1) Meredith f ound some lace at a price of 4.0 5/meter in Ireland that she like too much for it. The same lace in the Canada would sell for $5.99/yd. W (S1 0.498 E) (1 yard 3 ft)

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.

For making mathematical induction, we need:

An [tex]n_0[/tex] for which the relation holds true

if its true for [tex]n_i[/tex], then, is true for [tex]n_{i+1}[/tex]

the 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]

lets 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.

Answer:

number of way is [tex]\frac{59!}{50)!9!}[/tex]

Step-by-step explanation:

given data:

total number of problems 100

total points for each problem 5

let ten problems are

[tex]x_1, x_2,........., x_{10}[/tex]

according to the given information

[tex]x_1 +x_2 +.......+x_{10} = 100[/tex]

[tex]x_i \geq 5[/tex]

where i =1 + 10

so, number of way integer point can assign are

   [tex]^{(n+r-1)}C_{(r-1)}[/tex]

where

r = 10

[tex]n = 100 - 10\times 5 = 50[/tex]

so, we have

[tex]^{(59)}C_{9}[/tex]

[tex]\frac{59!}{59-9)!9!}[/tex]

number of way is [tex]\frac{59!}{50)!9!}[/tex]

Answer:

2.063X10^-4

0.999

0.833

3.26X10^-3

9.85x10^-3

Step-by-step explanation:

B="At least one component needs service during the warranty period"

C="All three components need service during the warranty period"

D="Only the receiver needs service during the warranty period"

E="Exactly one of the three components needs service during the warranty period"

P(A1)=0.0595

P(A2)=0.0598

P(A3)=0.058

a) P(A1∩A2∩A3)=P(A1)P(A2)P(A3)=0.0595*0.0598*0.058=2.063X10^-4

b) P(B)=1-P(A1∩A2∩A3)=1-2.063X10^-4=0.999

c) P(C)=P(A1'∩A2'∩A3')=P(A1')P(A2')P(A3')=0.9405*0.9402*0.942=0.833

d) P(D)=P(A1'∩A2∩A3)=P(A1')P(A2)P(A3)=0.9405*0.0598*0.058=3.26X10^-3

e) P(E)=P(A1'∩A2∩A3)+P(A1∩A2'∩A3)+P(A1∩A2∩A3')=3.26X10^-3 + 0.0595*0.9402*0.058+0.0595*0.0598*0.942=9.85x10^-3

Answer:

Step-by-step explanation:

Let M be the heights of men in the United States and W be the heights of women in the United States

Given that M is N(69.1, 2.9) and W (63.7, 2.7)

For a husband and wife we have average height as

[tex]\frac{x+y}{2}[/tex]=Z (say)

[tex]Mean =E(z) = \frac{1}{2} [E(M)+E(W)] = 66.4 inches[/tex]

Var (Z) =[tex]Var (\frac{x+y}{2} )=\frac{1}{4}[Var(x)+ Var(y)+2 cov (x,y)]\\[/tex]

=[tex]0.25[2.9^2+2.7^2+2*r*sx*sy]\\= 0.25(20.398)\\=5.0995[/tex]

Mean = 66.4" and std dev = 2.258

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.

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.

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.

Answer:

0=6

Step-by-step explanation:

-0.25yxyx0=6

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.

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.

Answer:

90 mi/h

Step-by-step explanation:

Given,

For first 30 miles, her speed is 30 miles per hour,

Let x be her speed in miles per hour for another 30 miles,

Since, here the distance are equal in each interval,

So, the average speed of the entire journey

[tex]=\frac{\text{Average speed for first 30 miles + Average speed for another 30 miles}}{2}[/tex]

[tex]=\frac{30+x}{2}[/tex]

According to the question,

[tex]\frac{30+x}{2}=60[/tex]

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

[tex]\implies x = 90[/tex]

Hence, she needs to go 90 miles per hour for remaining 30 miles.

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.

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.

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.

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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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

The child ingested approximately 20.18 milligrams of ferrous sulfate, which is determined by unit conversion.

Given data:

First, let's convert the 1.6 fluid ounces to millilitres:

[tex]1.6 \text{ fluid ounces} \times 29.5735 \text{ ml/oz} \approx 47.3176 \text{ ml}[/tex]

Now, calculate the amount of ferrous sulfate in milligrams:

[tex]\text{Concentration of ferrous sulfate} = \frac{2}{3} \text{ gr per 5 ml} \\= \frac{2}{3} \times \frac{1}{5} \text{ gr/ml}[/tex]

Since 1 grain (gr) is approximately equal to 64.79891 milligrams (mg), we'll convert the concentration to milligrams per millilitre (mg/ml):

[tex]\text{Concentration of ferrous sulfate} = \frac{2}{3} \times \frac{1}{5} \times 64.79891 \text{ mg/ml}[/tex]

To calculate the total amount of ferrous sulfate ingested:

[tex]\text{Amount of ferrous sulfate} = 47.3176 \text{ ml} \times \frac{2}{3} \times \frac{1}{5} \times 64.79891 \text{ mg/ml}[/tex]

Calculating this gives:

[tex]\text{Amount of ferrous sulfate} \approx 20.1817 \text{ mg}[/tex]

Therefore, the child ingested approximately 20.18 milligrams of ferrous sulfate.

Learn more about the unit conversion here:

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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]

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.

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.

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.

Answer:

a.Neither injective nor surjective

b.Injective but not surjective

c.Surjective but not injective

Step-by-step explanation:

Injective function:It is also called injective function.If f(x)=f(y)

Then, x=y

Surjective function:It is also called Surjective function.

If function is onto function then Range of function=Co-domain of function

a.We are given that

[tex]f:R\rightarrow R[/tex]

[tex]f(x)=x^2[/tex]

It is not injective because

f(1)=1 and f(-1)=1

Two elements have same image.

If function is one-to-one then every element have different image.

Function is not surjective because negative elements have not pre-image in R

Therefore, Co-domain not equal to range.

Given function neither injevtive nor surjective.

b.[tex]f:N\rightarrow N[/tex]

[tex]f(n)=n^2[/tex]

If [tex]f(n_1)=f(n_2)[/tex]

[tex]n^2_1=n^2_2[/tex]

[tex]n_1=n_2[/tex]

Because N={1,2,3,...}

Hence, function is Injective.

2,3,4,.. have no pre- image in N

Therefore, function is not surjective

Because Range not equal to co-domain.

Hence, given function is injective but not Surjective.

c.[tex]f:Z\times Z \rightarrow Z[/tex]

f(n,k)=n+k

It is not injective because

f(1,2)=1+2=3

f(2,1)=2+1=3

Hence, by definition of one-one function it is not injective.

For every element belongs to Z we can find pre- image in [tex]Z\times Z[/tex]

Hence, function is surjective.

For function (a) f(x) = x² and function (b) f(n) = n², neither is injective or surjective. Function (c) f(n, k) = n + k is both injective and surjective.

The functions injective and surjective are important concepts in mathematics that pertain to the type of mappings functions perform from one set to another. An injective, or one-to-one, function assigns distinct outputs to distinct inputs, while a surjective, or onto, function ensures that every element of the function's codomain is mapped to by at least one element of its domain.

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.

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.

Answer:

205.4 miles

Step-by-step explanation:

As per the question,

We have been provided that an airplane started flies 165 miles from point A  in the direction of 130 degrees and after that, it changes its direction to 245 degrees and travel to 80 miles again, which is drawn in the figure as below:

From the figure,

Let XY be the plane.

Now,

By using the cosine rule in triangle APR, we get

AP² = AR² + PR² - 2 × AR × PR × cos 115°

Now put the value from figure:

AP² = 165² + 80² - 2 × 165 × 80 × (-0.325)

AP² = 42205

AP = 205.4 miles.

Hence, the airplane is 205.4 miles far from point A.

Answer:

The frequency of the car is 0.0233743 laps/sec and 1.4024 laps were completed in 60 seconds

Step-by-step explanation:

We are given that he fastest ever qualifying lap was in 1987 by Bill Elliott, with a time of 42.782 seconds.

[tex]Frequency = \frac{1}{Time}[/tex]

[tex]Frequency = \frac{1}{42.782}[/tex]

[tex]Frequency = 0.0233743[/tex]

No. of laps completed in 42.782 seconds. = 1

No. of laps completed in 60 seconds. = [tex]\frac{1}{42.782} \times 60[/tex]

                                                                = [tex]1.4024[/tex]

Hence the frequency of the car is 0.0233743 laps/sec and 1.4024 laps were completed in 60 seconds

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.

Answer:

a) The variable of the study is: milligrams of nitrogen per liter of water.

This is the amount that needs to be measured and analyzed to reach conclusions in the study.

b) The variable is quantitative. The quantitative variables are those that represent quantities. This variables can be measured on a continuous or discrete scale. Then, all the variables that you can measure or count are quantitative variables(height of trees, number of passengers per car, wind speed, milligrams of nitrogen per liter, etc). On the other hand, qualitative variables are those that can’ t be measured, and they represent attributes, like apple colors (red, green), size of trousers (small, medium, large) and so on.

c) The population under study is the milligrams of nitrogen per liter of water that are in the entire lake. You can estimate the parameters of the population by taking samples (In the example, 28 samples are taken).

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.

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.

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).

Now, 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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