(v^(2)+11)/(v^(2)-3v-4) find all numbers for which the rational expression is undefined

Using the lengths of the time spent we can plot the histogram by the values in the table.

A histogram is a visual depiction of a frequency distribution with continuous classes that has been categorised.

To create a histogram, follow the steps shown below.

Mark the class intervals on the X-axis and the frequencies on the Y-axis to get started.

Both axes must have the exact same scales. Intervals between classes must be exclusive. Create rectangles with appropriate frequencies as the heights and bases acting as class intervals.

As the class limits are shown on the horizontal axis and the frequencies are shown on the vertical axis, a rectangle is constructed on each class interval.

If the intervals are identical, the height of each rectangle is proportional to the associated class frequency.

From the given information, we have the following:

Length of time (t)     Frequency

0 ≤ t < 10                    5

10 ≤ t < 25                  24

25 ≤ t < 30                 12

30 ≤ t < 50                 8

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The complete question is:

Answer:The father's position relative to the surface of the water is 4.2 meters below the surface of the water.

Step-by-step explanation:

What is an expression?

Expression in maths is defined as the collection of the numbers variables and functions by using signs like addition, subtraction, multiplication, and division.

The height relative to the water surface is calculated as:-

Height from iceberg to water surface = 0.5 meters

The height from iceberg to deep inside the water is = 4.7 meters

Father's height relative to the water surface is,

H= 4.7 - 0.5 = 4.2 meters.

The image of the condition is also attached with the answer below.

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Option:-

[tex] \: [/tex]

Given:-

[tex] \: [/tex]

Solution:-

[tex] \: [/tex]

[tex] \: [/tex]

[tex] \: [/tex]

[tex] \: [/tex]

[tex] \: [/tex]

or

[tex] \: [/tex]

[tex] \: [/tex]

[tex] \: [/tex]

[tex] \: [/tex]

━━━━━━━━━━━━━━━━━━━━━━━━━━

hope it helps! :)

The distance from the top of the cliff to the ship at C 373.4 m.

The angle of depression from the top of the cliff to the ship at C is 20.4 degrees

The distance BC = 200.2m

the distance is given as

[tex]\sqrt{130^{2} +350^{2} }[/tex]

= 373.4 m

The distance of the top of the cliff to the ship at c is 373.4 m

angle of depression is tan ∅ = 130 / 350

∅ = tan⁻¹  130 / 350

= 20.4 degrees

b. The distance BC

tan 33 = 130 / AB

AB = 130 / tan 33

= 200.2 m

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To solve for r, we can divide both sides of the equation by Pt: r = (A-P)/Pt

The formula A=P+Prt is used to calculate the total amount of money (A) after a certain period of time when a principal amount (P) is invested at a certain interest rate (r) for a certain amount of time (t). To solve for the specified variable r, we need to rearrange the formula and isolate r on one side of the equation. Here are the steps to do so:

Step 1: Subtract P from both sides of the equation to get:
A - P = P + Prt - P

Step 2: Simplify the right side of the equation to get:
A - P = Prt

Step 3: Divide both sides of the equation by Pt to get:
(A - P) / Pt = r

Step 4: Simplify the left side of the equation to get:
r = (A - P) / Pt

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The final answer is 2x^(2)-6x+21 with a remainder of -123. In polynomial division notation, this can be written as (2x^(2)-6x+21)-123/(x+5).

The polynomial division that Tiana is told to carry out is (2x^(3)+4x^(2)-9x-18)-:(x+5). To carry out this division, we can use long division.

Divide the first term of the dividend, 2x^(3), by the first term of the divisor, x. This gives us 2x^(2), which we write above the division symbol.
Multiply the divisor, x+5, by the first term of the quotient, 2x^(2), to get 2x^(3)+10x^(2).
Subtract this result from the dividend to get -6x^(2)-9x-18.
Repeat the process with the new dividend, -6x^(2)-9x-18. Divide the first term, -6x^(2), by the first term of the divisor, x, to get -6x, which we write above the division symbol next to 2x^(2).
Multiply the divisor, x+5, by the new term in the quotient, -6x, to get -6x^(2)-30x.
Subtract this result from the new dividend to get 21x-18.
Repeat the process with the new dividend, 21x-18. Divide the first term, 21x, by the first term of the divisor, x, to get 21, which we write above the division symbol next to 2x^(2)-6x.
Multiply the divisor, x+5, by the new term in the quotient, 21, to get 21x+105.
Subtract this result from the new dividend to get -123, which is our remainder.

So, the final answer is 2x^(2)-6x+21 with a remainder of -123. In polynomial division notation, this can be written as (2x^(2)-6x+21)-123/(x+5).

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Hence, a 19-gram sample would contain 2.5909 x 10⁵ germs.

How come we use measure?

We can compare unknown amounts to known values with the use of measurement. We can quantify the size, length, and speed of objects with the use of measurement. The final result won't be accurate without measurement.

We may start by calculating the bacterium to dirt sample weight ratio:

bacteria per gram = 1.5 x 10⁵ / 11 = 13636.36...

We are able to employ this ratio to determine how many bacteria are present in a 19-gram specimen:

bacteria = bacteria per gram x weight of sample

bacteria = 13636.36... x 19

bacteria = 259090.91...

We must write the number in scientific notation because it is more than 10⁵. This can be achieved by adding a factor of 10⁵ and shifting the decimal place five places to a left:

bacteria = 2.5909 x 10^5

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Answer:

B, D, E

Step-by-step explanation:

52= -4(2n+1)

= -8n-4

52= -8n-4

8n= -4-52

8n= -56

n= -7

-8n= 52+4

= 56

Answer:  The total number of milkshakes sold is 7 + 18 = 25.

The number of milkshakes sold with whipped cream is 7.

To find the percentage of milkshakes with whipped cream, we can use the formula:

percentage = (part/whole) x 100

Substituting the values, we get:

percentage = (7/25) x 100

percentage = 28

Therefore, 28% of the milkshakes had whipped cream.

Step-by-step explanation:

Jean-Luc's account will be worth approximately $740.55 after 5 years of continuous  at a compounding annual interest rate of 1.6%.

Compound interest is the addition of interest to the principal amount of a loan or investment. In other words, it is the interest earned on both the principal amount and any accumulated interest from previous periods.

Interest rate is the amount charged by a lender to a borrower for the use of money or the amount earned by an investor for lending money or investing in an asset. It is usually expressed as a percentage of the principal amount and is typically calculated on an annual basis.

In the given question,

The formula for continuous compounding is:

A = Pe^(rt)

where A is the amount of money in the account after t years, P is the principal amount (the initial deposit), r is the annual interest rate (as a decimal), and e is the mathematical constant approximately equal to 2.71828.

Plugging in the given values, we get:

A = 625 * e^(0.016*5)

Simplifying this expression, we get:

A = 625 * e^(0.08)

Using a calculator, we can evaluate this expression to get:

A ≈ $740.55

Therefore, Jean-Luc's account will be worth approximately $740.55 after 5 years of continuous compounding at an annual interest rate of 1.6%.

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Dijkstra's algorithm is a useful tool for finding the length of the shortest path between two vertices in a weighted graph.

To solve this problem, we can start by creating a distinguished vertex set Q that contains all of the vertices in the graph, and assigning a distance of ∞ to each vertex in the set. Then, we can assign the starting vertex, a, a distance of 0.

Next, we can choose the vertex in Q with the shortest distance from the source vertex, a. This vertex is then removed from the set Q, and all of its adjacent vertices are evaluated and added to the set Q if they are not already in it. We then assign each adjacent vertex the distance of the selected vertex, plus the weight of the edge connecting it to the selected vertex.

Finally, we continue this process until the destination vertex, z, is removed from the set Q. At this point, the length of the shortest path between vertices a and z can be found by looking at the distance of the destination vertex.

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The transformation of the equation y = -2 I x - 3 I + 3 from the parent function y = I x I includes a horizontal shift of 3 units to the right, a vertical stretch by a factor of 2, a reflection across the x-axis, and a vertical shift of 3 units up.

The transformation of the equation y = -2 I x - 3 I + 3 from the parent function y = I x I can be described as follows:

The parent function is shifted 3 units to the right. This is indicated by the "-3" inside the absolute value bars in the equation.
The parent function is vertically stretched by a factor of 2. This is indicated by the "-2" in front of the absolute value bars in the equation.
The parent function is reflected across the x-axis. This is indicated by the negative sign in front of the "2" in the equation.
The parent function is shifted 3 units up. This is indicated by the "+3" outside the absolute value bars in the equation.

In summary, the transformation of the equation y = -2 I x - 3 I + 3 from the parent function y = I x I includes a horizontal shift of 3 units to the right, a vertical stretch by a factor of 2, a reflection across the x-axis, and a vertical shift of 3 units up.

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The function y=10x(x+1) is already in factored form, with a=10, r=0, and s=-1.

A function in mathematics from a set X to a set Y allocates exactly one element of Y to each element of X. The set X is termed the domain of the function and the set Y is called the codomain of the function. Initially, functions represented the idealized relationship between varied quantities and other variables.

To see this, we can rewrite the function as f(x)=10(x-0)(x-(-1)), which matches the form f(x)=a(x-r)(x-s). Therefore, the values of a, r, and s are:

a = 10
r = 0
s = -1

It is important to note that the values of r and s are the x-intercepts of the function, or the values of x that make the function equal to zero. In this case, the x-intercepts are 0 and -1, which correspond to the values of r and s.

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Part A:  The cost of bowling will be the same at both centers if 4 games are bowled. Part B:  It will cost $10 to bowl at both Fannin Lanes and Fun Time Lanes for 4 games.

The cost of bowling games can vary depending on a number of factors, including the location, time of day, and day of the week. Generally, bowling alleys charge per game or per hour, with additional fees for shoe rentals and any other amenities or services.

The cost per game can range from a few dollars to upwards of $10 or more, depending on the location and other factors. Some bowling alleys also offer discounted rates for children, seniors, or members of certain groups.

Part A:

Let's assume that the number of games to be bowled is 'x'. The total cost of bowling at Fannin Lanes is 2.5x + 2, and at Fun Time Lanes it is 2x + 4. We need to find the number of games where the cost of bowling at both centers is the same.

2.5x + 2 = 2x + 4

0.5x = 2

x = 4

Therefore, the cost of bowling will be the same at both centers if 4 games are bowled.

Part B:

Using the value of 'x' found in Part A, we can find the cost of bowling at both centers.

At Fannin Lanes: 2.5(4) + 2 = 10

At Fun Time Lanes: 2(4) + 4 = 12

Therefore, it will cost $10 to bowl at both Fannin Lanes and Fun Time Lanes for 4 games.

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The complete question is:

Roland made 16 sales this week, while Vickie made 8 more sales than Roland.

An equation is an expression that shows the relationship between two or more numbers and variables. Equations can either be linear, quadratic, cubic and so on depending on the degree.

Let x represent the number of sales Roland has.

Roland earns a base of $49 plus $15 per sale. HenceL

Total earning by Roland = 15x + 49

Vickie made 8 more sales than Roland. Vickie earns a base of $145 plus $6 per sale.

Total earning by Vickie = 6(x + 8) + 145 = 6x + 193

Roland and Vickie are salespeople who earned the same amount this week, Therefore:

15x + 49 = 6x + 193

9x = 144

x = 16

Roland made 16 sales this week.

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The area of the circle which is drawn by tracing around the edge of the quarter on a sheet of paper is 0.78 in.²

All points in a plane that are at a specific distance from a specific point, the centre, form a circle. In other words, it is the curve that a moving point in a plane draws to keep its distance from a specific point constant.

Each of the four equally sized pieces that make up a circle is known as a quarter of a circle. The quadrant of the circle is the name given to each quarter. One-fourth of the total area of the circle is represented by the area of a quarter circle.

Given,

the diameter of the quarter  = 1 in.

When we trace around the edge of the quarter on a sheet of paper, we get a circle of diameter 1 in.

So the radius of the circle r = 1/2 = 0.5 inches

Area of the circle on the paper = πr² = 3.14 × 0.5² = 0.78 in.²

Therefore the area of the circle which is drawn by tracing around the edge of the quarter on a sheet of paper is 0.78 in.²

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If we consider the number line with rational number marked on it, then the number of parts to be present between 1 and 2 and -1 and -2 will be one part each.

A number line is a pictorial representation or drawing of numbers in which there are equal intervals between each number and is used to represent the real numbers on it. Real numbers are those numbers which may be positive or negative integers, rational numbers or irrational numbers. In general, a quantity which can be expressed as an infinite decimal expansion is called as a real number.

If we draw a number line and mark the points as 1, 2, 3,...,∞ and -∞,..., -3, -2, -1 on the right and left hand side of 0, then the intervals between each real number is equal to 1. Hence the number of parts between 1 and 2 is one part and -1 and -2 is one part. However, these parts may be subdivided into more parts to get more fine value (smaller value).

This value will be a fractional value between 1 and 2 or -1 and -2. Hence parts between two numbers on a number line can be infinity, but for sake of simplicity, one is taken in general case.

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Accοrding tο the given infοrmatiοn, nοne οf the expressiοns a), b), c), οr d) are equivalent tο (2-³.24) 2.

An algebraic expressiοn is a mathematical phrase that can cοntain numbers, variables, and mathematical οperatiοns such as additiοn, subtractiοn, multiplicatiοn, and divisiοn. Algebraic expressiοns are used tο represent quantities οr values that can vary, depending οn the values assigned tο the variables in the expressiοn. Algebraic expressiοns can be simple οr cοmplex, and they can have οne οr mοre variables.

We can simplify the expressiοn (2-³.24) 2 as fοllοws:

(2-³.24) 2 = (2-0.24) 2 = (1.76) 2 = 3.52

The expressiοn (2-³.24) 2 is equivalent tο 3.52.

Therefοre, nοne οf the expressiοns a), b), c), οr d) are equivalent tο (2-³.24) 2.

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(\mathrm{A}_{4}\) is larger than \(\mathrm{B}_{2}\)

Yes, \(\mathrm{A}_{4}\) is larger than \(\mathrm{B}_{2}\). To show this, we can use the following equation:

\[\mathrm{A}_{4} = a \cdot c \cdot \mathrm{H}\]

and

\[\mathrm{B}_{2} = b_{1} \cdot c_{1}\]

Given that \(a = 34\), \(c = 45\), \(\mathrm{H} = 194\), \(b_{1} = 7\), and \(c_{1} = 3\), then \(\mathrm{A}_{4} = 34 \cdot 45 \cdot 194 = 291930\) and \(\mathrm{B}_{2} = 7 \cdot 3 = 21\). As \(291930 > 21\), \(\mathrm{A}_{4}\) is larger than \(\mathrm{B}_{2}\).

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If a model a rocket is used to test materials that will be used to build the body of the actual rocket. The limitation of this model is: D. It is smaller than the actual rocket.

A model rocket is a scaled-down version of an actual rocket, which means it is smaller than the actual rocket. While a model rocket can be used to test materials that will be used to build the body of the actual rocket, there are limitations to using a model rocket for this purpose.

One of the main limitations is that the smaller size of the model rocket may not accurately represent the effects of a full-scale rocket launch on the materials.

In addition, there may be other differences between the model rocket and the actual rocket, such as the types of engines used, the payload carried, and the environmental conditions during the launch, which can also limit the applicability of the results obtained from testing the materials on the model rocket to the actual rocket.

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The perimeter of the inside of the track is approximately 354.36 meters.

What is the circumference?

In geometry, the circumference is the perimeter of a circle or ellipse.

To find the perimeter of the inside of the track, we need to find the length of the inside edge of the track and add it to twice the length of the semicircles at each end.

The inside edge of the track is formed by two parallel lines, each offset by a distance of 56 m from the outer edge. The length of this edge is equal to the length of the long straight edge minus twice the offset distance:

length of inside edge = 130 m - 2(56 m)

length of inside edge = 18 m

The length of each semicircle at each end is equal to half the circumference of a full circle with a radius equal to the width of the track (56 m). Therefore, the length of both semicircles is:

2 x (1/2 x 2π x 56 m) = 2π x 56 m

Adding the length of the inside edge to twice the length of the semicircles, we get:

perimeter of inside track = 18 m + 2π x 56 m

perimeter of inside track = 18 m + 112π m

the perimeter of the inside track ≈ 354.36 m (rounded to two decimal places)

Hence, the perimeter of the inside of the track is approximately 354.36 meters.

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Answer:

In the exponential function y = C(a)^x, the value of a is the base of the exponent, which determines how fast the function grows or decays. If a were equal to zero, then any value of x would result in y being equal to zero, which would make the function flat and unchanging. Therefore, a cannot be equal to zero in an exponential function, as this would result in an undefined or meaningless expression. Additionally, any number raised to the power of zero is equal to 1, so an exponential function with a base of zero would also violate this mathematical rule.

To further analyze the function f(x)=1.3x^(4)-5.7x^(2)+3.71 on the interval [-4,4,-8,14], we can find the critical points, relative extrema, and inflection points of the function.

First, let's find the critical points by taking the derivative of the function and setting it equal to zero:
f'(x)=5.2x^(3)-11.4x
0=5.2x^(3)-11.4x
0=x(5.2x^(2)-11.4)
x=0, x=±√(11.4/5.2)
So the critical points are x=0, x=±1.4656

Next, let's find the relative extrema by using the second derivative test:
f''(x)=15.6x^(2)-11.4
f''(0)=-11.4<0, so x=0 is a relative maximum
f''(1.4656)=11.4>0, so x=1.4656 is a relative minimum
f''(-1.4656)=11.4>0, so x=-1.4656 is a relative minimum

Finally, let's find the inflection points by setting the second derivative equal to zero:
0=15.6x^(2)-11.4
x=±√(11.4/15.6)
x=±0.8544

So the inflection points are x=±0.8544

Overall, the function f(x)=1.3x^(4)-5.7x^(2)+3.71 has a relative maximum at x=0, relative minimums at x=±1.4656, and inflection points at x=±0.8544 on the interval [-4,4,-8,14].

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The composition of the permutation (3 5 ) three times is (3 5) and the permutation (1 2 4) to the power of 2023 is (1 2 4)

The composition of the permutation (3 5 ) three times is (3 5) and the permutation (1 2 4) to the power of 2023 is (1 2 4).For part (c), we can find the composition of the permutation (3 5) three times by applying the permutation to itself three times. The first time we apply it, we get (5 3), the second time we get (3 5), and the third time we get (5 3) again. Since the permutation (5 3) is the same as the original permutation (3 5), the composition of the permutation (3 5) three times is (3 5).For part (d), we can find the permutation (1 2 4) to the power of 2023 by applying the permutation to itself 2023 times. Since the permutation (1 2 4) is a cycle of length 3, applying it to itself 3 times will give us the identity permutation (1 2 4)(1 2 4)(1 2 4) = (1)(2)(4). Therefore, we can reduce the exponent of 2023 to 2023 mod 3 = 1, and the permutation (1 2 4) to the power of 2023 is the same as the permutation (1 2 4) to the power of 1, which is (1 2 4).

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Answer:

m = -3/2

Step-by-step explanation:

Slope = rise/run or (y2 - y1) / (x2 - x1)

Points (4, − 6) and (− 2, 3)

We see the y increase by 9 and the x decrease by 6, so the slope is

m = - 9/6 = -3/2

So, the slope of the line is -3/2

Earl works 5 days a week, which is 50percentage of Monday through Friday. He works in the Tutoring center one day, cleans bathrooms one day, and has another day for unspecified work.

Earl works 5 days a week. He works in the Tutoring center one day, cleans bathrooms one day, and has another day for unspecified work. This means that he is working 50% of Monday through Friday. On Saturday and Sunday, he has no work. His work schedule is a great example of how one can balance work with free time. It also shows how one can make the most out of their working hours by doing different activities each day. It provides variety and prevents boredom from setting in. The days off also give Earl time to rest and relax, allowing him to be productive the other days. His weekly schedule is a great example of how one can efficiently manage their time.

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We use standard units when measuring an object because :

A. Standard units are used because they are relatively permanent.

B. Standard units enable communication over time.

C. Standard units enable communication over distance.

D. Standard units are exact.

We use standard units when measuring an object because they are relatively permanent and enable communication over time, distance, and exactness.

Standard units are used because they are relatively permanent. This means that they do not change over time and are consistent. Standard units enable communication over time. Since they are consistent, they allow for accurate communication of measurements even if they are taken at different times.

Standard units enable communication over distance. They allow for accurate communication of measurements even if the people communicating are in different locations. Standard units are exact. This means that they are precise and accurate, allowing for consistent measurements.

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The answers to each part about the scale factor is given above.

Scale factor is the ratio of final dimensions to the initial dimensions.

Mathematically, we can write -

K = {final dimension/initial dimension}

Given are the dimensions of the drawing as -

Original                        1.5 in.          3 yd

Enlargement               2.5 in.          5 yd

{ 1 } -

We can write the scale factor as -

K = 3/1.5

K = 2

{ 2 } -

Area of the scaled drawing -

A = 2.5 x 5

A = 2.5 x 180

A = 450 square inches

{ 3 } -

To find the area of the actual drawing, multiply the area of the scale drawing by the multiple of the scale factor.

Therefore, the answers to each part about the scale factor is given above.

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The sales tax that Bobbi paid was 10 dollars. Below, you will learn how to solve the problem.

To find the sales tax, we can simply subtract the cost of the text books from the total amount paid:

Sales tax = Total amount paid - Cost of text books

Sales tax = 130 dollars - 120 dollars
Sales tax = 10 dollars

Therefore, the sales tax that Bobbi paid was 10 dollars.

In mathematics, subtraction (also called subtraction) is an arithmetic operation that removes quantities from any kind of operation.

After subtracting, we will always have a smaller value than the one we had before.

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