To prove that the explicit formula for the nth term of a geometric sequence is given by an = a₀.rⁿ, we will use mathematical induction. Mathematical induction is a proof technique used to establish a statement or property for all natural numbers or integers greater than or equal to a starting value.
Step 1: Base Case
For n = 0, the formula gives a₀.r⁰ = a₀, which is the first term of the sequence. So, the formula holds true for the initial value.
Step 2: Inductive Hypothesis
Assume that the formula holds for some arbitrary value k, i.e., ak = a₀.rᵏ.
Step 3: Inductive Step
We need to prove that the formula also holds for k+1, i.e., a(k+1) = a₀.r^(k+1).
Using the definition of a geometric sequence, we have:
a(k+1) = r.aₖ
Now substitute the inductive hypothesis into the above equation:
a(k+1) = r.(a₀.rᵏ) = (r.a₀).rᵏ = a₀.r^(k+1)
Therefore, we have shown that if the formula holds for k, then it also holds for k+1.
Step 4: Conclusion
By the principle of mathematical induction, the explicit formula for the nth term of a geometric sequence, given by an = a₀.rⁿ, is true for all non-negative integers n.
Thus, we have proved the claim using mathematical induction.
The question should be:
Recall the following definition of an Geometric Sequence.
A sequence a₀, a₁, a₂, ... is called an geometric sequence if, and only if there is a constant r such that
ak = r.a_k_-1 where k≥ 1.
Also recall that we "claimed" the explicit formula for the nth term of the geometric sequence is given by
an = a₀.rⁿ where n≥ 0. Use mathematical induction to prove that our claim is true.
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Solve the equation and determine the value of t.
t + 38 = 420
t =
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
addition equality propertySubstraction equality propertymultiplication equality propertydivision equality propertytherefore
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!
Find the value of X.
Answer:
Angles of a triangle add to 180 degrees
42+58+x=180
x=80
A parent interest group is looking at whether birth order affects scores on the ACT test. It was suggested that, on average, first-born children earn lower ACT scores than second-born children. After surveying a random sample of 100100 first-born children, the parents’ group found that they had a mean score of 23.823.8 on the ACT. A survey of 250250 second-born children resulted in a mean ACT score of 24.124.1. Assume that the population standard deviation for first-born children is known to be 1.21.2 points and the population standard deviation for second-born children is known to be 2.42.4 points. Is there sufficient evidence at the 1%1% level of significance to say that the mean ACT score of first-born children is lower than the mean ACT score of second-born children? Let first-born children be Population 1 and let second-born children be Population 2.
Step 1 of 3 :
State the null and alternative hypotheses for the test. Fill in the blank below.
Step 2: What is the Test Statistic
Step 3: Do we reject or fail to reject the null hypothesis? Do we have sufficient or insufficient data?
Step 1: To tell the null and alternative hypotheses for the test:
Null hypothesis (H0): The mean ACT score of first-born children is equal to or greater than the mean ACT score of second-born children. μ1 ≥ μ2Alternative hypothesis (Ha): The mean ACT score of first-born children is lower than the mean ACT score of second-born children. μ1 < μ2What are the other stepsStep 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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Need help with this
Answer:
search it up
Step-by-step explanation:
The rank of the matrix of coefficients of a homogeneous system of m linear equations in n unknowns is never less than the rank of the augmented matrix. (a) Always true (b) Sometimes true (c) Never true, i.e., false (d) None of the above
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.
What is a linear equatiοn?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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anthony can run at the rate (in meters per minute) shown in the graph below) which of the following best describes anthony’s rate of speed
Step-by-step explanation:
no this is a test I cant hlep i will tell ms.jules you are searching for the answers
Graph y=2x+1
Can you please help.
Answer:
Here is the image you need to replicated
(っ◔◡◔)っ ♥ Hope It Helps ♥
Moving to another question will save this response. uestion 2 15 point Pretty Lady Cosmetic Products has an average production process time of 40 days. Finished goods are kept on hand for an average of 15 days before they are sold. Accounts receivable are outstanding an avera and the firm receives 40 days of credit on its purchases from suppliers. Assume net sales of $1,200,000 and cost of goods sold of $900,000. Determine the average investment in accounts receivable, inventories, and accounts payable. What would be the net financing need conside three accounts? *Note: To solve this problem, you will need to first find the Inventory Period, the Receivables Period, and the Payment Period. $153.054.79 $154,054.79 $152.054.79 $152,154.80
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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= a) Prove that the given function u(x, y) = -8x3y + 8xy3 is harmonic b) Find v, the conjugate harmonic function and write f(z).
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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what equation matches the graph?
Answer:
b
Step-by-step explanation:
Find a formula for the exponential function passing through the points (-3,1/2) and (3,32).
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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solve the equation
13. 9^3c + 1 = 27^3c - 1
In how many ways can 6 adults and 3 children stand together in a line so that no two children are next to each other?
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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How would u rewrite 4/5 and 1 2/3
Step-by-step explanation:
division fractions that's how u will rewrite them
Can someone help me with this problem
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 :)
how do i get all my points and brainlest back
.....um...i think u can't do this
Answer:
u should probably contact support or the creator of this site
Step-by-step explanation:
Is the spinner shown a fair spinner? Explain
why or why not.
5. Let T1 and T2 be two stop times with respect to the same filtration. Prove that me (T1, T2) and T₁ +T2 are also stopping times.
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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The blueprint for a fence shows a rectangle with the dimensions bin x
18in. The scale for the actual fence is 2in : 10ft. What is the length and
width of the actual fence?
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.
One die is blue and the other die is yellow. Both are rolled. What is the probability of getting the shm of 8?
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
Consider this right triangle.
Enter the ratio equivalent to sin (B)
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
Im giving brainliest
Answer:
> there you go
Step-by-step explanation:
Answer:
>
Step-by-step explanation:
Need help with this question
Answer:
Function A rate = 5/1
Function B rate = -4.5
Function A because its rate is positive while Function B's rate is negative
Leah works at an appliance store. She recorded her recent sales. washing machines 3 dishwashers 1 ovens 6 clothes dryers 5 What is the experimental probability that the next appliance Leah sells will be an oven? Write your answer as a fraction or whole number. P(oven)=
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!
What is the measure of the angle at the tail end of the kite? if the top area is where its pointed is 122?
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]
If a and b are the legs of a right triangle, and c is the hypotenuse, what is
the length of b if a = 6 and c = 18.5? (If necessary, round to the nearest
tenth)
Please help!!! I’ll do anything!!
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
d) Vegetable ghee is stored in a rectangular vessel of internal dimensions
60 cm x 10 cm x 45 cm. It is transferred into the identical cubical vessels. If the
internal length of each cubical vessel is 10 cm. how many vessels are required to
empty the rectangular vessel?
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
West Co.'s 2016 financial statements were authorized for issue by its directors on 28 April 2017 and the annual general meeting will be held on 1 June 2017. Slow moving inventories held at one of its warehouses were valued at its cost of $350.000 at 31 December 2016.On 4 April 2016,70% of this inventory was sold for $120,000 on which West Co.'s sales staff earned a commission of $10,000. On 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 utilised 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 Hazelnut Co leased out tangible non-current assets as operating leases. At 1 January 2016, the carrying amount of such assets was $40 million while the remaining useful life was 10 years. These assets were leased out on operating leases that have recently expired. On 31 December 2016, the company is undecided to whether sell or lease the assets to customers under operating leases but wants to recognise it 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. For each of the above situation: i. Identify the relevant MERS i. Explain the accounting treatment Prepare relevant journal entries (note: for the financial year-end is 31 December 2016)
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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Find the area of the shape shown below.
Answer:
42 units²
Step-by-step explanation: