a college student takes the same number of credits each semester. before beginning college, the student had some credits earned while in high school. after 2 semesters, the student had 45 credits, and after 5 semesters, the student had 90 credits. if c(t) is the number of credits after t semesters, write the equation that correctly represents this situation.

The Length of AB in square ABCD is 30 feet.

Since the squares ABCD and EFGH are similar, their corresponding sides are proportional, so we can set up the following relation:

AB/EF = 1/4

We can also use the fact that the ratio of the areas of two similar figures is equal to the square of the ratio of their corresponding sides. Therefore,

AB²/EF² = (Area of square ABCD)/(Area of square EFGH)

Substituting the given values:

AB²/EF² = (Area of square ABCD)/(14400)

Since the areas of squares are proportional to the square of their sides, we can write,

Area of square ABCD/Area of square EFGH = (AB/EF)²

Substituting this into the above equation and solving for AB, we get,

AB²/EF² = (AB/EF)²

AB² = (AB/EF)² * EF²

AB² = (1/4)² * 14400

AB² = 900

AB = 30 feet

Therefore, the length of the side AB of square ABCD is 30 feet.

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a. S = {(1, -1), (2, 1)}Let's begin by calculating the determinant of the matrix composed of the vectors of S, and checking if it is equal to 0. Because the two vectors are not colinear, they should span R2.|1 -1||2 1| determinant is not 0, therefore S spans R2. No geometric description is required for this example.

b. S = {(1, 1)} The set S contains one vector. A set containing only one vector cannot span a plane because it only spans a line. Therefore, S does not span R2. Geometric description: S spans a line that passes through the origin (0, 0) and the point (1, 1).c. S = {(0, 2), (1, 4)} Let's again begin by calculating the determinant of the matrix composed of the vectors of S, and checking if it is equal to 0.|0 2||1 4| determinant is 0, thus S does not span R2. In this scenario, S only spans the line that contains both vectors, which is the line with the equation y = 2x.

Geometric description: S spans a line that passes through the origin (0, 0) and the point (1, 2).

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The rate of change of the distance for limousine is less than the rate of change of the convertible.

How much a quantity changes over a specific time period or interval is the subject of the mathematical notion of rate of change. Several real-world occurrences are described using this basic calculus notion.

In mathematics, the ratio of a quantity change to a time change or other independent variable is used to indicate the rate of change. For instance, the rate at which a location changes in relation to time is called velocity, and the rate at which a velocity changes in relation to time is called acceleration.

The equation of the distance travelled by the convertible is given as:

y = 35x

The equation of the limousine can be calculated using the coordinates of the graph (1, 30) and (2, 60).

The slope is given as:

slope = (change in y) / (change in x) = (60 - 30) / (2 - 1) = 30

Using the point slope form:

y - 30 = 30(x - 1)

y = 30x

So the equation of the limousine is y = 30x.

Comparing the rates, that is the slope we observe that, the rate of change of the limousine is lower than the rate of change of the convertible.

Hence, the rate of change of the limousine is less than the rate of change of the convertible.

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The reason why the terms a and m have to be relatively prime integers is that it is the only way to make sure that ax≡1 (mod m) is solvable for x within the integers modulo m.

Theorem:"If a and m are relatively prime integers and m > 1, then an inverse of a modulo m exists. Furthermore, this inverse is unique modulo m. (That is, there is a unique positive integer a less than m that is an inverse of a modulo m and every other inverse of a modulo m is congruent to a modulo m.)"If a and m are relatively prime integers and m > 1, then an inverse of a modulo m exists. Furthermore, this inverse is unique modulo m. (That is, there is a unique positive integer a less than m that is an inverse of a modulo m and every other inverse of a modulo m is congruent to a modulo m.)The inverse of a modulo m is another integer, x, such that ax≡1 (mod m).

This theorem has an interesting explanation: if a and m are not co-prime, then there is no guarantee that ax≡1 (mod m) has a solution in Zm. The reason for this is that if a and m have a common factor, then m “absorbs” some of the factors of a. When this happens, we lose information about the congruence class of a, and so it becomes harder (if not impossible) to undo the multiplication by .This is the reason why the terms a and m have to be relatively prime integers.

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Answer:

To use an area model to determine the quotient of 556 and 16, we can divide a rectangle of area 556 into 16 equal parts. Each part will have an area of 556/16.

We can start by dividing 556 into 16 groups of 10 (160), and then into 16 groups of 3 (48). That leaves us with a remainder of 4.

So we have:

556 = 16 x 34 + 48 + 4

This shows that 556 can be written as 16 times some whole number (34) plus a remainder of 48 + 4/16.

Simplifying the remainder, we have:

48 + 4/16 = 48 + 1/4 = 48.25

Therefore, the quotient of 556 and 16 is:

556/16 = 34 1/4

The quotient of 556 and 16 using an area model can be determined by producing a rectangle with the total area of 556 and one side of 16. The length of the other side will be the quotient. In this case, the quotient is 34 3/4.

When asked to determine the quotient of 556 and 16 using the area model, one way to think of this is making a rectangle. The total area is 556 and one side is 16. The length of the other side will be the quotient.

Start by first estimating how many times 16 could fit into 556. Let's take 30 as an estimate, because 30*16 = 480, which is relatively close to 556. Draw a rectangle with the width of 16 and the length of 30.

Find the difference between the rectangle's area and 556. So, 556 - 480 = 76. Now, 76 is our remaining area to fill. 16 goes into 76 four more times, adding up to 64.

There is still a leftover area, which is 76-64 = 12. This is smaller than our width of 16. So, your final answer is 34 12/16 or 34 3/4 in simplest form.

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The answer to the question is 334 kittens.

Given that a cat gave birth to 333 kittens who each had a different mass between 147 g and 159 g. Then the cat gave birth to a 4th kitten with a mass of 57 g.

First of all, we will find out the range of the mass of kittens. The range is given as follows;Range = Maximum Value - Minimum Value Range = 159 g - 147 g Range = 12 g

Now, the cat gave birth to a 4th kitten with a mass of 57 g, we can say that the minimum value of kitten's mass is 57 g.So, the maximum value of kitten's mass can be calculated as follows;Maximum Value = 57 g + Range Maximum Value = 57 g + 12 g Maximum Value = 69 g Now, we can say that all kittens with a mass of 69 g or less would be born because the minimum value of kitten's mass is 57 g and the range of mass is 12 g.

Therefore, the answer to the question is 334 kittens.

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Answer: $22.50

Step-by-step explanation:

Let x be the cost per pound of the secret-formula coffee beans.

The total cost of the secret-formula beans is 12x dollars.

The total cost of the other beans is 15 × 18 = 270 dollars.

The total cost of the mix is (12 + 15) × 20 = 540 dollars.

Since the barista mixed 12 pounds of the secret-formula beans with 15 pounds of the other beans, the total weight of the mix is 12 + 15 = 27 pounds.

We can set up an equation based on the total cost of the mix:

12x + 270 = 540

Subtracting 270 from both sides:

12x = 270

Dividing both sides by 12:

x = 22.5

Therefore, the barista's secret-formula coffee beans cost $22.50 per pound.

By using differentials to estimate the maximum error when measuring the volume of the sphere if the possible error in measuring the radius is 0.5. It will be 32,572.03 in³. Which is option (d).

The possible error in measuring the radius of the sphere is 0.5 in

The formula for the volume of a sphere is given by V(r) = 4/3πr³

The volume of the sphere when r=72 in is given by V(72) = 4/3π(72)³

When r= 72 + 0.5 in= 72.5 in, the volume of the sphere can be calculated using the formula:

V(72.5) = 4/3π(72.5)³

The difference between these two volumes, V(72) and V(72.5), gives us the maximum error while measuring the volume of a sphere. It can be calculated as follows:

V(72.5) - V(72) = 4/3π(72.5)³ - 4/3π(72)³= 4/3π [ (72.5)³ - (72)³ ]= 4/3π [ (72 + 0.5)³ - 72³ ]= (4/3)π [ 3(72²)(0.5) + 3(72)(0.5²) + 0.5³ ]≈ (4/3)π [ 777.5 ]= 3.28 × 10⁴ in³

Therefore, the maximum error while measuring the volume of a sphere with a radius of 72 in, where the possible error in measuring the radius is 0.5 in, is approximately 3.28 × 10⁴ in³ or 32,572.03 in³. Therefore coorect option is (D).

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The average number of hours of sleep for working college students is between 6.28 and 6.72 hours of sleep each night

According to the given data,

Sample size n = 101

Sample mean x = 6.5

Standard deviation s = 2.14

Level of confidence C = 96%

Using a 96% confidence level, the value of t* for 100 degrees of freedom is 2.081, as given in the question.

Now, the formula for the confidence interval is:x ± (t* × s/√n)Here, x = 6.5, s = 2.14, n = 101, and t* = 2.081

Substituting the values in the above formula, we get:

Lower limit = x - (t* × s/√n) = 6.5 - (2.081 × 2.14/√101) = 6.28

Upper limit = x + (t* × s/√n) = 6.5 + (2.081 × 2.14/√101) = 6.72

Therefore, the 96% confidence interval for the average number of hours of sleep for working college students is between 6.28 and 6.72 hours of sleep each night.

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Anna should use 10 mL of the 72% salt solution and 20 mL of the 54% salt solution to make 30 mL of a 60% salt solution

Let's assume that Anna will use x mL of the 72% salt solution, and therefore she will use (30 - x) mL of the 54% salt solution (since the total volume is 30 mL).

To find out how much of each solution Anna should use, we can set up an equation based on the amount of salt in each solution.

The amount of salt in x mL of 72% salt solution is

= 0.72x

The amount of salt in (30 - x) mL of 54% salt solution is

= 0.54(30 - x)

To make a 60% salt solution, the total amount of salt in the final solution should be

0.6(30) = 18

So we can set up an equation

0.72x + 0.54(30 - x) = 1

Simplifying the equation

0.72x + 16.2 - 0.54x = 18

0.18x = 1.8

x = 10 ml

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Answer:

12

Step-by-step explanation:

i think its right

the triangles are similar, corresponding parts of the triangles are equal in measure. Thus, it is reasonable to conclude that [tex]AB ≅ DE.[/tex]

It is reasonable to conclude that [tex]AB ≅ DE[/tex]because triangles ABC and DEF are similar.

This means that corresponding parts of the two triangles are equal in measure. Specifically, ∠A and ∠D are equal in measure, as are ∠B and ∠E.

Therefore, the corresponding sides AB and DE are equal in measure.

A way to show that the two triangles are similar is by using the AA Similarity Postulate.

This postulate states that if two angles of one triangle are equal in measure to two angles of a second triangle, then the two triangles are similar. In this case, [tex]∠A ≅ ∠D and B ≅ ∠E[/tex], which means the two triangles are similar.

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Answer:

- 1 and 7

Step-by-step explanation:

to find h(0) substitute x = 0 into h(x)

h(0) = 4(0) - 1 = 0 - 1 = - 1

to find h2) substitute x = 2 into h(x)

h(2) = 4(2) - 1 = 8 - 1 = 7

Answer:

To find the point that partitions the line segment AB in a ratio of 1:3, we can use the following formula:

P = (3B + 1A) / 4

where P is the point that partitions the line segment in a ratio of 1:3, A and B are the endpoints of the line segment, and the coefficients 3 and 1 represent the ratio of the segment we are dividing.

Substituting the values, we get:

P = (3*(-1, -5) + 1*(-5, 3)) / 4

P = (-3, -7)

Therefore, the point that partitions the line segment AB in a ratio of 1:3 is (-3, -7).

Step-by-step explanation:

The Trapezoidal rule and Simpson's rule are two methods used to approximate the value of a definite integral. The Trapezoidal rule approximates the integral by dividing the region between the lower and upper limits of the integral into n trapezoids, each with a width h. The approximate value of the integral is then calculated as the sum of the areas of the trapezoids. The Simpson's rule is similar, except the region is divided into n/2 trapezoids and then the integral is approximated using the weighted sum of the area of the trapezoids.

For the given integral 4 x x2 1 0 dx, with n = 200, the Trapezoidal rule and Simpson's rule approximate the integral to be 7.4528 and 7.4485 respectively, rounded to four decimal places. The exact value of the integral is 7.4527. The difference between the exact and approximate values is very small, thus indicating that both the Trapezoidal rule and Simpson's rule are accurate approximations.

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Answer:

The 2nd equation is false.

Step-by-step explanation:

You don't even have to solve. DE is not 58, it's 40.

The 2nd equation is false.

Answer:

If the GM between √2 and 2√2 is a find the value of a.​

Step-by-step explanation:

To find the geometric mean between two numbers, we simply take the square root of their product.

In this case, we want to find the geometric mean between √2 and 2√2.

Their product is:

√2 * 2√2 = 2√4 = 2*2 = 4

So, the geometric mean between √2 and 2√2 is the square root of 4, which is:

√4 = 2

Therefore, the value of a is 2.

The three examples of contracts are:

Contracts are legal agreements between two or more parties that involve the exchange of goods, services, or money. They can be classified as unilateral or bilateral, express or implied, executed or executory.

Here are three examples of contracts that a person can be a part of:

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(a) The probability that exactly two cars finish the race is 0.0512.

(b) The probability that at most two cars finish the race is 0.05792.

(c) The probability that at least three cars finish the race is 0.94208.

(a) To determine the probability that exactly two cars finish the race, we have to use binomial distribution. In this case, we have n = 5 trials, and p = 0.8 is the probability that a car finishes the race (1 - 0.2). Using the binomial distribution formula:

P(X = k) = (nCk)(p^k)(1 - p)^(n - k)

Where X is the number of cars that finish the race, we get:

P(X = 2) = (5C2)(0.8²)(0.2)³= (10)(0.64)(0.008)= 0.0512

Therefore, the probability that exactly two cars finish the race is 0.0512.

(b) To determine the probability that at most two cars finish the race, we have to calculate the probabilities of 0, 1, and 2 cars finishing the race and add them up.

P(X ≤ 2) = P(X = 0) + P(X = 1) + P(X = 2)= (5C0)(0.8⁰)(0.2)⁵ + (5C1)(0.8¹)(0.2)⁴ + (5C2)(0.8²)(0.2)³= 0.00032 + 0.0064 + 0.0512= 0.05792

Therefore, the probability that at most two cars finish the race is 0.05792.

(c) To determine the probability that at least three cars finish the race, we can calculate the probability of 0, 1, and 2 cars finishing the race and subtract it from 1, which gives us the probability of at least three cars finishing the race.

P(X ≥ 3) = 1 - [P(X = 0) + P(X = 1) + P(X = 2)]= 1 - (0.00032 + 0.0064 + 0.0512)= 0.94208

Therefore, the probability that at least three cars finish the race is 0.94208.

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Answer:

[tex]{ \rm{s = 12( v_{0} + v_{1} )t}} \\ \\{ \boxed { \rm{t = \frac{s}{12(v_{0} + v_{1})} \: \: }}}[/tex]

A bin can hold 28 pounds. each toy car weighs 7 ounces., so the bin can hold 64 toy cars.

To determine the number of toy cars the bin can hold, we must first convert the weight limit of the bin and the weight of the toy cars to a uniform unit of measure.

We'll then divide the weight limit of the bin by the weight of one toy car. After that, we'll multiply the resulting value by the number of ounces in one pound (16).

Here's how to solve the problem:

1 pound = 16 ounces

Therefore, a bin that can hold 28 pounds can hold:28 × 16 = 448 Ounces

The weight of one toy car is 7 ounces.

Divide the weight limit of the bin (448 ounces) by the weight of one toy car (7 ounces):

448 ÷ 7 = 64

Therefore, the bin can hold 64 toy cars.

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The parking lot has a width of around [tex]0.937[/tex] meters.

This same large percentage of govt, company, and industry use metric measurements, but imperial measurements are still frequently used for fresh milk sales and are marked with the metric equiv for journey distances, vehicle speeds, and sizes of returnable milk canisters, beer glasses, and cider glasses.

100 centimeters make up one meter. Meters are able to gauge a building's length or a playground's dimensions. 1000 meters make up one kilometer.

[tex]3760 cm = 37.6 m[/tex]

Solve for the width,

[tex]area = (1/2) * (base1 + base2) * height[/tex]

where,

base1 [tex]= 73.4 m[/tex]

base2 [tex]= 37.6 m[/tex]

area [tex]= 2,930.4[/tex] square meters

Let's solve for the height first,

[tex]height = 2 * area / (base1 + base2)[/tex]

[tex]height = 2 * 2,930.4 / (73.4 + 37.6)[/tex]

[tex]height = 2 * 2,930.4 / 111[/tex]

[tex]height = 56.16 m[/tex]

We nowadays can apply the algorithm to determine the width.

[tex]width = (area * 2) / (base1 + base2) * height[/tex]

[tex]width = (2 * 2,930.4) / (73.4 + 37.6) * 56.16[/tex]

[tex]width = 5856.8 / 111 * 56.16[/tex]

[tex]width = 5856.8 / 6239.76[/tex]

[tex]width = 0.937[/tex]

Therefore, the width of the parking lot is approximately [tex]0.937[/tex] meters.

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a block slides along a track that descends through distance h. The track is frictionless except for the lower section. There the block slides to a stop in a certain distance d because of friction. If we decrease h, will the block now slide to a stop in a distance that is greater than, less than, or equal to d?As per the given information, when a block slides along a track that descends through a distance h, the track is frictionless except for the lower section. There the block slides to a stop in a certain distance d because of friction. Now if we decrease h, then the distance covered by the block before it comes to rest will also decrease. So the block will slide to a stop in a distance that is less than d. Hence the answer is less than d.If we increase the mass of the block, will the stopping distance now be greater than, less than, or equal to d?

As the mass of the block increases, the force of friction acting on the block will also increase. Hence the stopping distance will also increase. So the stopping distance now will be greater than d. Hence the answer is greater than d.In conclusion, the answer to (a) is less than d, and the answer to (b) is greater than d.

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Keenan's z-score is 0.71, rounded to the nearest hundredth.

The z-score measures how many standard deviations an individual's score is from the mean, and can be calculated using the formula:

z = (x - μ) / σ

where x is the individual's score, μ is the mean score, and σ is the standard deviation.

For Keenan's exam:

z = (80 - 77) / 4.2

z = 0.71

Therefore, Keenan's z-score is 0.71, rounded to the nearest hundredth.

Rounding to the nearest hundredth means the rounding of any decimal number to its nearest hundredth value. In decimal, hundredth means 1/100 or 0.01. For example, the rounding of 2.167 to its nearest hundredth is 2.17.

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Solve the equation [tex]7/11=18/x+1[/tex] we find the solution is [tex]x = 27.2857[/tex]

Your example is an equation since an equation is any statement with an equals sign. Equations are frequently utilized for mathematical equations since mathematicians like equal signs. A set of instructions for achieving a certain result is called an equation.

A number, a constant, or a mix of numbers, variables, and operation symbols make up an expression. Two expressions joined by such an assignment operator form an equation.

we can cross-multiply,

[tex]7(x+1) = 11(18)[/tex]

Expanding the left side,

[tex]7x + 7 = 198[/tex]

Subtracting [tex]7[/tex] from both sides,

[tex]7x = 191[/tex]

Dividing both sides by [tex]7[/tex],

[tex]x = 191/7[/tex]

Therefore, the solution to the proportion is

[tex]x = 27.2857[/tex]

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Answer:

[tex] |x - 300| \leqslant 10[/tex]

The value of the double integral using the easier order, ydA bounded by y = x − 6; x = y² is 125/12.

The double integral, indicated by ', is mostly used to calculate the surface area of a two-dimensional figure. By using double integration, we may quickly determine the area of a rectangular region. If we understand simple integration, we can easily tackle double integration difficulties. Hence, first and foremost, we will go over some fundamental integration guidelines.

Given, the double integral ∫∫yA and the region y = x-6 and x = y²

y = x-6

x = y²

y² = y +6

y² - y - 6 = 0

y² - 3y +2y - 6 = 0

(y-3) (y+2) = 0

y = 3 and y = -2

[tex]\int\int\limits_\triangle {y} \, dA\\ \\[/tex]

= [tex]\int\limits^3_2 {y(y+6-y^2)} \, dx \\\\\int\limits^3_2 {(y^2+6y-y^3)} \, dx \\\\(\frac{y^3}{3} + 3y^2-\frac{y^4}{4} )_-_2^3\\\\\frac{63}{4} -\frac{16}{3} \\\\\frac{125}{12}[/tex]

The value for the double integral is 125/12.

Integration is an important aspect of calculus, and there are many different forms of integrations, such as basic integration, double integration, and triple integration. We often utilise integral calculus to determine the area and volume on a very big scale that simple formulae or calculations cannot.

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Answer:

P = [tex]2y^{2}[/tex] + 4xy +4

Q = [tex]-3y^{2}[/tex] + 7 -3xy

R = -3xy +8

Step-by-step explanation:

Answer:

142 sq cm

Step-by-step explanation:

A= 2(lh + wh + lw)

2(7*3+5*3+7*5)

2(21+15+35)

2(71)

A= 142 sq cm

The parabolic trajectory of the falling prey can be described by the equation y = 228 – x2/57, where y is the height above the ground and x is the horizontal distance traveled in meters. In this case, the prey was dropped at a height of 228 m and flying at 19 m/s. To calculate the total distance traveled by the prey, we can use the equation for the parabola to solve for x.

We can rearrange the equation y = 228 – x2/57 to solve for x, which gives us[tex]x = √(57*(228 – y))[/tex]. When the prey hits the ground, the height (y) is 0. Plugging this into the equation for x, we can calculate that the total distance traveled by the prey is[tex]x = √(57*(228 - 0)) = √(57*228) = 84.9 m.\\[/tex] Expressing this answer to the nearest tenth of a meter gives us the final answer of 84.9 m.

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