what is the domain of the function

Answer:

2/5

Step-by-step explanation:

Add up everything which would be 15, then put the number of ovens their was as the numerator. Then you'd have 6/15 so then just simplify!

Answer:

[tex]t=382[/tex]

Step-by-step explanation:

[tex]\mathrm{Subtract\:}38\mathrm{\:from\:both\:sides}[/tex]

[tex]t+38-38=420-38[/tex]

[tex]=382[/tex]

Answer:

[tex] \displaystyle t = 382[/tex]

Step-by-step explanation:

we are given a equation

we want to figure out t

to do so

recall that

therefore

if we substract 38 from both sides it won't affect the equality

so

[tex] \displaystyle t + 38 - 38 = 420 - 38 \\ \therefore \: t = 382[/tex]

let's verify it

[tex] \displaystyle382 + 38\stackrel{?}{=}420[/tex]

simplify addition:

[tex] \displaystyle420\stackrel{ \checkmark}{=}420[/tex]

hence, verified!

Answer:

> there you go

Step-by-step explanation:

Answer:

>

Step-by-step explanation:

The function f(z) = u(x, y) + iv(x, y) simplifies to f(z) = -8x³y + 8xy³.

The function f(z) corresponding to the given function u(x, y) is f(z) = -8x³y + 8xy³.

To prove that the function u(x, y) = -8x³y + 8xy³ is harmonic, we need to show that it satisfies the Laplace's equation:

∇²u = ∂²u/∂x² + ∂²u/∂y² = 0

Let's calculate the second partial derivatives of u with respect to x and y:

∂u/∂x = -24x²y + 8y³

∂²u/∂x² = -48xy

∂u/∂y = -8x^3 + 24xy²

∂²u/∂y²= 48xy

Now, substitute these values into the Laplace's equation:

∂²u/∂x² + ∂²u/∂y² = -48xy + 48xy = 0

Since the Laplace's equation is satisfied, the function u(x, y) = -8x³y + 8xy³ is harmonic.

b) To find the conjugate harmonic function v, we need to find a function v(x, y) such that f(z) = u(x, y) + iv(x, y) is analytic, where z = x + iy.

Since the given function u(x, y) = -8x³y + 8xy³ is a real-valued function, the imaginary part of the conjugate harmonic function v(x, y) must be zero. Therefore, v(x, y) = 0, and the conjugate harmonic function is purely real.

Hence, the function f(z) = u(x, y) + iv(x, y) simplifies to f(z) = -8x³y + 8xy³.

Therefore, the function f(z) corresponding to the given function u(x, y) is f(z) = -8x³y + 8xy³.

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The exponential function passing through the points (-3, 1/2) and (3, 32) can be represented by the equation [tex]f(x) = (2^{(5/6)}) * (2^{(x/6)})[/tex].

To find the exponential function passing through the given points, we can start by assuming the general form of an exponential function, [tex]f(x) = a * (b^x)[/tex], where a and b are constants to be determined. Plugging in the coordinates (-3, 1/2) into this equation gives us [tex]1/2 = a * (b^{(-3)})[/tex], and plugging in (3, 32) gives us [tex]32 = a * (b^3)[/tex].

Now, we can solve this system of equations to find the values of a and b. Taking the ratio of the two equations, we get [tex](1/2) / 32 = (a * (b^{(-3)})) / (a * (b^3))[/tex], which simplifies to [tex]1/64 = 1/b^6[/tex]. Solving for b, we find [tex]b = 2^{(1/6)[/tex].

Substituting this value back into either of the original equations, we can solve for a. Using the equation [tex]1/2 = a * (2^{(-3/6)})[/tex], we find [tex]a = 2^{(5/6)[/tex].

Therefore, the exponential function passing through the given points is [tex]f(x) = (2^{(5/6)}) * (2^{(x/6)})[/tex].

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Step-by-step explanation:

division fractions that's how u will rewrite them

Answer:

This example is not a random example due to the fact that it is only questioning the eighth-graders. Random sample means that the sample is chosen simply randomly and everyone has an equal opportunity to be apart of the sampling.

Step-by-step explanation:

There are 168 ways for 6 adults and 3 children to stand together in a line with no adjacent children.

we can treat the 6 adults as distinct entities and place them in the line first. There are 6! (factorial) ways to arrange the adults. Now, we have 7 spaces between the adults (including the ends) where the children can be placed. We need to choose 3 out of these 7 spaces for the children, which can be done in 7C3 ways (7 choose 3). Therefore, the total number of arrangements is 6! × 7C3 = 720 × 35 = 16800. However, we need to exclude the cases where two or more children are adjacent. By considering the possible positions of two adjacent children, there are 48 such cases. Subtracting these cases from the total gives us 16800 - 48 = 16752. Therefore, there are 16752 ways for 6 adults and 3 children to stand together in a line with no adjacent children.

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Identification of the relevant MERS: Slow moving inventories (cost $350,000) held at West Co.'s warehouseTotal borrowing of $12 million for the construction of a building Non-current tangible assets held by Hazelnut CoThe accounting treatment for each of the above situations and relevant journal entries are as follows:

Accounting Treatment for Slow moving inventories held at West Co.'s warehouse. The slow-moving inventories held at West Co.'s warehouse will be valued at its cost of $350,000 at 31 December 2016. Since 70% of the inventory was sold, the remaining inventory will be valued at 30% of $350,000 = $105,000The accounting entry to record the sale of the inventory and the commission earned is as follows :Accounting Treatment for Slow moving inventories soldDateAccountTitlesDebitCredit4 April 2016Cost of Goods Sold 84,000Slow moving Inventories 84,000(70% x $350,000)Sales Commission 10,000Sales 10,000(70% x $120,000)Accounting Treatment for Non-current assets held by Hazelnut CoOn 31 December 2016, Hazelnut Co is undecided whether to sell or lease out the non-current assets, but it wants to recognize them as held for sale. The fair value less selling costs of the asset is $37 million, and the value in use is estimated at $38 million. Since the fair value less selling costs is lower than the value in use, the recoverable amount of the asset is lower than the carrying amount, and the asset will be written down.The accounting entry to write down the asset and recognize it as held for sale is as follows:Accounting Treatment for Non-current assets held by Hazelnut CoDateAccountTitlesDebitCredit31 December 2016Loss on Impairment 2,000,000Non-current Assets Held for Sale 2,000,000(Write down of non-current asset to fair value less selling costs of $37m)As of 31 December 2016, Hazelnut Co will recognize depreciation expense for the non-current assets for the year. The accounting entry to recognize the depreciation expense is as follows:Accounting Treatment for Depreciation of Non-current assets held by Hazelnut CoDateAccountTitlesDebitCredit31 December 2016Depreciation Expense 4,000,000Accumulated Depreciation - Non-current Assets 4,000,000(10% x $40,000,000)iii. Accounting Treatment for Total borrowing of $12 million for the construction of a buildingOn 1 January 2018, an entity borrowed $12 million for the construction of a building that is estimated to cost $30 million. Only $5 million was utilized during the year, and the balance of $7 million was invested temporarily in a fixed deposit which yielded an interest of 5% per annum. Total interest accrued on the borrowing of $12 million for the year 2016 was $0.84 million.On 31 December 2016, the entity will recognize the interest expense for the interest accrued on the borrowing of $12 million.The accounting entry to record the interest expense is as follows:Accounting Treatment for Total borrowing of $12 millionDateAccountTitlesDebitCredit31 December 2016Interest Expense 840,000Interest Payable 840,000(5% x $12,000,000)Thus, the accounting treatment for slow-moving inventories sold and non-current assets held by Hazelnut Co includes the recognition of losses due to impairment. The accounting treatment for total borrowing of $12 million includes the recognition of interest expense.

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The relevant MERS for each situation are as follows:

i. West Co.: Inventory

ii. The entity: Borrowing

iii. Hazelnut Co.: Assets

The accounting treatments and journal entries for each situation are given below:

i. West Co.: Inventory

The accounting treatment for slow-moving inventory valuation is to record the inventory’s cost of $350,000 on the balance sheet. When 70% of the inventory is sold for $120,000, the sales amount of $120,000 is recognized on the income statement, while the commission amount of $10,000 is recognized as selling expenses.

The journal entry is as follows:

Income statement: Dr. Cash $120,000Cr.

Revenue $120,000

Dr. Commission expense $10,000Cr.

Cash $10,000

Balance sheet:

Dr. Inventory $245,000Cr.

Cost of goods sold $245,000

ii. The entity: Borrowing

When an entity borrows funds, it records the principal amount on the balance sheet as a liability and recognizes interest expenses as the cost of borrowing. The interest amount for the year 2016 was $0.84 million. Therefore, the journal entry is as follows:

Balance sheet:

Dr. Cash $5,000,000

Dr. Fixed deposit $7,000,000 Cr.

Borrowing $12,000,000

Income statement:

Dr. Interest expense $0.84 million Cr.

Cash $0.84 million

iii. Hazelnut Co.: Assets

When an asset is recognized as held for sale, it is carried at the lower of the fair value less selling costs or value in use. The carrying amount of the asset is reduced to its fair value less selling costs.

The journal entry is as follows:

Balance sheet:

Dr. Provision for impairment $1 million Cr.

Tangible non-current assets $1 million

Balance sheet:

Dr. Tangible non-current assets $3 million Cr.

Provision for impairment $3 million

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T1 and T2 are stop times, both of these events belong to Ft, their union also belongs to Ft. Hence, we can conclude that T1 + T2 is also a stop time.

We are given two stop times T1 and T2 with respect to the same filtration.

We are to prove that the maximum and the sum of T1 and T2, i.e., max(T1, T2) and T1 + T2 are also stop times.  

Let us consider the stop time T1.

This means that the event {T1 ≤ t} belongs to the sigma-algebra Ft, for all t≥0.

Similarly, let us consider the stop time T2.

This means that the event {T2 ≤ t} belongs to the sigma-algebra Ft, for all t≥0.

We are to prove that max(T1, T2) is also a stop time.

We can do so by considering the following event:{max(T1, T2) ≤ t}.

If T1 ≤ T2, then this event reduces to {T2 ≤ t} which belongs to Ft.

Similarly, if T2 ≤ T1, then this event reduces to {T1 ≤ t} which also belongs to Ft.

Thus, we can conclude that max(T1, T2) is a stop time.

We are to prove that T1 + T2 is also a stop time.

We can do so by considering the following event:{T1 + T2 ≤ t}.

This event can be expressed as:{T1 ≤ t − T2} ∪ {T2 ≤ t − T1}.

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Answer:

search it up

Step-by-step explanation:

Answer:

[tex]\boxed {\boxed {\sf sin(B)= \frac {21}{29}}}[/tex]

Step-by-step explanation:

Sine is the ratio of the opposite side to the hypotenuse.

[tex]sin \theta= \frac {opposite}{hypotenuse}[/tex]

We want to find the sine of angle B. The side AC, which measures 21, is opposite angle B.

The side AB, which measures 29, is the hypotenuse because it is the longest side and opposite the right angle.

[tex]opposite= 21\\hypotenuse=29[/tex]

Substitute the values into the formula.

[tex]sinB= \frac {21}{29}[/tex]

This ratio cannot be reduced further, so it is the final answer.

The sine of B is 21/29

Answer: x = 33°

Step-by-step explanation: A triangle's angles should all equal 180, so we can set up an equation to help us.

106 + 41 + x = 180

Add 106 and 41

147 + x = 180

Subtract 147 on both sides.

x = 33

And there you go :)

Step 1: To tell the null and alternative hypotheses for the test:

Step 2: The test statistic for this scenario is the z-score, calculated as:

Step 3 involves comparing the chosen significance level (1% in this instance) with the test statistic to decide whether to reject or accept the null hypothesis. In case the test statistic falls within the critical region, the null hypothesis is rejected, but if it doesn't, the null hypothesis is not rejected.

We can also determine the p-value related to the test statistic and assess its compatibility with the selected level of significance. If the significance level exceeds the p-value, we do not reject the null hypothesis, but if the p-value is lower, then we reject the null hypothesis.

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Answer:

5 /36

Step-by-step explanation:

Sample space :

Number of faces^number of dice = 6^2 = 36

Sum of 8 on a roll of 2 dies :

Probability = required outcome / Total possible outcomes

Required outcome = number of times a sum of 8 is obtained on a roll of 2 dies = 5

Total possible outcomes = sample space = 36

P(obtaining a sum of 8) = 5 /36

27 vessels

Step-by-step explanation:

get the volume of the rectangular vessel 60cm ×10cm ×45cm giving you 27000cm3.Find volume of one of the cube 10cm×10cm×10cm giving 1000cm3.

if 1 cube =1000cm3

? = 27000cm3

27000cm3× 1cube

1000cm3

di

vide then get your answer as 27 vessels

Step-by-step explanation:

no this is a test I cant hlep i will tell ms.jules you are searching for the answers

Answer:

I believe the answer would be m = -28

Step-by-step explanation:

To remove the demoninater, multiple both sides by 7

12*7 = 84 so now you have -3 = 84

Divide 84 by -3, giving you -28

.....um...i think u can't do this

Answer:

u should probably contact support or the creator of this site

Step-by-step explanation:

Answer:

Angles of a triangle add to 180 degrees

42+58+x=180

x=80

The net financing need considering the three accounts is approximately $185,753.42.

To determine the average investment in accounts receivable, inventories, and accounts payable, we need to calculate the Inventory Period, Receivables Period, and Payment Period.

Inventory Period:

The Inventory Period is the average number of days it takes for finished goods to be sold. In this case, the average production process time is given as 40 days, and finished goods are kept on hand for an average of 15 days before they are sold. Therefore, the Inventory Period is the sum of these two periods:

Inventory Period = Production Process Time + Days Kept on Hand

Inventory Period = 40 days + 15 days

Inventory Period = 55 days

Receivables Period:

The Receivables Period is the average number of days it takes for accounts receivable to be collected. It is given that accounts receivable are outstanding for an average of 40 days. Therefore, the Receivables Period is 40 days.

Payment Period:

The Payment Period is the number of days the firm receives credit on its purchases from suppliers. It is given that the firm receives 40 days of credit. Therefore, the Payment Period is 40 days.

Now, we can calculate the average investment in accounts receivable, inventories, and accounts payable.

Average Investment in Accounts Receivable:

Average Investment in Accounts Receivable = (Net Sales / 365) * Receivables Period

Average Investment in Accounts Receivable = ($1,200,000 / 365) * 40

Average Investment in Accounts Receivable ≈ $131,506.85

Average Investment in Inventories:

Average Investment in Inventories = (Cost of Goods Sold / 365) * Inventory Period

Average Investment in Inventories = ($900,000 / 365) * 55

Average Investment in Inventories ≈ $142,191.78

Average Investment in Accounts Payable:

Average Investment in Accounts Payable = (Cost of Goods Sold / 365) * Payment Period

Average Investment in Accounts Payable = ($900,000 / 365) * 40

Average Investment in Accounts Payable ≈ $87,945.21

Finally, to calculate the net financing need, we subtract the average investment in accounts payable from the sum of the average investment in accounts receivable and inventories:

Net Financing Need = Average Investment in Accounts Receivable + Average Investment in Inventories - Average Investment in Accounts Payable

Net Financing Need = $131,506.85 + $142,191.78 - $87,945.21

Net Financing Need ≈ $185,753.42

Therefore, the net financing need considering the three accounts is approximately $185,753.42.

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Answer:

The angle is 58 degrees

Step-by-step explanation:

Given

See attachment for kite

Required

The angle at the tail end

Represent this angle with x.

From the attached kite, we have:

1 angle = 122

2 angles = right-angled

So, we have:

[tex]x + 122 +90+90 = 360[/tex] --- sum of angles in a kite

[tex]x + 302 = 360[/tex]

Solve for x

[tex]x =- 302 + 360[/tex]

[tex]x =58^\circ[/tex]

Answer:

The width of the actual fence is 90ft

Step-by-step explanation:

If 2in = 10ft, you want to find the scale factor. To do this, you need to divide 10/2, so for every 5ft, it is 1in. Now, you would take how many inches you have and multiply it by 5. (in our case, 18in)

18 x 5 = 90.

So the actual fence length is 90ft.

Answer:

42 units²

Step-by-step explanation:

Answer:

Here is the image you need to replicated

(っ◔◡◔)っ ♥ Hope It Helps ♥

The cοrrect answer is (c) Never true, i.e., false. "The rank οf the matrix οf cοefficients οf a hοmοgeneοus system οf m linear equatiοns in n unknοwns is never less than the rank οf the augmented matrix" is nοt always true.

A linear equatiοn is an algebraic equatiοn that represents a straight line when graphed οn a Cartesian cοοrdinate plane. It is an equatiοn in which the highest pοwer οf the variable(s) is 1.

The rank οf the matrix οf cοefficients οf a hοmοgeneοus system οf m linear equatiοns in n unknοwns represents the maximum number οf linearly independent equatiοns in the system. It gives us infοrmatiοn abοut the dimensiοn οf the sοlutiοn space.

On the οther hand, the rank οf the augmented matrix οf the system includes bοth the matrix οf cοefficients and the cοlumn vectοr οf cοnstants. It represents the maximum number οf linearly independent rοws in the augmented matrix.

In general, the rank οf the matrix οf cοefficients can be less than, equal tο, οr greater than the rank οf the augmented matrix. It depends οn the specific system οf equatiοns.

Therefοre, the statement "The rank οf the matrix οf cοefficients οf a hοmοgeneοus system οf m linear equatiοns in n unknοwns is never less than the rank οf the augmented matrix" is nοt always true.

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