The courtyard outside East Middle School is in the shape of a rectangle and half-circle.
What is the approximate area of the courtyard?

The distance the duck fly in all is 99 miles

From the question, we have the following parameters that can be used in our computation:

The distance covered in all can be calculated as

Distance = The sum of the product of speed and time

Substitute the known values in the above equation, so, we have the following representation

Distance = 18 * 3 + 15 * 2

Evaluate the products

This gives

Distance = 54 + 45

Evaluate the sum

Distance = 99 miles

Hence, the distance is 99 miles

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Answer:

$32.70 ÷ 5

Each case is $6.54.

Step-by-step explanation:

$3270 for 5 cases, meaning you would split $32.70 into 5.

$32.70 ÷ 5 phone cases = $6.54

This also means each case is $6.54. To prove this, multiply by 5.

Answer: 2x+1

                  x

Step-by-step explanation:

The {-4, 0, 1} represents members of the domain of the graphed function.

What is the domain of the function?

A function is a mathematical object that accepts input, appears to apply a rule to it, and returns the result.

A function can be thought of as a machine that requires in a number, performs some operation(s), and then outputs the result.

The domain of a function is the collection of all its inputs. Its codomain is the set of possible outputs.

The range refers to the outputs which are actually used.

Domain: {-4, 0, 1}.

Simply list the domain as -4 < x < 2, which would imply ALL values between -4 and 2 inclusive.    

Yes, this is a function. No x-values repeat, and it passes the Diagonal Line Test for functions.

Hence, the {-4, 0, 1} represents members of the domain of the graphed function.

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After integration, the required function f is (2x³ - sin (x) + Cx + D).

What is the integration of 'xⁿ' and 'sin (x)'?

[tex]\int {x^{n} } \, dx = \frac{x^{n+1} }{n+1} + C\\\\\\\int {sinx} \, dx = -cosx + C[/tex]

Given, f''(x) = 12x + sin x

Therefore,

[tex]\int {f''(x)} \, dx \\\\=\int {f'(x)} \, dx \\\\\\= \int{12x + sin x} \, dx + C\\\\= 6x^{2} - cosx + C\\[/tex]

Again, f'(x) = 6x - cos (x) + C

Therefore,

[tex]\int {f'(x)} \, dx\\ \\=\int {f(x)} \, dx \\\\= \int {6x^{2} - cosx + C } \, dx \\\\= 2x^{3} - sinx + Cx + D[/tex]

Therefore, the required function is (2x³ - sin (x) + Cx + D).

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Answer: 1.0017805194 ounces

Step-by-step explanation:

To change 28.4 g to oz divide the mass in grams by 28.349523125. The 28.4 grams to ounces formula is [oz] = [28.4] / 28.349523125. Thus, we get (all results rounded to 2 decimals):

28.4 g to oz = about 1 oz

28.4 g to ounces =  about 1 ounces

28.4 grams to ounces = about  1 oz

Twenty-eight point four grams to ounces equal 1 international avoirdupois ounces, the unit used to measure grocery products (mass) in retail stores in the United States for example.

Answer:

You just need to think about this a little. There is a number, that you don't know, x, and if you square it ([tex]x^2[/tex]), that value will be equal to 72 more than the unknown number itself, so you put that into an equation.

Step-by-step explanation:

[tex]x^2=x+72\\x^2-x-72=0\\[/tex]

Apply quadratic formula (you can look it up)

[tex]x_1=\frac{-(-1)+\sqrt{(-1)^2+(-4*1*-72} )}{2(1)} \\\\x_2=\frac{-(-1)-\sqrt{(-1)^2+(-4*1*-72} )}{2(1)} \\\\[/tex]

There are two answers to the quadratic formula.

The rest should be easy.

Radius = TW

Diameter = TX

Chord = UV

The length of TX is 6 units.

A circle is a two-dimensional figure with a radius and circumference of 2 x pi x r.

The area of a circle is given as πr².

We have,

From the figure:

Diameter = TX

Radius = TW or WX or WY

Chord = UV

WY = 3 units

TX = TW + WX = 3 + 3 = 6 units.

Thus,

TX is 6 units.

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The input value for the expression is 1 and the remainder is -15.

The Remainder Theorem begins with an unnamed polynomial p(x), where "p(x)" simply means "some polynomial p with variable x". The Theorem then discusses dividing that polynomial by some linear factor x a, where an is simply a number.

Given polynomial is 5x⁴⁵ - 3x¹⁷ +2x⁴ - 19 and is divided by x + 1. Find the value of the polynomial at 1 and then calculate the remainder by synthetic solution.

F(x) = 5x⁴⁵ - 3x¹⁷ +2x⁴ - 19

F(1) = 5 - 3 + 2 - 19

F(1) = -19 + 4

F(1) = -15

The remainder will be calculated by synthetic calculation,

1 __5 ___-3___2_____-19

 |

 | ______5___ 2______4          

      5         2      4          -15

The remainder of the given expression is -15.

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Note that at any time during business hours, about 7 shoppers, on average, are waiting in the checkout line to make a purchase at the Good Deals Store. This is solved using Little Law.

Little's Law asserts that the long-term average number of individuals in a stable system, L, is equal to the long-term average effective arrival rate,λ, multiplied by the average duration a customer spends in the system, W.

To compute,

First, let's make sure that all the variables use the same time unit.

Thus, if there are 84 shoppers who are making a purchase per house, then there will be: 84/60 minutes

= 1.4

This means that  λ (rate) = 1.4 and the average time (W) is 5 minutes

Using Little's Law which states that the queuing formula is:

L = λW

Note that:

λ = 1.4 (computed)

W = 5 mintues (given)

Hence, the Average number of people in queue per time is:

L = 1.4 x 5

L = 7 shoppers.

Thus, it is correct to state that at any time during work hours, about 7 shoppers, on average, are waiting in the checkout line to make a purchase at the Good Deals Store.

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Full Question:

If shoppers enter a store at an average rate of r shoppers per minute and each stays in the store for an average time of T minutes, the average number of shoppers in the store, N, at any one time is given by the formula N=rT. This relationship is known as Little's law.

The owner of the Good Deals Store estimates that during business hours, an average of 3 shoppers per minute enter the store and that each of them stays an average of 15 minutes. The store owner uses Little's law to estimate that there are 45 shoppers in the store at any time.

Little's law can be applied to any part of the store, such as a particular department or the checkout lines. The store owner determines that, during business hours, approximately 84 shoppers per hour make a purchase and each of these shoppers spend an average of 5 minutes in the checkout line. At any time during business hours, about how many shoppers, on average, are waiting in the checkout line to make a purchase at the Good Deals Store?

the total number of license plate is 17576000 and In this the number ways to place 'A' in step 1 is = '1'1, and the number of steps '0' in final steps id '1' thus the total answer is 67600

In mathematics, there are two alternative methods for dividing up a collection of items into subsets: combinations and permutations. Any order may be used by a combination to list the subset's elements. An ordered list of a subset's components is called a permutation.

There are [tex]26^3[/tex] = 17576, ways to perform step 1-3, and there are [tex]10^3[/tex] = 1000, ways to perform remaining steps .

the total number of license plate is 17576000

In this the number ways to place 'A' in step 1 is = '1'1, and the number of steps '0' in final steps id '1'

thus the total answer is 67600

26 X 25 X 24 X 10 X 9 X 8 = 11232000

24 X 10 X 9 X 8 = 17280

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We need to use Right Angled Triangle Properties.

PR≈15 and QR≈12 is the length of the sides.

What is a Right Angled Triangle?

According to the definition of a right triangle, a triangle is referred to as a right-angled triangle or simply a right triangle if one of its angles is a right angle (90o). Triangle PQR is a right triangle in the illustration, and its components are its base, height, and hypotenuse. Here, QR is the base, PQ is the height, and PR is the hypotenuse. The essential side of a right triangle is its hypotenuse, which is also its largest side and sits across from the triangle's right angle.

In a right triangle we have: (Hypotenuse)2 = (Base)2 + (Altitude)2

Area of a right triangle = (1/2 × base × height) square units.

PR is the Hypotenuse of the Triangle.

PQ=9

Angle ∡PQR=37° and Angle ∡RPQ=53°

So Sin(theta)=opposite side/Hypotenuse

⇒Sin(37°)=9/Hypotenuse

⇒Hypotenuse=9/0.6018

⇒Hypotenuse=14.955

PR⇒14.955

We know that (Hypotenuse)2 = (Base)2 + (Altitude)2

⇒(14.955)2=(9)2+(QR)2

QR=11.9438

PR≈15 and QR≈12

We need to use Right Angled Triangle Properties.

PR≈15 and QR≈12 is the length of the sides.

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Applying the definition of congruent triangles, the values of x and y are:

x = 2; y = 6

Length of CR = 8 units.

If two triangles are congruent to each other, based on the CPCTC, all their corresponding angles, and their corresponding sides will be equal to each other.

Given that triangles NEW and CAR are congruent to each other, therefore:

NE = CA

NW = CR

EW = AR

Given the following measures:

NE = 11

EW = 12

AR = 4y - 12

CA = 4x + 3

NW = x + y

Therefore:

NE = CA

11 = 4x + 3

Solve for x:

11 - 3 = 4x

8 = 4x

8/4 = x

x = 2

EW = AR

12 = 4y - 12

12 + 12 = 4y

24 = 4y

24/4 = y

y = 6

NW = CR = x + y

Plug in the values of x and y:

CR = 2 + 6

CR = 8 units.

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The weight of harsh on the moon if he weighs 54 pounds on earth is 9 pounds

Weight varies directly with gravity

Let

Weight of a person = w

Gravity = g

So,

w = k × g

Where,

k = constant of proportionality

If w = 180 pounds and g = 30 pounds

w = k × g

180 = k × 30

180 = 30k

divide both sides by 30

k = 180/30

k = 6

If w = 54 pounds g = ?

w = k × g

54 = 6 × g

54 = 6g

divide both sides by 6

g = 54/6

g = 9 pounds

Therefore, harsh weighs 9 pounds on the moon.

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Considering the definition of a line, the equation of the line that passes through the point (10, -9) and has a slope of -2 is y= -2x +11.

A linear equation o line can be expressed in the form y = mx + b

where

In this case, you know:

Substituting the value of the slope m and the value of the point y=mx+b, the value of the ordinate to the origin b is obtained as follow:

-9= -2×10 + b

-9= -20 + b

-9 + 20= b

11= b

Finally, the equation of the line is y= -2x +11.

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The Venn diagram has been drawn and the product of LCM and GCF are the same as 24 x 36.

A Venn diagram is a pictorial representation of data represented in a circle.

The intersection parts of the circle represent the commonality of both sets.

As per the given, 24 and 36

The factors of 24 are as 2³ × 3

The factors of 36 are as 2² × 3²

The common factors 2² × 3

The LCM of 24 and 36 is 72 and the GCF is 12.

LCM x GCF = 24 x 12

72 x 12 = 24 x 12

864 = 864

Hence "After drawing a Venn diagram, it was discovered that the LCM and GCF product equals 24 x 36.".

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a) The value of x is 42.6 when k =16/3 is directly proportional to y.

b) The value of x is -56 when k is -7  is directly proportional to y.

What kind of variation is one where x and y are directly proportional?

x varies directly as y

x α y

x = ky

a) x=6 ; y = 32

x = ky

6 = k(32)

k = 16/3

if y = 8

then,

x = (16/3) * 8

= 128/3

x = 42.6

Hence, the value of x is 42.6 when k =16/3 is directly proportional to y.

b) x= 14 and y = -2

x = ky

14 = k(-2)

k = -7

if y = 8

then,

x = ky

x= (-7) 8

x = -56

Hence, the value of x is -56 when k is -7  is directly proportional to y.

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The joint relative frequency for those students who don't like to run, but however enjoy cycling, is C. 32%.

The joint frequency for students who like cycling, but do not like running, can be found by the formula :
= Number of students who enjoy cycling but don't enjoy running / Number of students in total

Number of students who enjoy cycling but don't enjoy running = 64

Number of students in total = 200

The joint frequency is :

= 64 / 200 x 100%

= 0.32 x 100 %

= 32 %

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In the aforementioned scenario, the choice you must include in your response is to find a common denominator.

One can refer to a fraction as being in its lowest term or in its simple form when the numerator and denominator are known to have no common factors besides 1. In general, any number of equal pieces is represented by a fraction, which is a portion of the entire. A fraction, such as one-half, eight-fifths, or three-quarters, indicates the number of parts of a particular size when it is used in daily speech.

The common denominator will be between 10 and 20, and there is a common numerator as well. The common denominator will be 20 because both can divide it and 10 is less than 20.

Finding a common denominator is the solution you should choose in the scenario mentioned above.

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The Divergence Test is inconclusive because lim ak = ∞

The Divergence Test is used to determine whether a given infinite series converges or diverges.

In order to use the Divergence Test, we must examine the limit of the terms of the series as n approaches infinity.

In the given series, the limit of the terms as n approaches infinity is not defined. Therefore, the Divergence Test is inconclusive.

The Divergence Test is inconclusive because lim ak = ∞

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∠U is congruent ∠S because all right angles are congruent.

If it is possible to superimpose one geometric figure on the other so that their entire surface coincides, that geometric figure is said to be congruent, or to be in the relation of congruence. When two sides and their included angle in one triangle are equal to two sides and their included angle in another, two triangles are said to be congruent. This concept of congruence appears to be based on the idea of a "rigid body," which may be moved without affecting the internal relationships between its components.

Given

∠U ≅ ∠S

All right angles are congruent.

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Answer:

natural exponential

natural

quizlet

Answer:

b. natural exponential

c. natural

Step-by-step explanation:

Given function:

[tex]f(x)=e^x[/tex]

The given exponential function is called the:

The base e is called the:

The number "e" occurs naturally in math and the physical sciences.

It is the base rate of growth shared by all continually growing processes, and so is called the natural base.

It is an irrational number and named after the 18th century Swiss mathematician, Leonhard Euler, and so is often referred to as "Euler's number".

Alex's speed in the first half of the time is 12 miles per hour

Information from the problem include:

For the first half of the time, Alex biked a 48-mile trail in 4 hours.

he biked twice as fast as he did the second half of the time.

Average speed is calculated using the ratio of distance to time spent

this is represented by the formula

= distance / time

The speed for the first half is  solved as below

= 48 mile / 4 hours

= 12 mile / hours

= 12 mph

speed for the first half is 12 mph

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Answer:

15.5

Step-by-step explanation:

Pepe= 2/3 x 12 = 8

Paula= 1/4 x 30 = 7.5

8+7.5=15.5

Answer:

Step-by-step explanation: 9 multiplied by .95 = $8.55 before tax

Answer:

The area of the circle is 113.04 square feet.

Step-by-step explanation:

A = 3.14 × (6)² ft

A = 3.14 × 36 ft

A = 113.04 square feet.

Thus, The area of the circle is 113.04 square feet.

Answer:

28.274 square feet

Step-by-step explanation:

Multiply the radius by pi

Answer: a=3 and b=1

Step-by-step explanation:

Let p(x)=x 2 +ax+2b

If x+2 is a factor of p(x)

Then by factor theorem

p(−2)=0

⇒(−2)

2

+a(−2)+2b=0

⇒4−2a+2b=0

⇒a−b=2   ---(i)

Also given that a+b=4    ---(ii)

Adiing (i) and (ii) we get

2a=6

⇒a=3

putting a=3 in (i) we get

a−b=2

⇒3−b=2

⇒b=1

Therefore we get a=3 and b=1

Answer:

answer is 2.5

Step-by-step explanation:

f cannot have a jump discontinuity at [tex]$x_0 \in(a, b)$[/tex] and  [tex]$$ \lim _{x \uparrow x_0} f(x) < \lim _{x \mid x_0} f(x) .$$[/tex]

f cannot have a removable discontinuity at [tex]$$x_0 \in(a, b) $$[/tex] and [tex]\alpha=\lim _{x \rightarrow x_0} f(x) < f\left(x_0\right)[/tex]

Let f be a function defined on (a, b) satisfies intermediate value property.

Claim: f ca not have removable on jump discontinuity.

Suppose f has a jump discontinuity at [tex]$x_0 \in(a, b)$[/tex]

We take [tex]$\theta$[/tex] such that

[tex]$$\lim _{x \rightarrow x_0} f(x) < \theta < \lim _{x \downarrow x_0} f(x) \text { and } \theta \neq f\left(x_0\right)$$[/tex]

Now there exist [tex]$\delta > 0$[/tex] such that [tex]$f(x) < \theta$[/tex] for all [tex]$x \in\left[x_0-\delta, x_0\right)$[/tex] and [tex]$f(x) > \theta$[/tex] for all [tex]$x \in\left(x_0, x_0+\delta\right]$[/tex]

Now [tex]$f\left(x_0-\delta\right)[/tex][tex]< \theta < f\left(x_0+\delta\right)$[/tex] for all [tex]$x \in\left[x_0-\delta, x_0+\delta\right] \backslash\left\{x_2\right\}$[/tex] and [tex]$f\left(x_0\right) \neq \theta$[/tex].

Therefore the point [tex]$\theta$[/tex] has no preimage under f

that is, there does not exists [tex]$y \in\left[x_0-\delta, x_0+\delta\right][/tex] for which

[tex]$$f(y)=\theta[/tex] because [tex]\left\{\begin{array}{l}y=x_0 \Rightarrow f(y) \neq \theta \\y > x_0 \Rightarrow f(y) > \theta \\y < x_0 \Rightarrow f(y) < \theta\end{array}\right.$$[/tex]

Therefore f does not satisfies intermediate value property on [tex]$\left[x_0-\delta, x_0+\delta\right]$[/tex],

Hence f does not satisfies IVP on (a, b) which is not possible because we assume f satisfies IVP on (a, b),

Therefore f can not have a jump discontinuity.

Suppose f has a removable point of discontinuity at [tex]$x_0 \in(a, b)$[/tex],

Let [tex]$\alpha=\lim _{\alpha \rightarrow x_0} f(x)$[/tex],

Let [tex]\alpha < f\left(x_0\right)$[/tex] so [tex]$f\left(x_0\right)-\alpha > 0$[/tex].

Now [tex]$\lim _{x \rightarrow x_0} f(x)=\alpha$[/tex] then [tex]\exists$ \delta > 0$[/tex] such that

[tex]$$\begin{aligned}& |f(x)-\alpha| < \frac{f\left(x_0\right)-\alpha}{2} \text { for all } x \in\left\{x_0-\delta, x_0-\alpha\right]-\left\{x_0\right\} \\& \Rightarrow \quad f(x) < \alpha+\frac{f\left(x_0\right)-\alpha}{2} \text { for all } x \in\left[x_0-\delta, x_0+\delta\right]-\left\{x_0\right\}\end{aligned}$$[/tex]

So [tex]$f(x) < \frac{f\left(x_0\right)+\alpha}{2}$[/tex] for all [tex]$x \in\left[x_0-\delta, x_0\right]-\left\{x_0\right\}$[/tex]

Now [tex]$f\left(x_0\right) > \alpha$[/tex].

And  [tex]$f(x) < \frac{f\left(x_0\right)+\alpha}{2} < f\left(x_0\right)$[/tex] for all [tex]$x \in\left[\left(x_0 \delta, x_0\right)\right.$[/tex]

Let [tex]$\mu=\frac{f\left(x_0\right)+\alpha}{2}$[/tex].

Then there does not exist [tex]$e \in\left[x_0-\delta, c\right]$[/tex] such that [tex]$f(c)=\mu$[/tex]

Because for [tex]$e=x_0 \quad f(e) > \mu$[/tex]

                for [tex]$c < x_0 \quad f(c) < \mu$[/tex].

Therefore f does not satisfy IVP on [tex]$\left[x_0-\delta_1 x_0\right]$[/tex] which contradict our hypothesis,

therefore [tex]$\alpha \geqslant f\left(x_0\right)$[/tex]

Let [tex]$\alpha > f\left(x_0\right)$[/tex]. so [tex]$\alpha-f\left(x_0\right) > 0$[/tex]

[tex]$\lim _{x \rightarrow x_0} f(x)=\alpha$[/tex]

Then [tex]\exists $ \varepsilon > 0$[/tex] such that

[tex]$|f(x)-\alpha| < \frac{\alpha-f\left(x_0\right)}{2}$[/tex] for all [tex]$\left.x \in\left[x_0-\varepsilon_0 x_0+\varepsilon\right]\right\}\left\{x_i\right\}$[/tex]

[tex]$\Rightarrow f(x) > \alpha-\frac{\alpha-f\left(x_0\right)}{2}$[/tex] for all [tex]$x \in\left[x_0-\varepsilon_1, x_0+\varepsilon\right] \backslash\left\{x_0\right\}$[/tex]

[tex]$\Rightarrow f(x) > \frac{\alpha+f\left(x_0\right)}{2}$[/tex] for all [tex]$x \in\left[x_0-\varepsilon_1, x_0\right)$[/tex]

Now [tex]$f\left(x_0\right) < \alpha$[/tex]

The [tex]$f(x) > \frac{f\left(x_0\right)+\alpha}{2} > f\left(x_0\right)$[/tex].

So [tex]$f\left(x_0\right) < \frac{f\left(x_0\right)+\alpha}{2} < f(x)$[/tex] for all [tex]$x \in\left[x_0 \varepsilon, \varepsilon_0\right)$[/tex]

Let [tex]$\eta=\frac{f\left(x_e\right)+\alpha}{2}$[/tex]

Then there does not exist [tex]$d \in\left[x_0-\varepsilon, x_0\right]$[/tex] such that [tex]$f(d)=\xi$[/tex].

Because if [tex]$d=x_0, f(d)=f\left(x_0\right) < \eta$[/tex] if [tex]$d E\left[x_0-\varepsilon, x_0\right)$[/tex]

Then [tex]$f(d) > \eta$[/tex]

Therefore f does not satisfies IVP on [tex]$\left[x_0-\varepsilon, x_0\right]$[/tex] which contradict olio hypothesis.

Therefore [tex]$\alpha \leq f\left(x_0\right)$[/tex] (b) From (a) and (b) it follows [tex]$\alpha=f\left(x_0\right)=\lim _{x \rightarrow x_0} f(x)$[/tex]. Therefore f can not have a removable discontinuous

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