Calc Limits. Attatched question.

The winning average of the Varsity football team is a non-terminating decimal.

The winning average of the Junior Varsity football team is a terminating decimal.

Which team had a better season?  Varsity team

How is the winning average calculated?

a ) Part A

1. Team Varsity

Number of  total matches  won = 8

Number of total matches lost = 3

Total number of matches = 11

The winning average  [tex]=\frac{\text{total number of matches won}}{\text{total matches}}[/tex]

                                     =[tex]\frac{8}{11} \\\\[/tex]

                                     = 0.72727

       0.72727 is a non-terminating decimal

2. Team Junior Varsity

Number of  total matches  won = 7

Number of total matches lost = 3

Total number of matches = 10

The winning average  [tex]=\frac{\text{total number of matches won}}{\text{total matches}}[/tex]

                                      =[tex]\frac{7}{10} \\\\[/tex]

                                       = 0.7

             0.7 is a terminating decimal.

The winning average of the Varsity football team is a non-terminating decimal.

The winning average of the Junior Varsity football team is a terminating decimal.

b) Part B

Which team had a better season?  Varsity team

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By interpreting the vertex form of the equation of parabola, the focus of the curve is equal to F(x, y) = (2, 0). (Correct choice: D)

In this problem we find the equation of a parabola in standard form, which has to be rearranged into its vertex form in order to determine the coordinates of its focus.

The focus of a parabola is a point outside the curve such that the least distance from any point of the parabola and the least distance between that the point on the parabola and directrix are the same.

First, rearrange the polynomial into its vertex form by algebraic handling:

y = (1 / 8) · (x² - 4 · x - 12)

y + (1 / 8) · 16 = (1 / 8) · (x² - 4 · x - 12) + (1 / 8) · 16

y + 2 = (1 / 8) · (x² - 4 · x + 4)

y + 2 = (1 / 8) · (x - 2)²

y + 2 = [1 / (4 · 2)] · (x - 2)²

Second, determine the vertex and the distance between the vertex and the focus:

The parabola has a vertex of (h, k) = (2, - 2) and vertex-to-focus distance of 2.

Third, determine the coordinates of the focus:

F(x, y) = (h, k) + (0, p)

F(x, y) = (2, - 2) + (0, 2)

F(x, y) = (2, 0)

The focus of the parabola is equal to F(x, y) = (2, 0).

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Answer:

$1.15

Step-by-step explanation:

750 + 400 = 1,150

1/1000 * 1,150 = 1,150/1000 = $1.15 expected value of one ticket.

Since he paid more than $1.15 for the ticket he has a bad bet.

The coordinates of the midpoint of a,b is given as;

The set of letters in the word 'woodpecker' using the most concise method is {c, d, e, k, o, p, r, w}

From the question, we are to write the set of letters of the given word.

The word is 'woodpecker'

We are to write the set using the listing (roster) method or the set builder notation.

The roster method or listing method is a way to show the elements of a set by listing the elements inside of brackets

Set builder notation is a mathematical notation for describing a set by enumerating its elements, or stating the properties that its members must satisfy

The set builder notation is not a suitable method to list the elements of the given word.

The most concise method to list the elements of the given word, 'woodpecker', is the listing (roster) method.

Thus,

Using the listing (roster) method,

The set of letters of the given word is {c, d, e, k, o, p, r, w}

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Answer:

c

Step-by-step explanation:

2-2/5=12/5

12/5=

2-2/5 is the mixed fraction for the number 12/5

well, the key and you are absolutely correct, so you're golden

[tex]\stackrel{reciprocals}{\cfrac{3}{2}\cdot \cfrac{8}{1}\cdot \cfrac{1}{5}}\implies \cfrac{3}{5}\cdot \cfrac{8}{2}\cdot \cfrac{1}{1}\implies \cfrac{3}{5}\cdot \cfrac{4}{1}\cdot \cfrac{1}{1}\implies \cfrac{12}{5}\implies \cfrac{10+2}{5} \\\\\\ \cfrac{10}{5}+\cfrac{2}{5}\implies 2+\cfrac{2}{5}\implies 2\frac{2}{5} ~~ \checkmark[/tex]

The simplified expression of (-7)^2 x (-7)^5 x (-7)^-9 is (-7)^-2

Expressions are mathematical statements that are represented by variables, coefficients and operators

The expression is given as

(-7)^2 x (-7)^5 x (-7)^-9

The base of the above expression are the same

i.e. Base = -7

This means that we can apply the law of indices

When the law of indices is applied, we have the following equation:

(-7)^2 x (-7)^5 x (-7)^-9 = (-7)^(2 + 5 - 9)

Evaluate the sum in the above equation

So, we have

(-7)^2 x (-7)^5 x (-7)^-9 = (-7)^(7 - 9)

Evaluate the difference in the above equation

So, we have

(-7)^2 x (-7)^5 x (-7)^-9 = (-7)^-2

Hence, the simplified expression of the expression given as (-7)^2 x (-7)^5 x (-7)^-9 is (-7)^-2

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The large box weighs 45 pounds and the small box weighs 35 pounds.

Let the weight of the small box = x

Let the weight of large box = y

The combined weight of a large box and a small box is 80 pounds. The truck is transporting 65 large boxes and 55 small boxes. If the truck is carrying a total of 4850 pounds in boxes. This will be illustrated as:

x + y = 80 ...... i

55x + 65y = 4850 .... ii

From equation i x = 80 - y

This will be put into equation ii

55x + 65y = 4850

55(80 - y) + 65y = 4850

4400 - 55y + 65y = 4850

10y = 4850 - 4400

10y = 450

y = 450 / 10 = 45

Large box = 45 pounds

Since x + y = 80

x = 80 - 45 = 35

Small box = 35 pounds.

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Perfoming each operation, we have:

Answer:

Step-by-step explanation:

Given two planes flying in opposite directions are 810 miles apart after 3 hours, and the first is 30 mph slower than the second, you want the speed of each plane.

Let s represent the speed of the slower plane. Then faster plane will have a speed of (s+30). The distance between the planes increases at a rate equal to the sum of their speeds. Distance is the product of speed and time, so we have ...

  distance = speed × time

  810 = (s + (s+30)) × 3

Dividing the equation by 3, we get ...

  270 = 2s +30

  240 = 2s . . . . . . subtract 30

  120 = s . . . . . . . divide by 2

  150 = s+30 . . . the speed of the faster plane

The speed of the first plane is 120 mph; the speed of the second plane is 150 mph.

The rate of the two planes flying in opposite direction was found to be

given data

The first plane is flying 30 mph slower than the second plane.

time = 3 hours

distance = 810 miles

let the rate of the faster plane be x

then rate if the slower plane will be x - 3

rate of both planes

= x + x - 30

= 2x - 30

Finding the rate of each plane

rate of both planes = total distance / total time

2x - 30 = 810 / 3

2x - 30 = 270

2x = 270 + 30

2x = 300

x = 150

Then the slower plane = 150 - 30 = 120 mph

Hence the rate of the faster plane is 150 mph and the rate of the slower plane is 120 mph

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Answer:

Part 1 is B, Part 2 is D

If you want an explanation I can give one :)

When Seema used compatible numbers to estimate the product of (–25.31)(9.61), her estimate is A. -250.

From the information, it should be noted that Seema used compatible numbers to estimate the product of (–25.31)(9.61).

It should be noted that -25.31 when rounded will be -25.

It should be noted that 9.61 when rounded will be 10.

Therefore, the multiplication will be:

= -25 × 10

= -250.

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Seema used compatible numbers to estimate the product of (–25.31)(9.61). What was her estimate?

-250

-240

240

250

The quadratic equation in standard form that contains the points (5, - 6), (- 7, 0) and (3, 0) is y = - (1 / 4) · x² - x + 21 / 4.

Herein we find the equation of a parabola that contains a points and its x-intercepts (two real roots). According to fundamental theorem of algebra, we can derive a quadratic function with real coefficients if we know three points of the parabola. The procedure is shown below.

First, use the quadratic function in product form and substitute on all known variables to determine the lead coefficient:

y = a · (x + 7) · (x - 3)

- 6 = a · (5 + 7) · (5 - 3)

- 6 = a · 12 · 2

- 6 = 24 · a

a = - 6 / 24

a = - 1 / 4

Second, expand the quadratic function into its standard form:

y = - (1 / 4) · (x + 7) · (x - 3)

y = - (1 / 4) · (x² + 4 · x - 21)

y = - (1 / 4) · x² - x + 21 / 4

The quadratic equation in standard form is y = - (1 / 4) · x² - x + 21 / 4.

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Answer:

Can't see sh## ur photo is crazy low quality

Answer:

please provide options

Step-by-step explanation:

The value of  k = -0.079

The Fahrenheit temperature of the turkey, to the nearest degree, after 5.5 hours =  T(5.5) = 159.865

What is Temperature?

Temperature is a numerical expression of how hot a substance or radiation is. There are three different types of temperature scales: those that are defined in terms of the average, like the SI scale;

Newton's law of cooling states that T=Ta + (To-Ta)e-kt

T = Ta + (To-Ta)e-kt

Ta is the object's ambient temperature.

To equal the starting temperature.

t = the number of hours.

after t hours, temperature equals T.

decay constant is k.

According to the formula,

To = 70 and

Ta = 325,

therefore

(To -Ta) = 70 - 325.

(To -Ta)           = -255

T = 325 + (-255 e^kt) or

325 -255e^kt.

After 2.5 hours, the turkey reaches a temperature of 116°F.

In other words,

T(2.5) = 116 F.

Consequently,

116 = 325 - 255e^k2.5

Following that,

116 - 325 = - 255e^2.5k

Now,

-209 = -255e^2.5k

e^2.5k = -209/-255

Using the function 'ln,'

now

ln(e2.5k) = ln(209 / 255)

Then, 2.5k = -0.198929

The value of k is now given by k = -0.198929 / 2.5, which is -0.079571.

The result is k = -0.079.

After 5.5 hours, the temperature is

T(5.5) = 325 - 255e^(-0.079 x 5.5)

Then, = 325 - 255e^(-0.4345)

Now,

         = 325 - (165.1350) (165.1350)

T(5.5) = 159.865 is the result.

Consequently, T(5.5) = 159.865 degrees Fahrenheit is the turkey's exact temperature after 5.5 hours.

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All the options occurred as a result of Roman expansion following the Punic Wars except; B: It allowed many Romans to buy large farming estates

The three Punic Wars between Carthage and Rome took place over about a century, starting in 264 B.C. and it ended with the event of the destruction of Carthage in the year 146 B.C.

Now, at the time the First Punic War broke out, Rome had become the dominant power throughout the Italian peninsula, while Carthage–a powerful city-state in northern Africa–had established itself as the leading maritime power in the world. The First Punic War commenced in the year 264 B.C. when Rome expressed interference in a dispute on the island of Sicily controlled by the Carthaginians. At the end of the war, Rome had full control of both Sicily and Corsica and this meant that the it emerged as a naval and a land power.

In the Second Punic War, the great Carthaginian general Hannibal invaded Italy and scored great victories at Lake Trasimene and Cannae before his eventual defeat at the hands of Rome’s Scipio Africanus in the year 202 B.C. had to leave Rome to be controlled by the western Mediterranean as well as large swats of Spain.

In the Third Punic War, we saw that Scipio the Younger led the Romans by capturing and destroying the city of Carthage in the year 146 B.C., thereby turning Africa into yet another province of the mighty Roman Empire.

Thus, we can see that  the cause of the Punic wars is that the Roman republic grew, so they needed to expand their territory by conquering other lands, including Carthage.

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The equations that have x as a solution are 15x - 60 = 120 and 3x = 24.

The equation is as follows:

3x - 12 = 24

The equations that also has x as the solution can be found as follows:

Let's use the law of multiplication equality to find a solution that has x as the solution.

The multiplication property of equality states that when we multiply both sides of an equation by the same number, the two sides remain equal.

Multiply both sides of the equation by 5

3x - 12 = 24

Hence,

15x - 60 = 120

By adding a number to both sides of the equation, we can get same solution for x.

3x - 12 = 24

add 12 to both sides of the equation

3x - 12 + 12 = 24 + 12

3x = 24

Therefore, the two solution are 15x - 60 = 120 and 3x = 24

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We can find the x and y-intercept by substituting zero for x and y respectively.Part A

iven the eequation below;

When x=0

when y=0

Answer 1

Part B

When x=0

When y=0

Answer 2:

Answer:

Step-by-step explanation:

it is my first time doing dis so it is 12.

Answer:

I think it's D because it can't be factorized

The  three equivalent ratios are given below .

What are equivalent ratios?

Ratios that can be streamlined or decreased to the same value are said to be equivalent. In other words, if one ratio can be written as a multiple of the other, then they are said to be equivalent. The ratios 1:2 and 4:8, 3:5 and 12:20, 9:4 and 18:8, and others are some examples of analogous ratios. To put it another way, two ratios are said to be equivalent if one of them can be written as the multiple of the other. Therefore, we must multiply the two values (antecedent and consequent) by the same number in order to obtain the equivalent ratio of another ratio. This approach is comparable to the approach for determining equivalent fractions.

1:9 , 99:11 , 9:11 , 1:99

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Answer:

a) x³

b)y⁵

Step-by-step explanation:

go on play store and download symbolab it can help you

The dragon would take 2.16 seconds to eat a human weighing 90 kilograms.

Fraction number consists of two parts, one is the top of the fraction number which is called the numerator and the second is the bottom of the fraction number which is called the denominator.

Let x seconds it would take to eat a human weighing 90kilograms

500 kg →  12 seconds

90 kg →  x

500/90 = 12/x

x = (90×12)/500

x = 2.16 seconds

Thus, the dragon would take 2.16 seconds to eat a human weighing 90 kilograms.

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Graph both equations. The coorinates of the point where the graphs intersect is the solution to the system of equations.

To graph them, notice that each equation corresponds to a line. A straight line can be drawn if two points on that line are given. Replace two different values of x into each equation to find its corresponding value of y, then, plot the coordinate pairs (x,y) to draw the lines.

First equation:

For x=2 and x=5 we have that:

Then, the points (2,1) and (5,7) belong to the line:

Second equation:

For x=0 and x=6 we have:

Then, the points (0,4) and (6,2) belong to the line:

Solution:

The lines intersect at the point (3,3).

Then, the solution for this system of equations, is:

The expression for this scenario is:

Now substitute the value of t into the elevation expression above, to get the elevation reached after 3 hours.

Let x be her saving accounts balance before the $200 deposit

So;

x + 200 = 450

or

x = 450 -200

Answer:

Maximum length = 30 cm

Step-by-step explanation:

Perimeter of a rectangle = 2 × (length + width)

According to the question,

2 × (length + width) < 71 cm (It can be 70 cm at maximum)

length + width < 71/2 cm

length + width < 36 cm

Since, width = 5 cm,

length + 5 cm < 36 cm

length < 36 - 5 cm

length < 31 cm

Therefore, the maximum length can be 30 cm

A rather lengthy solution using a neat method I just learned relying on complex analysis.

First observe that

[tex]e^{-x^2} \cos(2x^2) = \mathrm{Re}\left[e^{-x^2} e^{i\,2x^2}\right] = \mathrm{Re}\left[e^{a x^2}\right][/tex]

where [tex]a=-1+2i[/tex].

Normally we would consider the integrand as a function of complex numbers and swapping out [tex]x[/tex] for [tex]z\in\Bbb C[/tex], but since it's entire and has no poles, we cannot use the residue theorem right away. Instead, we introduce a new function [tex]g(z)[/tex] such that

[tex]f(z) = \dfrac{e^{a z^2}}{g(z)}[/tex]

has at least one pole we can work with, along with the property (1) that [tex]g(z)[/tex] has period [tex]w[/tex] so [tex]g(z)=g(z+w)[/tex].

Now in the complex plane, we integrate [tex]f(z)[/tex] along a rectangular contour [tex]\Gamma[/tex] with vertices at [tex]-R[/tex], [tex]R[/tex], [tex]R+ib[/tex], and [tex]-R+ib[/tex] with positive orientation, and where [tex]b=\mathrm{Im}(w)[/tex]. It's easy to show the integrals along the vertical sides will vanish as [tex]R\to\infty[/tex], which leaves us with

[tex]\displaystyle \int_\Gamma f(z) \, dz = \int_{-R}^R f(z) \, dz + \int_{R+ib}^{-R+ib} f(z) \, dz = \int_{-R}^R f(z) - f(z+w) \, dz[/tex]

Suppose further that our cooked up function has the property (2) that, in the limit, this integral converges to the one we want to evaluate, so

[tex]f(z) - f(z+w) = e^{a z^2}[/tex]

Use (2) to solve for [tex]g(z)[/tex].

[tex]\displaystyle f(z) - f(z+w) = \frac{e^{a z^2} - e^{a(z+w)^2}}{g(z)} = e^{a z^2} \\\\ ~~~~ \implies g(z) = 1 - e^{2azw} e^{aw^2}[/tex]

Use (1) to solve for the period [tex]w[/tex].

[tex]\displaystyle g(z) = g(z+w) \iff 1 - e^{2azw} e^{aw^2} = 1 - e^{2a(z+w)w} e^{aw^2} \\\\ ~~~~ \implies e^{2aw^2} = 1 \\\\ ~~~~ \implies 2aw^2 = i\,2\pi k \\\\ ~~~~ \implies w^2 = \frac{i\pi}a k[/tex]

Note that [tex]aw^2 = i\pi[/tex], so in fact

[tex]g(z) = 1 + e^{2azw}[/tex]

Take the simplest non-zero pole and let [tex]k=1[/tex], so [tex]w=\sqrt{\frac{i\pi}a}[/tex]. Of the two possible square roots, let's take the one with the positive imaginary part, which we can write as

[tex]w = \displaystyle -\sqrt{\frac\pi{\sqrt5}} e^{-i\,\frac12 \tan^{-1}\left(\frac12\right)}[/tex]

and note that the rectangle has height

[tex]b = \mathrm{Im}(w) = \sqrt{\dfrac\pi{\sqrt5}} \sin\left(\dfrac12 \tan^{-1}\left(\dfrac12\right)\right) = \sqrt{\dfrac{\sqrt5-2}{10}\,\pi}[/tex]

Find the poles of [tex]g(z)[/tex] that lie inside [tex]\Gamma[/tex].

[tex]g(z_p) = 1 + e^{2azw} = 0 \implies z_p = \dfrac{(2k+1)\pi}2 e^{i\,\frac14 \tan^{-1}\left(\frac43\right)}[/tex]

We only need the pole with [tex]k=0[/tex], since it's the only one with imaginary part between 0 and [tex]b[/tex]. You'll find the residue here is

[tex]\displaystyle r = \mathrm{Res}\left(\frac{e^{az^2}}{g(z)}, z=z_p\right) = \frac12 \sqrt{-\frac{5a}\pi}[/tex]

Then by the residue theorem,

[tex]\displaystyle \lim_{R\to\infty} \int_{-R}^R f(z) - f(z+w) \, dz = \int_{-\infty}^\infty e^{(-1+2i)z^2} \, dz  = 2\pi i r \\\\ ~~~~ \implies \int_{-\infty}^\infty e^{-x^2} \cos(2x^2) \, dx = \mathrm{Re}\left[2\pi i r\right] = \sqrt{\frac\pi{\sqrt5}} \cos\left(\frac12 \tan^{-1}\left(\frac12\right)\right)[/tex]

We can rewrite

[tex]\cos\left(\dfrac12 \tan^{-1}\left(\dfrac12\right)\right) = \sqrt{\dfrac{5+\sqrt5}{10}}[/tex]

so that the result is equivalent to

[tex]\sqrt{\dfrac\pi{\sqrt5}} \cos\left(\dfrac12 \tan^{-1}\left(\dfrac12\right)\right) = \boxed{\sqrt{\frac{\pi\phi}5}}[/tex]