2a[/tex]

Replacing in 2)

[tex]5a - 1 - 2a = 2[/tex]

[tex]3a = 3[/tex]

[tex]a = \frac{3}{3}[/tex]

[tex]a = 1[/tex]

[tex]b = -1 - 2a = -1 - 2 = -3[/tex]

The equation of the line is:

[tex]y = x - 3[/tex]

Answer:

Step-by-step explanation:

Let's say that each side of the square is of length [tex]s[/tex]. The area of this square would then be:

[tex]A = s^{2}[/tex]

If we increase the length of each side by 50%, then the length becomes [tex]1.5s[/tex], which will result in the area being:

[tex]A = (1.5s)(1.5s)[/tex]

[tex]A = 2.25s^{2}[/tex]

This means the area has increased by [tex]1.25[/tex] the original amount, or 125%.

Answer:

Part 1.

In 12 hours 30 units of  Humulin R insulin in 300 mL is to be infused.

So, per hour = [tex]\frac{30}{12}= 2.5[/tex] units are to be infused.

Part 2.

Now 30 units Humulin R insulin in 300 mL of normal saline (NS).

So, 2.5 units will be in : [tex]\frac{300\times2.5}{30}= 25[/tex] mL

Hence. 2.5 units Humulin R insulin in 25 mL of normal saline (NS).

Answer:

MNP is a Congruent Triangle according to Side-Side-Side Congruence.

Step-by-step explanation:

Whenever we talk about Euclidean Geometry and Congruence of Triangles. We are taking into account that in any given plane, three given points, in this case, M, N, and P is and its segments between two points make up a Triangle.

In this case MNP and an identical one and PNM

To be called congruent, it's necessary to have the same length each side and when it comes to angles, congruent angles have the same measure.

A postulate, cannot be proven since it's self-evident. And there's one that fits for this case which says

"Every SSS (Side-Side-Side) correspondence is a congruence"

So this is why we can classify MNP as Side-Side-Side congruence since its segments are the same size MN, NP, and MP for both of them. The order of the letters does not matter.

Answer:

a) The patient weighs 75.15kg = 75.1kg, rounded to the nearest tenth.

b) The patient should 75.1mg a day of the drug.

c) The patient should receive 0.37mL per dose, rounded to the nearest hundreth.

Step-by-step explanation:

These problems can be solved by direct rule of three, in which we have cross multiplication.

a. How many kilograms does the patient weigh?

The problem states that patient weighs 167lb. Each lb has 0.45kg. So:

1 lb - 0.45kg

167 lb - xkg

[tex]x = 167*0.45[/tex]

[tex]x = 75.15[/tex]kg

The patient weighs 75.15kg = 75.1kg, rounded to the nearest tenth.

b. How many milligrams should the patient receive per day?

The drug has 1mg/kg. The patient weighs 75.1kg. So

1 mg - 1 kg

x mg - 75.1kg

[tex]x = 75.1[/tex]mg

The patient should 75.1mg a day of the drug.

c. How many milliliters should the patient receive per dose?

The drug is SubQ, q12h. This means that the drug is administered twice a day, so there are 2 doses. 75.1mg of the drug are administered a day. so:

2 doses - 75.1mg

1 dose - xmg

[tex]2x = 75.1[/tex]

[tex]x = \frac{75.1}{2}[/tex]

[tex]x = 37.5[/tex] SubQ, q12h

For each dose, the patient should receive 37.5mg. Each 40mg of the drug has 0.4mL. So:

40mg - 0.4ml

37.5mg - xmL

[tex]40x = 0.4*37.4[/tex]

[tex]x = \frac{0.4*37.4}{40}[/tex]

[tex]x = 0.374mL[/tex]

The patient should receive 0.37mL per dose, rounded to the nearest hundreth.

Answer:

Angle A = 60º and Angle B = 30º

Step-by-step explanation:

We know that A and B are complementary angles, therefore ∠A + ∠B =90. (+)

On the other hand, ∠A is half as large as ∠B; this can be written algebraically as [tex]A=\frac{B}{2}[/tex]. (*)

If we substitute (*) in (+) we get:

[tex]\frac{B}{2}+B = 90\\\\ \frac{B+2B=180}{2} \\\\ B+2B=180\\\\ 3B=180\\\\ B=\frac{180}{3}\\\\ B=60[/tex]

And now we substitute the value of B in (+) and we get:

∠A+60 = 90

∠A = 90-60

∠A = 30

Answer:

98w +115(18-w)

Step-by-step explanation:

The fees per credit hour at Westside are $98.

The fees per credit hour at Pinewood are $115.

The total amount of hours is 18, this means the hours at Westside plus the hours at Pinewood add up to 18.

The problem asks us to name "w" the credit hours at Westside, therefore, the credit hours at Pinewood would be "18 - w"

Combining this information, we have that the combined total amount  he paid would be:

(Fee per hour at Westside)(hours at Westside) + (Fee per hour at Pinewood)(hours at Pinewood) = Total

⇒98w + 115(18 - w) = Total

Final answer:

The expression for the combined total cost of Ravi's classes is 98w + 115(18 - w), where w represents the number of credit hours taken at Westside College, and (18 - w) represents the number of credit hours taken at Pinewood College.

Explanation:

To calculate the combined total cost of Ravi's classes at Westside Community College and Pinewood Community College, we can create an expression based on the number of credit hours he is taking at each school. If he is taking w credit hours at Westside, where the cost is $98 per credit hour, the total cost at Westside would be 98w. Since Ravi is taking a total of 18 credit hours at both schools combined, he would be taking (18 - w) credit hours at Pinewood, where the cost is $115 per credit hour. The total cost at Pinewood would therefore be 115(18 - w).

The expression for the combined total amount paid for Ravi's classes would be the sum of the costs for each school:

Total Paid = 98w + 115(18 - w)

In this type of problems what we have to do is unit conversion. In order to do so we need all the equivalences which we will be mentioning during the explanation of the problem:

First of all the answer is asked to be in acre-feet and we can see the data we are getting from the rain is in [tex]in*Km^{2}[/tex] not even a volume unit.

To calculate the volume of poured rain we need to have both numbers in the same units, we will convert [tex]Km^{2}[/tex] to [tex]in^{2}[/tex] using the equivalence [tex]1 Km^{2}=1550001600in^{2}[/tex] like this:

[tex]101Km^{2}*\frac{1550001600in^{2} }{1Km^{2}}=156550161600in^{2}[/tex]

it is possible now to calculate the volume ([tex]Volume_{cuboid}=Area*Height[/tex]) like this:

[tex]Volume_{cuboid}=156550161600in^{2}*2.5in=391375404000in^{3}[/tex]

Now we just need to convert this volume to acre-feet and we will do so using the equivalence [tex]1acre-foot=751271680in^{3}[/tex] like this:

[tex]391375404000in^{3}*\frac{1acre-foot}{751271680in^{3}}=5199.50403658 acre-feet[/tex]

5199.50403658 acre-feet would be the answer to our problem

Final answer:

Explaining the impact of selling additional sandwiches on MTA's profits.

Explanation:

To prepare the schedule showing the impact on MTA's profits:

Calculate the total income from selling 800 additional sandwiches: $2.85 x 800 = $2,280.

Calculate the total cost of producing 800 sandwiches: $5 x 800 = $4,000.

Subtract the cost from the income to find the profit impact: $2,280 - $4,000 = lower profit.

Explanation (150 words): Adding 800 sandwiches at a cost of $5 each while selling at $2.85 results in a loss, impacting MTA's profits negatively. This shift reduces the overall profitability due to the increased production costs outweighing the revenue generated from the additional sandwich sales.

Answer: The total purchase price is $ 29,032

Step-by-step explanation:

Hi, to solve this problem you have to solve the percentages of each taxes first.

So :

The next step is adding the taxes results to the speedboat cost.

so:

speedboat = $24,500

speedboat + taxes = $24,500 + $1,347.5 +$3,185 = $29,032

The total purchse price for the speedboat is $29,032.

Answer:

I think that what you are trying to show is:  If [tex]s[/tex] is irrational and [tex]r[/tex] is rational, then [tex]r+s[/tex]  is rational. If so, a proof can be as follows:

Step-by-step explanation:

Suppose that [tex]r+s[/tex] is a rational number. Then [tex]r[/tex] and [tex]r+s[/tex] can be written as follows

[tex]r=\frac{p_{1}}{q_{1}}, \,p_{1}\in \mathbb{Z}, q_{1}\in \mathbb{Z}, q_{1}\neq 0[/tex]

[tex]r+s=\frac{p_{2}}{q_{2}}, \,p_{2}\in \mathbb{Z}, q_{2}\in \mathbb{Z}, q_{2}\neq 0[/tex]

Hence we have that

[tex]r+s=\frac{p_{1}}{q_{1}}+s=\frac{p_{2}}{q_{2}}[/tex]

Then

[tex]s=\frac{p_{2}}{q_{2}}-\frac{p_{1}}{q_{1}}=\frac{p_{2}q_{1}-p_{1}q_{2}}{q_{1}q_{2}}\in \mathbb{Q}[/tex]

This is a contradiction because we assumed that [tex]s[/tex] is an irrational number.

Then [tex]r+s[/tex] must be an irrational number.

Final answer:

The height of the pile after 3 weeks is 43 181/192 inches.

Explanation:

To find the height of the pile after 3 weeks, we need to add the increase in height for each week. The initial height of the pile is 17 3/4 inches. For each week, the height increases by 8 7/12 inches.

So after 1 week, the height is (17 3/4 + 8 7/12) inches.

After 2 weeks, the height is [(17 3/4 + 8 7/12) + 8 7/12] inches.

And after 3 weeks, the height is [((17 3/4 + 8 7/12) + 8 7/12) + 8 7/12] inches.

Let's calculate:

So, after 3 weeks, the height of the pile of newspapers is 43 181/192 inches.

Answer:

[tex] (x-4)^2+(y+5)^2=90[/tex]

Step-by-step explanation:

The equation of a circle of radius r, centered at the point (a,b) is

[tex](x-a)^2+(y-b)^2=r^2[/tex]

We already know the center is at [tex](4,-5)[/tex], we are just missing the radius. To find the radius, we can use the fact that the circle passes through the point (7,4), and so the radius is just the distance from the center to this point (see attached image). So we find the distance by using distance formula between the points (7,4) and (4,-5):

radius[tex]=\sqrt{(7-4)^2+(4-(-5))^2}=\sqrt{3^2+9^2}=\sqrt{90}[/tex]

And now that we know the radius, we can write the equation of the circle:

[tex] (x-4)^2+(y-(-5))^2=\sqrt{90}^2[/tex]

[tex] (x-4)^2+(y+5)^2=90[/tex]

Answer:

You must pick at least 13 pennies to be sure of getting at least five from the same year.

Step-by-step explanation:

You have:

12 1967 pennies

7 1868 pennies

11 1971 pennies.

how many must you pick to be sure of getting at least five pennies from the same year?

This value is the multiplication of the number of different pennies by the antecessor of the number of pennies you want, added by 1.

So

You have 3 differennt pennies

You want to get at least five from the same year.

[tex]3*4 + 1 = 13[/tex]

You must pick at least 13 pennies to be sure of getting at least five from the same year.

For example, if you pick 12 pennies, you can have four from each year.

Adding three values, 13 is the smallest number that you need at least one term of the addition being equal or bigger than 5.

To be sure of getting at least five pennies from the same year, you need to pick 35 pennies in total.

To ensure that you get at least five pennies from the same year, you need to consider the worst-case scenario. In this case, the worst-case scenario is where you pick pennies from each of the three different years first before getting five from the same year. So, you need to pick the maximum number of pennies from each year first before reaching the desired goal. The maximum number of pennies you need to pick is:

12 + 7 + 11 + 5 = 35 pennies

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Answer:

[tex]m\angle B=m\angle C=120^{\circ}[/tex]

[tex]m\angle A=m\angle D=60^{\circ}[/tex]

Step-by-step explanation:

Trapezoid ABCD is isosceles trapezoid, because AB = CD (given). In isosceles trapezoid, angles adjacent to the bases are congruent, then

Since BK ⊥ AD, the triangle ABK is right triangle. In this triangle,  AB = 8, AK = 4. Note that the hypotenuse AB is twice the leg AK:

[tex]AB=2AK.[/tex]

If in the right triangle the hypotenuse is twice the leg, then the angle opposite to this leg is 30°, so,

[tex]m\angle ABK=30^{\circ}[/tex]

Since BK ⊥ AD, then BK ⊥ BC and

[tex]m\angle KBC=90^{\circ}[/tex]

Thus,

[tex]m\angle B=30^{\circ}+90^{\circ}=120^{\circ}\\ \\m\angle B=m\angle C=120^{\circ}[/tex]

Now,

[tex]m\angle A=m\angle D=180^{\circ}-120^{\circ}=60^{\circ}[/tex]

Final answer:

The number 4320 is an abundant number because the sum of its proper divisors exceeds 4320. All positive multiples of 6 greater than 6 are abundant because their smallest divisor, 6, is the sum of its proper divisors, and they have additional divisors that increase the sum.

Explanation:

To determine if the number 4320 is perfect, abundant, or deficient, we must compare the sum of its proper divisors (excluding the number itself) with the number. A perfect number is equal to the sum of its divisors. A number is abundant if the sum of its divisors is greater than the number, and it is deficient if the sum is less.

For 4320, the divisors are 1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 16, 18, 20, 24, 27, 30, 32, 36, 40, 45, 48, 54, 60, 64, 72, 80, 90, 96, 108, 120, 135, 144, 160, 180, 192, 216, 240, 270, 288, 320, 360, 432, 480, 540, 576, 720, 864, 1080, and 1440. Their sum is greater than 4320, so it is an abundant number.

Regarding positive multiples of 6 greater than 6 being abundant, the smallest divisor of such a multiple is always 6, which is already the sum of its divisors (1, 2, and 3). Since there are additional divisors beyond 1, 2, and 3, the sum of divisors must exceed the number, making it abundant.