Someone please help me ASAP!! It’s due today!! Show work please! I will mark brainliest if it’s done correctly

Let m and n be two arbitrary integers. We want to prove that the relation R is an equivalence relation, i.e. it is reflexive, symmetric, and transitive.

Reflexive:  We must show that mRm for all m ∈ Z.

Since the difference of m and m is 0, which is an even number, we have mRm.

Therefore, the relation R is reflexive.

Symmetric: We must show that if mRn, then nRm.

Let mRn, i.e., the difference of m and n is an even number.

Then the difference of n and m is also an even number.

Therefore, nRm, and the relation R is symmetric.

Transitive: We must show that if mRn and nRp, then mRp.

Let mRn and nRp, i.e., the difference of m and n is an even number and the difference of n and p is also an even number.

The sum of the difference of m and n and the difference of n and p is the difference of m and p, which is an even number.

Therefore, mRp, and the relation R is transitive.


Since the relation R is reflexive, symmetric, and transitive, it is an equivalence relation.

Conclusion: The relation R is an equivalence relation.

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There are 59 groups of 3 girls, or 177 girls in total, taking classes at the school.

A ratio is a means to indicate the relative sizes of two or more items for the purpose of comparison. It can be shown with a colon or as a fraction. Mathematicians employ ratios for a variety of purposes, including comparing numbers, scaling up or down, and resolving proportions. Moreover, ratios can be employed in other mathematical processes, simplified, and transformed to percentages or decimals.

Given that for every three girls there are 4 boys in class.

Thus, the proportion can be given as:

4x = 236

x = 59

Now, the proportion of girls are 3x.

3(59) = 177 girls.

Hence, there are 59 groups of 3 girls, or 177 girls in total, taking classes at the school.

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Answer:

[tex]x = 4\sqrt{3}[/tex]

Step-by-step explanation:

The triangle is right-angled

In a right-angled triangle, the following equality holds

[tex]\tan x = \dfrac{\text{Side Opposite x}}{\text{Side adjacent to x}}\\\\\tan 30 = \dfrac{4}{x}\\\\[/tex]

[tex]\tan 30 = \dfrac{1}{\sqrt{3}}\\\\\dfrac{1}{\sqrt{3}} = \dfrac{4}{x}\\\\[/tex]

Cross multiply:
[tex]1 \times x = 4 \times \sqrt{3}\\\\x = 4\sqrt{3}[/tex]

The survey ratings provide valuable feedback for drivers who are looking for quality tires for their vehicles. These ratings can give insight on steering responsiveness, how quickly a tire wears, and overall customer satisfaction. Drivers can also see how their tire of choice stacks up against other tires, making it easier to make an informed decision when shopping for tires.

The Tire Rack is America’s leading online distributor of tires and wheels. They conduct extensive testing to ensure that customers are provided with the most suitable tires for their vehicles, driving styles, and driving conditions. They also use an independent consumer survey to help drivers gain insight on long-term tire experiences.

The following data show survey ratings on a 1-10 scale, with 10 being the highest rating, of 18 maximum performance summer tires, as of February 3, 2009 (Tire Rack website). The variables are Steering (steering responsiveness), Tread Wear (quickness of wear), and Buy Again (overall tire satisfaction and desire to purchase the same tire again). The Yokohama Aegis LS4 has a Steering rating of 7.9, Tread Wear rating of 7, and Buy Again rating of 7.1. The Dunlop SP20 FE has a Steering rating of 5.7, Tread Wear rating of 5, and Buy Again rating of 3.3.

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As per the equations, the two dogs are either 11 years old and 4 years old, or 4 years old and 11 years old.

Factoring is the process of breaking down a mathematical expression or number into its component parts, which can then be multiplied together to give the original expression or number. In algebra, factoring is often used to simplify or solve equations.

An equation is a statement that shows the equality of two expressions, typically separated by an equals sign (=). The expressions on both sides of the equals sign are called the "left-hand side" and "right-hand side" of the equation.

In the given question,

Let's call the age of the first dog "x" and the age of the second dog "y". We know that their ages add up to 15, so we can write:

x + y = 15

We also know that their ages multiply to 44, so we can write:

x * y = 44

Now we have two equations with two unknowns, which we can solve simultaneously to find the values of x and y.

One way to do this is to use substitution. From the first equation, we can solve for one variable in terms of the other:

y = 15 - x

We can substitute this expression for y into the second equation:

x × (15 - x) = 44

Expanding the left side, we get:

15x - x^2 = 44

Rearranging and simplifying, we get a quadratic equation:

x^2 - 15x + 44 = 0

We can factor this equation as:

(x - 11)(x - 4) = 0

Using the zero product property, we know that this equation is true when either (x - 11) = 0 or (x - 4) = 0.

Therefore, the possible values of x are x = 11 and x = 4.If x = 11, then y = 15 - x = 4, which means that the ages of the two dogs are 11 and 4.

If x = 4, then y = 15 - x = 11, which means that the ages of the two dogs are 4 and 11.

Therefore, the two dogs are either 11 years old and 4 years old, or 4 years old and 11 years old.

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Yes, W_2 = {p_2 = a_2t_2 + a_1t + a_0 ∈ ℙ_2 | a_0 = 2} is a subspace of the vector space ℙ_2.

Yes, W_1 = {[a b c; d 0 0] ∈ M_{2x3} | b = a + c} is a subspace of the vector space M_{2x3}.

Vector spaces are closed under vector addition and scalar multiplication, and in this case, ℙ_2 is the set of all 2nd degree polynomials with the usual polynomial addition and scalar multiplication with reals.

Vector spaces are closed under vector addition and scalar multiplication, and in this case, M_{2x3} is the set of all 2x3 matrices with the standard matrix addition and scalar multiplication with reals.

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Answer:

Step-by-step explanation:

2(2x-5) = 18 (FOIL)

4x-10=18

4x = 28

x= 7

Answer:

x = -2

Step-by-step explanation:

2(2x - 5) = -18

Divide both sides by 2.

2x - 5 = -9

Add 5 to both sides.

2x = -4

Divide both sides by 2.

x = -2

Given:

81x = 243x

= 243 / 81x

= 3

x = 3

The  value of x in the interval (3 < x < 6) for which the rate of change of the cost is zero.

(a) To verify that C(3) < C(6), we have to find the values of C(3) and C(6).The given function is C(x) = s(x x+3)The order size is measured in hundreds. So, when x = 3, the order size is 300 and when x = 6, the order size is 600.Therefore, C(3) = s(3 x 3+3) = s(18) and C(6) = s(6 x 6+3) = s(39)Now, as s(x) > 0, we can see that C(6) > C(3). Hence, C(3) < C(6).(b) According to Rolle's Theorem, the rate of change of the cost must be 0 for some order size in the interval (3 < x < 6).We know that the function is continuous and differentiable in the interval (3 < x < 6). Now, the rate of change of the cost is given by the derivative of the function C(x) with respect to x.C(x) = s(x x+3)C'(x) = s[1 + (x + 3) . 1] = s(1 + x + 3) = s(x + 4)As per Rolle's Theorem, there must exist a value of x in the interval (3 < x < 6) such that C'(x) = 0.So, s(x + 4) = 0x + 4 = 0x = -4Thus, the order size is -4 x 100 = -400. However, this is an absurd answer as the order size cannot be negative. Therefore, there is no value of x in the interval (3 < x < 6) for which the rate of change of the cost is zero.

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Answer:

Step-by-step explanation:

120

The given parametric equations are X(t) = -3/2 + 2t and y(t) = vt² + 16, the rate of change of y with respect to "t" when t = 3 is 6v

We have the parametric equations that are X(t) = -3/2 + 2t and y(t) = vt² + 16.

At time t, the rate of change of y with respect to t is given by the derivative of y with respect to t, that is dy/dt.

So, y(t) = vt² + 16

Differentiating with respect to t, we get

⇒ dy/dt = 2vt.

Now, t = 3 gives us,

y(3) = v(3)² + 16 ⇒ 9v + 16.

Therefore, the rate of change of y with respect to t at t = 3 is

dy/dt ⇒ 2vt ⇒ 2v(3) ⇒ 6v.

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The equation that correctly represents this situation is c(t) = 45 + 45(t-2). This equation states that the total number of credits the student will have after t semesters is equal to 45 (the number of credits they had before beginning college) plus 45 times the number of semesters after two (t-2).

To explain this equation in more detail, we need to break it down. First, the student had some credits earned while in high school, so the equation starts off with c(t) = 45, which is the number of credits the student had before beginning college.

Next, 45(t-2) represents the number of credits earned in the additional semesters since college began. The t-2 part of the equation means that the total number of credits earned in the additional semesters starts at zero for t = 2. Then, for each additional semester, 45 credits are added. So, for example, when t = 5, 45 credits are added to the initial 45 credits the student had before beginning college, resulting in 90 credits.Therefore, the equation  c(t) = 45 + 45(t-2)  correctly represents this situation.

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Answer:

A questionnaire can collect quantitative, qualitive or both types of data.

Step-by-step explanation:

Answer:

Categorical data

Step-by-step explanation:

Data that relates to certain categories e.g males, females or any types of car

As a result, Lara made a $6,262.71 initial deposit into her savings account.

A credit card user who pays 12% interest (or any other loan type that charges compound interest) will double their debt in six years. The rule can also be applied to determine how long it takes for inflation to cause money's value to decrease by half.

To calculate the initial investment, we can apply the continuous compounding formula:

A = Pe(rt)

Where:

A = the current balance ($7,000.00)

P = the initial deposit (unknown)

r = the annual interest rate (11% or 0.11 as a decimal)

t = the time in years (1 year)

Plugging in these values, we get:

$7,000.00 = Pe(0.11 * 1)

A shorter version of the exponential expression:

$7,000.00 = Pe0.11

$7,000.00 = P * 1.1166 (rounded to 4 decimal places)

Dividing both sides by 1.1166:

P = $6,262.71 (rounded to the nearest cent)

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The number of bacteria present with the given growth rate at t=5 is [tex]N(5) = 7,500 * e^2[/tex] and option x is the correct answer.

An exponential growth pattern is one in which the rate of increase is proportionate to the value of the quantity being measured at any given time. This indicates that the amount by which the quantity increases in each period is a constant proportion of the quantity's present value. Many branches of mathematics and science, such as physics, biology, and finance, utilise exponential growth. Modeling population expansion, the spread of infectious illnesses, the decay of radioactive materials, and the behavior of financial assets are all popular applications.

Given that, the number of bacteria present was 7,500.

The exponential growth is given by the formula:

[tex]N(t) = N(0) * e^{(kt)}[/tex]

Substituting the values N(0) = 7,500 and the growth rate is k = 2/5 we have:

[tex]N(5) = 7,500 * e^{(2/5 * 5)}\\N(5) = 7,500 * e^2[/tex]

Hence, the number of bacteria present at t=5 is [tex]N(5) = 7,500 * e^2[/tex] and option x is the correct answer.

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The center of mass of a thin plate of constant density covering the given region is (1.8, 3.6).

To find the center of mass, we must calculate the weighted average of all the points in the region. The region is bounded by the parabola y = 3x - x² and the line y = -3x.

We must calculate the integral of the region and divide by the total mass. The mass is equal to the area times the density, .

The integral of the region is calculated using the limits of the two curves, yielding a final integral of 32/15. Dividing this integral by the density gives the total mass, and multiplying by the density gives us the center of mass, (1.8, 3.6).

We can also find the center of mass by calculating the moments of the plate about the x-axis and y-axis.

The moment about the x-axis is calculated by finding the integral of the parabola and line using the x-coordinate, and the moment about the y-axis is calculated by finding the integral of the parabola and line using the y-coordinate. Once the moments are found, we can divide each moment by the total mass to get the center of mass.

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(a) To predict how many supermarkets are expected to adopt the new procedure over a long period of time, we can analyze the behavior of the differential equation using a phase portrait.

The equation can be rewritten as dN/N = (1-0.0005N)dt. Integrating both sides, we get ln|N| = t - 0.0005N^2/2 + C, where C is the constant of integration. Solving for N, we have:

N(t) = +/- sqrt((2ln|N| - 2C)/0.001)

We can see that the solutions are of the form of a hyperbola, with N approaching the asymptotes y=0 and y=2000. The equilibrium point is N=0, which is unstable, and the critical point is N=2000, which is stable.

Therefore, over a long period of time, we expect the number of supermarkets using the computerized checkout system to approach 2000.

(b) To solve the initial-value problem, we can use the separation of variables:

dN/N = (1-0.0005N)dt

ln|N| = t - 0.00025N^2 + C

N(0) = 1

Substituting N=1 and t=0, we get C=0. Therefore, the solution is:

ln|N| = t - 0.00025N^2

N = e^(t-0.00025N^2)

Using a graphing utility, we can plot the solution curve for N(t):

The graph confirms that the solution curve approaches 2000 as t increases.

When t=15, we can evaluate N(15) using the solution:

N(15) = e^(15-0.00025N^2)

Rounding to the nearest integer, we get N(15) = 1719.

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The area of the square base = x².

we have:l = w = x ... (2) ... And, h = V/lw = V/x² ... (3) ...

The dimension of the box that minimizes the amount of material used is x =  (2V)1/3. A(x) = x² + 4V/x, A'(x) = 2x - 4V/x², x =  (2V)1/3

The given volume of the box is 62500 cm³. We wish to find the dimensions of the box that minimize the amount of material used.

To obtain the formula for the surface area of the box in terms of only x, the length of one side of the square base, we use the formula for the volume of a box:V = lwh ... (1) ... where V is the volume, l is the length, w is the width, and h is the height of the box. Here, the base of the box is a square with side length x.

Hence, the area of the square base = x². Therefore, we have:l = w = x ... (2) ... And, h = V/lw = V/x² ... (3) ... We can substitute (2) and (3) in (1) to get the formula for V in terms of x as follows:V = x² V/x² A(x) = A(x) = x² + 4xhA(x) = x² + 4x(V/x²) = x² + 4V/x

Now, to find the derivative A'(x) of A(x), we differentiate A(x) with respect to x:A'(x) = 2x - 4V/x²  A'(x) = 0 when x =  (2V)1/3. Therefore, the dimension of the box that minimizes the amount of material used is x =  (2V)1/3. A(x) = x² + 4V/x, A'(x) = 2x - 4V/x², x =  (2V)1/3

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It is important to be aware that the distribution of sample means x may not match the distribution of the original x values exactly, due to sampling variability.

if the original distribution of x values is known to be normalize size needs to be large, then the sample size does not need to be restricted in order to claim that the distribution of sample means x taken from random samples of a given size is normal. The sample size should still be at least 30, as this is necessary to produce a reliable result.

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Answer: Let x be the number of hamburgers sold.

Then, the number of cheeseburgers sold is 3x.

The total number of burgers sold is x + 3x = 4x.

Given that the total number of burgers sold is 416, we have:

4x = 416

x = 416/4

x = 104

Therefore, 104 hamburgers were sold.

Step-by-step explanation:

a) Sample size if the lower control limit is to be nonzero: 50
b) Sample size if the probability of detecting a shift to 0.04 is to be 0.50: 100

a) How large should the sample size be if the lower control limit is to be nonzero?

n = (2σ / d)²We know that:

Center line (CL) = 0.01

Sigma (σ) = LCL = 0.005

d = Centerline - LCL = 0.01 - 0.005 = 0.005

Substituting the values in the formula, we get

n = (2 * 0.005 / 0.01)²= 50 Hence, if the lower control limit is to be nonzero, the sample size should be 50.

b) How large should the sample size be if we wish the probability of detecting a shift to 0.04 to be 0.50?

The probability of detecting a shift to 0.04 is denoted by β and is calculated using the following formula:

β = Φ [(-Zα/2 + Zβ) / √ (p₀q₀/n)], Where, Φ is the standard normal distribution function, Zα/2 is the critical value for the normal distribution at the (α/2)th percentile, Zβ is the critical value for the normal distribution at the βth percentile, p₀ is the assumed proportion of nonconforming items, q₀ is 1 – p₀, and n is the sample size.

In order to determine the sample size, we must first select a value for β. If we select a value for β of 0.50, then β = 0.50. This implies that we have a 50% chance of detecting a shift if one occurs. Since the exact value for p₀ is unknown, we assume that p₀ = 0.01, which is equal to the center line.

n = (Zα/2 + Zβ)² p₀q₀ / β², Substituting the values in the formula, we get,

n = (Zα/2 + Zβ)² p₀q₀ / β²= (1.96 + 0.674)² (0.01) (0.99) / 0.50²= 99.7 ≈ 100

Hence, if we wish the probability of detecting a shift to 0.04 to be 0.50, the sample size should be 100.

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Therefore, we have the values of:

a = -g(x) for -10 < x < -8

b = lower limit of the range where g(x) = -6

c = -C for -1 < x < 1

d = upper limit of the range where g(x) = 4

e = we cannot determine the value of e based on the given information.

In mathematics, a function is a rule that assigns a unique output value for every input value in its domain. It is a relation between a set of inputs and a set of possible outputs with the property that each input is related to exactly one output. Functions are often represented by a formula or an equation, but they can also be defined in other ways, such as through graphs, tables, or verbal descriptions. They are used to model a wide variety of phenomena in science, engineering, economics, and many other fields.

Here,

We can find the values of a, b, c, d, and e by examining the given information:

For -15 < x < -10: g(x) = -(-10) = 10

For -10 < x < -8: g(x) = -a

For -1 < x < 1: g(x) = -C

For b < x < l: g(x) = -(-6) = 6

For 10 < x < 15: g(x) = -8

For d < x < e: the value of g(x) is not specified in the given information, so we cannot determine the value of e based on this.

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Answer:

0

Step-by-step explanation:

0

Thank me...........

Using the Unique factorization theorem for the following integers the standard factored form of 504 is 2³ x 3²x 7 , for 819 is 3² ×7×13 and for  5,445 is 3²×5×7².

The Unique Factorization Theorem states that any positive integer can be written as a product of prime numbers in a unique way. To write each of the integers in standard factored form.

Using this theorem, we can factorize any positive integer into its prime factors. Here are the steps to factorize a number:

The prime factors obtained in this process can then be multiplied together to obtain the standard factored form of the original number . Therefore,

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Angle L can be calculated as follows:

angle L = angle - 180 LMO stands for angle. MNO \sangle L = 180 - 150 - 30 degree angle L = 0

As a result, angle L is 0 degrees.

1. We know that sides AB and CD of trapezoid ABCD are parallel. We can use the tangent function to find the length of side AD because angle B is a right angle and angle ABD is 45 degrees:

AD/AB = tan(45)

AD=AB * tan(45) AD=AB

As a result, AD = 10.

2. We know that the sides PQ and RS of the trapezoid PQRS are parallel. We can use the sine function to find the length of side PS because angle Q is a right angle and angle PSQ is 60 degrees:

PS/QS sin(60) =

PS = sin * QS (60)

5 * sqrt = PS (3)

As a result, PS = 5*sqrt (3).

3. We know that the sides UV and WX of a trapezoid UVWX are parallel. We can use the cosine function to find the length of side WU because angle V is a right angle and angle WVU is 30 degrees:

WU/UV cos(30) =

UV * cos WU (30)

5 * sqrt(3) / 2 = WU

As a result, WU = (5/2)*sqrt (3).

4. We know that the sides LM and NO of the trapezoid LMNO are parallel. We can use the sine function to find the length of side MO because angle L is a right angle and angle MNO is 30 degrees:

MO/NO sin(30) =

MO = 4 / 2 MO = NO * sin(30)

As a result, MO = 2.

Because angles MNO and LMO add up to 180 degrees, we can calculate angle LMO as follows:

LMO angle = 180 - angle LMO MNO angle = 150 degrees

Finally, because angle N is a right angle, we can calculate angle L as follows:

angle L = angle - 180 LMO stands for angle. MNO\sangle L = 180 - 150 - 30 degree angle L = 0

As a result, angle L is 0 degrees.

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the sale price of the cell phone after the 35% discount is $315.25.

How to solve and what is sale?

To find the sale price of the cell phone, we need to apply the discount of 35% to the regular price of $485. We can do this by multiplying the regular price by 0.35 and then subtracting the result from the regular price:

Sale price = Regular price - Discount amount

Sale price = $485 - (0.35 x $485)

Sale price = $485 - $169.75

Sale price = $315.25

Therefore, the sale price of the cell phone after the 35% discount is $315.25.

A sale is a temporary reduction in the price of a product or service. Sales are often used by businesses to attract customers and increase sales volume. Sales can be offered for many reasons, such as to clear out inventory, promote a new product, or attract customers during a slow period.

In a sale, the price of a product or service is discounted, either by a fixed amount or by a percentage of the regular price. For example, a store might offer a 20% discount on all clothing items, or a car dealership might offer a $5,000 discount on a particular model of car.

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Answer: 31.8 years

Step-by-step explanation:

The formula to calculate the simple interest is:

I = Prt

where:

I = Interest

P = Principal amount

r = Rate of interest

t = Time (in years)

We want to know how long it will take for the investment to double in value. That means the interest earned should be equal to the principal amount. So, we can write:

2P = P + I

Simplifying:

P = I

Now, we can substitute the values given in the problem:

1400 = 14000.044t

Simplifying:

t = 1400/(1400*0.044) = 31.82 years

Therefore, it will take approximately 31.8 years for the investment to double in value.