(×+13)²=(×+12)²+(×-5)²
Give it to me please :(​​

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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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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The journal entry is given in answer part b

A discount coupon is defined as a certificate that entitles the customer to avail of a reduction in the invoice price if the terms mentioned on it are met. It is a technique applied by the sellers to expand sales.

Given that, Clarks Inc., a shoe retailer, sells boots in different styles. In early November, the company starts selling “SunBoots” to customers for $70 per pair. When a customer purchases a pair of SunBoots, Clarks also gives the customer a 30% discount coupon for any additional future purchases made in the next 30 days. Customers can’t obtain the discount coupon otherwise. Clarks anticipates that approximately 20% of customers will utilize the coupon, and that on average those customers will purchase additional goods that normally sell for $100.

a) Two performance obligations are associated with contract. The first one is related to the sun boots, and the other one is associated with discount coupon provided. The discount coupon would be considered a separate obligation because it gives a certain right to the customers that they would not have received if the sale was not made to them.

b) Journal entry =

Account title and explanation            Debit ($)                    Credit ($)

Cash (1000×$70)                                    $70000                  

Sales Revenue ($70,000-$6000)                                           $64000

Deferred Revenue                                                                    $6000

Now,

Computation of the deferred revenue:

Deferred revenue = pairs sold × Average purchase price × Discount × Redemption of coupon

= 1000 × 100 × 30% × 20%

= $6000

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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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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

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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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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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".

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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Answer: 2x+1

                  x

Step-by-step explanation:

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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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: 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:

My sister is

18

years old.

Step-by-step explanation:

s = m + 8        Eq. 1

s + m = 28      Eq. 2

s = my sister age

m = my age

Replacing Eq. 1 in Eq. 2:

(m+8) + m = 28

2m + 8 = 28

2m = 28 - 8

2m = 20

m = 20/2

m = 10

From Eq. 1:

s = m + 8

s = 10 + 8

s = 18

Check:

from Eq. 2:

s + m = 28

18 + 10 = 28

Answer:

[tex]x=1[/tex]

Step-by-step explanation:

[tex]2^{2x}-2^2 2^x-12=0 \\ \\ (2^x)^2-4(2^x)-12=0[/tex]

Factoring the quadratic in [tex]2^x[/tex],

[tex](2^x+6)(2^x-2)=0 \\ \\ 2^x=-6,2 \\ \\ 2^x=2 \implies x=1 \\ \\ 2^x=-6 \implies x \notin \mathbb{R} \\ \\ \therefore x=\log_{2} (6)[/tex]

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.

The remainder term Rn is equal to (-2) raised to the power of n multiplied by e raised to the power of -2c, divided by n+1 factorial, multiplied by (x-2) raised to the power of n, for some c between x and 2.

The remainder term Rn in the nth order Taylor polynomial centered at a for the given function f(x)=e^-2x, a=2 is equal to (-2) raised to the power of n multiplied by e raised to the power of -2c, divided by n+1 factorial, multiplied by (x-2) raised to the power of n, for some c between x and 2. This remainder term represents the difference between the approximate value of the function given by the Taylor polynomial and the actual value of the function. By taking the limit as n approaches infinity, the remainder term can be used to determine the exact value of the function. The remainder term is an important concept in calculus as it is used to determine the accuracy of a given approximation.

For n=5, R5(x) = (-2)^5 e^-2c/(5+1)! (x-2)^5 for some c between x and 2.

Substituting n=5, x=7 and c=4, we have:

R5(7) = (-2)^5 e^-8/(6)! (7-2)^5

R5(7) = (-2)^5 e^-8/(720) (25)

R5(7) = (-2)^5 e^-8/18000

R5(7) = (-2)^5 e^-8/18000

R5(7) = -1.388888888 x 10^-9

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Answer:

C

Step-by-step explanation:

Multiply z times itself. Then multiply z times that first answer, to get z^3.

There's a formula for cubing a binomial but even if you don't remember it (like me) you can still find z^3.

But remember that i^2 = -1

This is related to sqrt(-1)= i, a definition for the imaginary number.

On the graph, go left or right for the real part and up or down for the imaginary part.

see image.

Answer:

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

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 formula to find the arc length of a circle = θ/360 ×2××r

The formula to find the area of a sector= θ/360××r²

The only thing missing is "θ"

The arc length is given to be 15cm so basically θ/360×2×3.14×12 = 15

θ= 360×15 ÷ 2×3.14×12 = 5400÷ 75.36= 71.65

now that we found "θ" we just continue with the formula to find the area of a sector (θ/360×r²)

=71.65×3.14×12²= 90cm

(it's a bit complicated but i hope this helped i tried to simplify it as much as i could)

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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Answer:

[tex]\frac{x}{x+2}+\frac1x=1\\\rightarrow \frac{x^2+(x+2)}{(x^2+2x}=1\\\rightarrowx^2+x+2=x^2+2x\\\rightarrowx+2=2x\\\rightarrow x=\boxed2[/tex]

Only one solution, which is two. The answer should be C. You could obviously see that if there is no squaring then there will be no extraneous solutions.

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?

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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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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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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Answer:

  see below for the chart

  64 pages in 1 hour

Step-by-step explanation:

You want to fill in the chart showing the analysis of the problem statement saying HP can read 16 pages in 1/4 hour. The problem statement is requesting the number of pages he can read in 1 hour.

See below for a filled-in chart.

Analysis of the bottom line tells you that the right column is 4 times the middle column.

  1 = 4 × 1/4

  ?? = 4 × 16 = 64

Harry Potter can read 64 pages in one hour.

∠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:

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.