Fully factorise w² - 144

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 equation of the line with the given properties is y - 17 = 3/2(x - 18)

Linear equations are equations that have constant average rates of change.

The points are given as

(8, 17)

The slope is given as

Slope = 3/2

Calculate the slope of the points using

m = (y2 - y1)/(x2 - x1)

Where

Slope = m = 3/2

(x1, y1) = (8, 17)

So, we have

3/2 = (y - 17)/(x - 8)

Cross multiply in the above equation

So, we have

y - 17 = 3/2(x - 18)

A linear equation is represented as

y = m(x - x1) + y1

In point-slope form, we have

y - y1 = m(x - x1)

By comparing y - y1 = m(x - x1) and y - 17 = 3/2(x - 18), we can see that

y - 17 = 3/2(x - 18) is in point-slope form

Hence, the equation in point-slope form of the line that passes through the point (8, 17) is y - 17 = 3/2(x - 18)

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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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The algebraic expression for two less than the quotient of 15 and a number is 15x - 2.

What is algebraic expression ?

At least one variable and one operation must be present in an algebraic expression (addition, subtraction, multiplication, division). One such algebraic expression is 2(x + 8y).

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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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go on play store and download symbolab it can help you

In the picture, There is graph with a triangle SNZ. The reflected triangle of y=3 as (1,2),(5,3) and (5,5).

Given that,

In the picture, There is graph with a triangle.

The triangle is SNZ.

We have to find the reflection across y=3.

We have to draw a line on y=3.

On the line y=3,

Z point is there so it will be same that is (5,3).

Now, the point S is on (5,1)

Here, y is 1 that is 3+2=5

So, we take the reflected S point as (5,5)

Now, the point N is on (1,4)

Here, y is 4 that is 3-1=2

So, we take the reflected N point as (1,2).

Therefore, we get the reflected triangle of y=3 as (1,2),(5,3) and (5,5).

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Answer:

Explanation:

Given the system of equations:

To

Answer:

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

Answer:

please provide options

Step-by-step explanation:

Answer:

Step-by-step explanation:

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

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:

a) x³

b)y⁵

Step-by-step explanation:

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.

Answer:

16%

Step-by-step explanation:

A salesperson earns a commission of $624 for selling $3900 in merchandise. Find the rate of commission:

624 is ?% of 3,900

624/3,900 = 16% of 3,900

check answer:

3,900 * 0.16 = 624

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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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

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

So;

x + 200 = 450

or

x = 450 -200

Perfoming each operation, we have:

The statement is (a) [tex]E_{xy}[/tex] Q(x, y); (b) -Q(x, y); (c) [tex]E_{x}[/tex] (Q(x, jeopardy) ∩ Q(x, Wheel of fortune)); (d) ∀[tex]_{y}[/tex][tex]E_{x}[/tex]Q(x, y); (e) [tex]E_{x}E_{y}[/tex](x ≠ y) (Q(x, jeopardy) ∩ Q(y, jeopardy)).

Simply put, probability is the likelihood that something will occur. When we don't know how an event will turn out, we can discuss the likelihood or likelihood of several outcomes. Statistics is the study of events that follow a probability distribution.

Given that,

Q(x, y) be the statement "student x has become a contestant on quiz event y",

where the domain for x consists of all students in your major and for y consists of all quiz shows on television.

a) [tex]E_{xy}[/tex] Q(x, y)

b) -Q(x, y)

c) [tex]E_{x}[/tex] (Q(x, jeopardy) ∩ Q(x, Wheel of fortune))

d) ∀[tex]_{y}[/tex][tex]E_{x}[/tex]Q(x, y)

e)  [tex]E_{x}E_{y}[/tex](x ≠ y)(Q(x, jeopardy) ∩ Q(y, jeopardy)).

Therefore, the statements are above.

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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.

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:

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:

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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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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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

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]

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

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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