Therefore, we have proved that if A ⊂ B, then inf A ≥ inf B.
Let A, B be nonempty subsets of R that are bounded below. We have to prove that if A ⊂ B, then inf A ≥ inf B.
Let's begin the proof:
We know that since A is a non-empty subset of R and is bounded below, therefore, inf A exists.
Similarly, since B is a non-empty subset of R and is bounded below, therefore, inf B exists. Also, we know that A ⊂ B, which means that every element of A is also an element of B. As a result, we can conclude that inf B ≤ inf A because inf B is less than or equal to each element of B and since each element of B is an element of A, therefore, inf B is less than or equal to each element of A as well.
Therefore, we have proved that if A ⊂ B, then inf A ≥ inf B.
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The doubling period of a bacterial population is 20 minutes. At time \( t=80 \) minutes, the bacterial population was 60000 . What was the initial population at time \( t=0 \) ? Find the size of the b
The size of the bacterial population at time t=100 is 120,000.Since the doubling period of the bacterial population is 20 minutes, this means that every 20 minutes, the population doubles in size. Let's let N be the initial population at time t=0.
After 20 minutes (i.e., at time t=20), the population would have doubled once and become 2N.
After another 20 minutes (i.e., at time t=40), the population would have doubled again and become 4N.
After another 20 minutes (i.e., at time t=60), the population would have doubled again and become 8N.
After another 20 minutes (i.e., at time t=80), the population would have doubled again and become 16N.
We are given that at time t=80, the population was 60,000. Therefore, we can write:
16N = 60,000
Solving for N, we get:
N = 60,000 / 16 = 3,750
So the initial population at time t=0 was 3,750.
Now let's find the size of the bacterial population at time t=100 (i.e., 20 minutes after t=80). Since the population doubles every 20 minutes, the population at time t=100 should be double the population at time t=80, which was 60,000. Therefore, the population at time t=100 should be:
2 * 60,000 = 120,000
So the size of the bacterial population at time t=100 is 120,000.
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If the population of a certain region is now 6.7 billion people and if it continues to grow at an annual rate of 1.3% compounded continuously, how long (to the nearest year) would it take before there is only 1 square yard of land per person in the region? (The region contains approximately 1.61 x 10¹ square yards of land.) Which equation could be used to find the number of years it would take before there is only 1 square yard of land per person in the region? (Type an equation using t as the variable. Type an exact answer in terms of e. Use scientific notation. Use the multiplication symbol in the math palette as needed. Use integers or decimals for any numbers in the equation. Do not simplify.) How long would take before there is only 1 square yard of land per person in the region? years (Round to the nearest integer as needed.)
It would take approximately 37 years before there is only 1 square yard of land per person in the region.
To solve this problem, we can use the formula for continuous compound interest, which can also be applied to population growth:
[tex]A = P * e^(rt)[/tex]
Where:
A = Final amount
P = Initial amount
e = Euler's number (approximately 2.71828)
r = Growth rate
t = Time
In this case, the initial population (P) is 6.7 billion people, and the final population (A) is the population at which there is only 1 square yard of land per person.
Let's denote the final population as P_f and the final amount of land as A_f. We know that A_f is given by 1.61 x 10¹ square yards. We need to find the value of P_f.
Since there is 1 square yard of land per person, the total land (A_f) should be equal to the final population (P_f). Therefore, we have:
A_f = P_f
Substituting these values into the formula, we get:
[tex]A_f = P * e^(rt)[/tex]
[tex]1.61 x 10¹ = 6.7 billion * e^(0.013t)[/tex]
Simplifying, we divide both sides by 6.7 billion:
[tex](1.61 x 10¹) / (6.7 billion) = e^(0.013t)[/tex]
Now, to isolate the exponent, we take the natural logarithm (ln) of both sides:
[tex]ln[(1.61 x 10¹) / (6.7 billion)] = ln[e^(0.013t)][/tex]
Using the property of logarithms, [tex]ln(e^x) = x,[/tex]we can simplify further:
[tex]ln[(1.61 x 10¹) / (6.7 billion)] = 0.013t[/tex]
Now, we can solve for t by dividing both sides by 0.013:
[tex]t = ln[(1.61 x 10¹) / (6.7 billion)] / 0.013[/tex]
Calculating the right side of the equation, we find:
t ≈ 37.17
Therefore, it would take approximately 37 years before there is only 1 square yard of land per person in the region.
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Question 10 Write the equation in slope-intercept form of the line with a slope of -5 passing through (-4, 22). y= Submit Question G
y=-5x+2 is the equation in slope-intercept form of the line with a slope of -5 passing through (-4, 22).
The slope of the line is the ratio of the rise to the run, or rise divided by the run. It describes the steepness of line in the coordinate plane.
The slope intercept form of a line is y=mx+b, where m is slope and b is the y intercept.
The given slope is -5.
Let us find the y intercept.
22=-5(-4)+b
22=20+b
Subtract 20 from both sides:
b=2
So equation is y=-5x+2.
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The function f(x) = (x - tan x)/ {x^{3}} has a hole at the point (0, b). Find b.
To find the value of b for the function f(x) = (x - tan(x))/x^3 at the point (0, b), we need to evaluate the limit of the function as x approaches 0. By applying the limit definition, we can determine the value of b.
To find the value of b, we evaluate the limit of the function f(x) as x approaches 0. Taking the limit involves analyzing the behavior of the function as x gets arbitrarily close to 0.
Using the limit definition, we can rewrite the function as f(x) = (x/x^3) - (tan(x)/x^3). As x approaches 0, the first term simplifies to 1/x^2, while the second term approaches 0 because tan(x) approaches 0 as x approaches 0. Therefore, the limit of the function f(x) as x approaches 0 is 1/x^2.
Since we are interested in finding the value of b at the point (0, b), we evaluate the limit of f(x) as x approaches 0. The limit of 1/x^2 as x approaches 0 is ∞. Therefore, the value of b at the point (0, b) is ∞, indicating that there is a hole at the point (0, ∞) on the graph of the function.
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A steep mountain is inclined 74 degree to the horizontal and rises to a height of 3400 ft above the surrounding plain. A cable car is to be installed running to the top of the mountain from a point 920 ft out in the plain from the base of the mountain. Find the shortest length of cable needed. Round your answer to the nearest foot.
The shortest length of cable needed is ft
The shortest length ( hypotenuse) of cable needed is approximately 3500 ft (rounded to the nearest foot).
To find the shortest length of cable needed, we can use trigonometry to calculate the hypotenuse of a right triangle formed by the height of the mountain and the horizontal distance from the base of the mountain to the cable car installation point.
Let's break down the given information:
- The mountain is inclined at an angle of 74 degrees to the horizontal.
- The mountain rises to a height of 3400 ft above the surrounding plain.
- The cable car installation point is 920 ft out in the plain from the base of the mountain.
We can use the sine function to relate the angle and the height of the mountain:
sin(angle) = opposite/hypotenuse
In this case, the opposite side is the height of the mountain, and the hypotenuse is the length of the cable car needed. We can rearrange the equation to solve for the hypotenuse:
hypotenuse = opposite/sin(angle)
hypotenuse = 3400 ft / sin(74 degrees)
hypotenuse ≈ 3500.49 ft (rounded to 2 decimal places)
So, the shortest length of cable needed is approximately 3500 ft (rounded to the nearest foot).
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Consider the integer numbers in set U={0,1,2,3,4,5,6,7,8,9,10}. Let : P={0,2,4,6,8,10}
Q={x∣0≤x≤6}
R={x∣3
i) Find: (P∪Q)−(Q∩R) ii) List the ordered pairs in the relation S on the set (Q∩R), where S={(a,b), if a+b>=11}
i) (P∪Q)−(Q∩R)={4, 6, 8, 10, 5}
ii) The ordered pairs in the relation S on the set (Q∩R) are {(2,3), (3,2), (3,3)}.
i) We need to find (P∪Q)−(Q∩R).
P∪Q is the union of sets P and Q, which contains all the elements in P and Q. So,
P∪Q={0, 2, 4, 6, 8, 10, 1, 3, 5, 6}
Q∩R is the intersection of sets Q and R, which contains only the elements that are in both Q and R. So,
Q∩R={0, 1, 2, 3}
Therefore,
(P∪Q)−(Q∩R)={4, 6, 8, 10, 5}
ii) We need to list the ordered pairs in the relation S on the set (Q∩R), where S={(a,b), if a+b[tex]\geq[/tex]11}.
(Q∩R)={0, 1, 2, 3}
To find the ordered pairs that satisfy the relation S, we need to find all pairs (a,b) such that a+b[tex]\geq[/tex]11.
The pairs are:
(2, 3)
(3, 2)
(3, 3)
So, the ordered pairs in the relation S on the set (Q∩R) are {(2,3), (3,2), (3,3)}.
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A bond paying $20 in semi-annual coupon payments with an current
yield of 5.25% will sell at:
Therefore, the bond will sell at approximately $761.90.
To determine the selling price of the bond, we need to calculate the present value of its cash flows.
The bond pays $20 in semi-annual coupon payments, which means it pays $40 annually ($20 * 2) in coupon payments.
The current yield of 5.25% represents the yield to maturity (YTM) or the required rate of return for the bond.
To calculate the present value, we can use the formula for the present value of an annuity:
Present Value = Coupon Payment / YTM
In this case, the Coupon Payment is $40 and the YTM is 5.25% or 0.0525.
Present Value = $40 / 0.0525
Calculating the present value:
Present Value ≈ $761.90
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If 9 people will attend a lunch and 3 cans of juice should be
provided per person, how many total cans of juice are needed?
3 cans
27 cans
12 cans
18 cans
The total of 27 cans of juice are needed for the lunch.
We multiply the total number of lunch attendees by the average number of juice cans per person to determine the total number of cans of juice required.
How many people attended the lunch? 9 juice cans per person: 3
Number of individuals * total number of juice cans *Cans per individual
Juice cans required in total: 9 * 3
27 total cans of juice are required.
For the lunch, a total of 27 cans of juice are required.
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Question 1 Calculator For the function f(x) = 5x² + 3x, evaluate and simplify. f(x+h)-f(x) h Check Answer ▼ || < >
The solution to the given problem is `f(x + h) - f(x) / h = 10x + 5h + 3` and the slope of the given function `f(x) = 5x² + 3x` is `10x + 5h + 3`.
To evaluate and simplify the function `f(x) = 5x² + 3x`, we need to substitute the given equation in the formula for `f(x + h)` and `f(x)` and then simplify. Thus, the given expression can be expressed as
`f(x + h) = 5(x + h)² + 3(x + h)` and
`f(x) = 5x² + 3x`
To solve this expression, we need to substitute the above values in the above mentioned formula.
i.e., `
= f(x + h) - f(x) / h
= [5(x + h)² + 3(x + h)] - [5x² + 3x] / h`.
After substituting the above values in the formula, we get:
`f(x + h) - f(x) / h = [5x² + 10xh + 5h² + 3x + 3h] - [5x² + 3x] / h`
Therefore, by simplifying the above expression, we get:
`= f(x + h) - f(x) / h
= (10xh + 5h² + 3h) / h
= 10x + 5h + 3`.
Thus, the final value of the given expression is `10x + 5h + 3` and the slope of the function `f(x) = 5x² + 3x`.
Therefore, the solution to the given problem is `f(x + h) - f(x) / h = 10x + 5h + 3` and the slope of the given function `f(x) = 5x² + 3x` is `10x + 5h + 3`.
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2. Given that the linear system Ax=b has a particular solution p. Show that for every solution y of Ax=b, there is a solution v of the homogeneous linear system Ax=0 such that y=p+v. Hint: Consider y−p.
This proves that for every solution y of Ax = b, there is a solution v of the homogeneous linear system Ax = 0 such that y = p + v.
Given that the linear system Ax = b has a particular solution p.
We are supposed to show that for every solution y of Ax = b, there is a solution v of the homogeneous linear system Ax = 0 such that y = p + v.
Hint: Consider y - p.
To prove this, we can consider the difference between the two solutions y and p and take that as our solution v of Ax = 0.
Since p is a solution to Ax = b,
it follows that Ap = b.
Since y is also a solution to Ax = b,
it follows that Ay = b.
We can subtract the two equations to get:
Ay - Ap = 0 which gives us:
A(y - p) = 0
So, the solution to Ax = 0 is y - p,
which means that there exists some vector v such that Av = 0 and y - p = v.
Therefore, we have y = p + v where v is a solution of Ax = 0.
Hence, this proves that for every solution y of Ax = b, there is a solution v of the homogeneous linear system Ax = 0 such that y = p + v.
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Convert the given measurements to the indicated units using dimensional analysis. (Round your answers to two decimal places.) (a) 310ft=yd (b) 3.5mi=ft (c) 96 in =ft (d) 2100yds=mi Additional Materials /2 Points] FIERROELEMMATH1 11.2.005. Use a formula to find the area of the triangle. square units
The solutions are
(a) 310 ft is equivalent to 103.33 yd.
(b) 3.5 mi is equivalent to 18,480 ft.
(c) 96 in is equivalent to 8 ft.
(d) 2,100 yds is equivalent to 1.19 mi.
To convert measurements using dimensional analysis, we use conversion factors that relate the two units of measurement.
(a) To convert 310 ft to yd, we know that 1 yd is equal to 3 ft. Using this conversion factor, we set up the proportion: 1 yd / 3 ft = x yd / 310 ft. Solving for x, we find x ≈ 103.33 yd. Therefore, 310 ft is approximately equal to 103.33 yd.
(b) To convert 3.5 mi to ft, we know that 1 mi is equal to 5,280 ft. Setting up the proportion: 1 mi / 5,280 ft = x mi / 3.5 ft. Solving for x, we find x ≈ 18,480 ft. Hence, 3.5 mi is approximately equal to 18,480 ft.
(c) To convert 96 in to ft, we know that 1 ft is equal to 12 in. Setting up the proportion: 1 ft / 12 in = x ft / 96 in. Solving for x, we find x = 8 ft. Therefore, 96 in is equal to 8 ft.
(d) To convert 2,100 yds to mi, we know that 1 mi is equal to 1,760 yds. Setting up the proportion: 1 mi / 1,760 yds = x mi / 2,100 yds. Solving for x, we find x ≈ 1.19 mi. Hence, 2,100 yds is approximately equal to 1.19 mi.
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Let X={1,3,5} and Y={s,t,u,v}. Define f:X→Y by the following arrow diagram. a. Write the domain of f and the co-domain of f. b. Find f(1),f(3), and f(5). c. What is the range of f ? 17. Define vertex set V, edge set E, order, size and degree sequence.
The domain of f is X and the co-domain of f is Y And f(1) = s, f(3) = t, f(5) = u. The range of f is {s, t, u}.
a. The domain of function f is X, which consists of the elements {1, 3, 5}. The co-domain of f is Y, which consists of the elements {s, t, u, v}.
b. Evaluating f(x) for each element in the domain, we have:
f(1) = s
f(3) = t
f(5) = u
c. The range of f represents the set of all possible output values. From the given information, we can see that f(1) = s, f(3) = t, and f(5) = u. Therefore, the range of f is the set {s, t, u}.
In graph theory, a graph consists of a vertex set V and an edge set E. The order of a graph is the number of vertices in the vertex set V. The size of a graph is the number of edges in the edge set E. The degree sequence of a graph represents the degrees of its vertices listed in non-increasing order.
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A.
Translate each sentence into an algebraic equation.
1.A number increased by four is twelve.
2.A number decreased by nine is equal to eleven.
3. Five times a number is fifty.
4. The quotient of a number and seven is eight.
5. The sum of a number and ten is twenty.
6. The difference between six and a number is two.
7. Three times a number increased by six is fifteen.
8. Eight less than twice a number is sixteen.
9. Thirty is equal to twice a number decreased by four.
10. If four times a number is added to nine, the result is forty-nine
To translate each sentence into an algebraic equations are:
1. x + 4 = 12, 2. x - 9 = 11. 3. 5x = 50, 4. x / 7 = 8, 5. x + 10 = 20, 6. 6 - x = 2, 7. 3x + 6 = 15, 8. 2x - 8 = 16, 9. 30 = 2x - 4, 10. 4x + 9 = 49
1. A number increased by four is twelve.
Let's denote the unknown number as "x".
Algebraic equation: x + 4 = 12
2. A number decreased by nine is equal to eleven.
Algebraic equation: x - 9 = 11
3. Five times a number is fifty.
Algebraic equation: 5x = 50
4. The quotient of a number and seven is eight.
Algebraic equation: x / 7 = 8
5. The sum of a number and ten is twenty.
Algebraic equation: x + 10 = 20
6. The difference between six and a number is two.
Algebraic equation: 6 - x = 2
7. Three times a number increased by six is fifteen.
Algebraic equation: 3x + 6 = 15
8. Eight less than twice a number is sixteen.
Algebraic equation: 2x - 8 = 16
9. Thirty is equal to twice a number decreased by four.
Algebraic equation: 30 = 2x - 4
10. If four times a number is added to nine, the result is forty-nine.
Algebraic equation: 4x + 9 = 49
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8) In Germany gas costs 0.79 Euros for a liter of gas. Convert this price from Euros per liter to dollars per gallon. ( \( 3.79 \mathrm{~L}=1 \mathrm{gal}, \$ 1.12=1 \) Euro)
The cost of gas in Germany is $0.239/gal.
A conversion factor is a numerical value used to convert one unit of measurement to another. It is a ratio derived from the equivalence between two different units of measurement. By multiplying a quantity by the appropriate conversion factor, express the same value in different units.
Conversion factors:1 gal = 3.79 L1€ = $1.12
convert the cost of gas from €/L to $/gal.
Using the conversion factor: 1 gal = 3.79 L
1 L = 1/3.79 gal
Multiply both numerator and denominator of
€0.79/L
with the reciprocal of
1€/$1.12,
which is
$1.12/1€.€0.79/L × $1.12/1€ × 1/3.79 gal
= $0.79/L × $1.12/1€ × 1/3.79 gal
= $0.239/gal
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1) David makes clay vases in the shape of right triangular prisms, as shown, then paints them bright colours. A can of spray paint costs $5.49 and covers 2 m 2
. How much will it cost David to paint the outer surface of 15 vases, including the bottom, with three coats of paint? Assume the vases do not have lids. [6]
The cost for David to paint the outer surface of 15 vases, including the bottom, with three coats of paint is $4,005.30First, we need to calculate the surface area of one vase:
Cost of painting 15 vases = 15 × $2.03 = $30.45But this is only for one coat. We need to apply three coats, so the cost of painting the outer surface of 15 vases, including the bottom, with three coats of paint will be:Cost of painting 15 vases with 3 coats of paint = 3 × $30.45 = $91.35The cost of painting the outer surface of 15 vases, including the bottom, with three coats of paint will be $91.35.Hence, the : The cost for David to paint the outer surface of 15 vases, including the bottom, with three coats of paint is $4,005.30.
Height of prism = 12 cmLength of base = 24 cm
Width of base = 24 cmSlant
height = hypotenuse of the base triangle = `
sqrt(24^2 + 12^2) =
sqrt(720)` ≈ 26.83 cmSurface area of one vase = `2 × (1/2 × 24 × 12 + 24 × 26.83) = 2 × 696.96` ≈ 1393.92 cm²
Paint will be applied on both the sides of the vase, so the outer surface area of one vase = 2 × 1393.92 = 2787.84 cm
We know that a can of spray paint covers 2 m² and costs $5.49. Converting cm² to m²:
1 cm² = `10^-4 m²`Therefore, 2787.84 cm² = `2787.84 × 10^-4 = 0.278784 m²
`David wants to apply three coats of paint on each vase, so the cost of painting one vase will be:
Cost of painting one vase = 3 × (0.278784 ÷ 2) × $5.49 = $2.03
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Powers can undo roots, and roots can undo powers. True or false? Any number with an exponent of 0 is equal to 0. True or false?
Rachel bought a meal and gave an 18% tip. If the tip was $6.30 and there was no sales tax, how much did her meal cost?
The statement "Powers can undo roots, and roots can undo powers" is generally false.
Rachel's meal cost $35. This was determined by dividing the tip amount of $6.30 by the tip percentage of 18%.
To find out how much Rachel's meal cost, we can start by calculating the total amount including the tip. We know that the tip amount is $6.30, and it represents 18% of the total cost. Let's assume the total cost of the meal is represented by the variable 'x'.
So, we can set up the equation: 0.18 * x = $6.30.
To isolate 'x', we need to divide both sides of the equation by 0.18: x = $6.30 / 0.18.
Now, we can calculate the value of 'x'. Dividing $6.30 by 0.18 gives us $35.
Therefore, Rachel's meal cost $35.
In summary, Rachel's meal cost $35. This was determined by dividing the tip amount of $6.30 by the tip percentage of 18%.
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Find the inverse function of f(x)=15+³√x f−1(x)=
Answer:
f−1(x) = (x - 15)³
Step-by-step explanation:
f(x)=15+³√x
And to inverse the function we need to switch the x for f−1(x), and then solve for f−1(x):
x =15+³√(f−1(x))
x- 15 =15+³√(f−1(x)) -15
x - 15 = ³√(f−1(x))
(x-15)³ = ( ³√(f−1(x)) )³
(x - 15)³= f−1(x)
f−1(x) = (x - 15)³
2. Find the domain of the function f(x) = a. (-[infinity]0,00) b. x 1 c. x = -2 d. x = 2,-1 e. x = -2,1 1 x²+x-2
The correct answer is a. (-∞, +∞), which represents all real numbers.
The collection of values for x that define the function, f(x) = x2 + x - 2, must be identified in order to identify its domain.
Polynomials are defined for all real numbers, and the function that is being presented is one of them. As a result, the set of all real numbers, indicated by (-, +), is the domain of the function f(x) = x2 + x - 2.
As a result, (-, +), which represents all real numbers, is the right response.
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The rate constant for the beta decay of thorium-234 is 2.881 x
10-2 day-1. What is the half-life of this
nuclide?
Therefore, the half-life of Thorium-234 is approximately 24.1 days.
Given that the rate constant for the beta decay of thorium-234 is 2.881 x 10-2 day-1.
We are to find the half-life of this nuclide.
A rate constant is a proportionality constant that links the concentration of reactants to the rate of the reaction. It is denoted by k. It is always specific to a reaction and is dependent on temperature.
A half-life is the time taken for half of the radioactive atoms in a sample to decay. It is denoted by t1/2.
To find the half-life, we use the following formula:
ln (2)/ k = t1/2,
where k is the rate constant given and ln is the natural logarithm.
Now, substituting the given values,
ln (2)/ (2.881 x 10-2 day-1) = t1/2t1/2 = ln (2)/ (2.881 x 10-2 day-1)≈ 24.1 days
Therefore, the half-life of Thorium-234 is approximately 24.1 days.
The half-life of thorium-234 is approximately 24.1 days.
The half-life of a nuclide is the time taken for half of the radioactive atoms in a sample to decay. It is denoted by t1/2. It is used to determine the rate at which a substance decays.
The rate constant is a proportionality constant that links the concentration of reactants to the rate of the reaction. It is denoted by k. It is always specific to a reaction and is dependent on temperature.
The formula used to find the half-life of a nuclide is ln (2)/ k = t1/2, where k is the rate constant given and ln is the natural logarithm.
Given the rate constant for the beta decay of thorium-234 is 2.881 x 10-2 day-1, we can use the above formula to find the half-life of the nuclide.
Substituting the given values,
ln (2)/ (2.881 x 10-2 day-1) = t1/2t1/2 = ln (2)/ (2.881 x 10-2 day-1)≈ 24.1 days
Therefore, the half-life of Thorium-234 is approximately 24.1 days.
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Let f : Z × Z → Z × Z be a function defined by g(m, n) = 2 − n, 3 + m). a. Carefully prove that f is injective (one-to-one). Important! In each step of your proof make sure it is clear whether what is written is something you are assuming, something you are about to prove, or something that follows from a previous step. If any variables appear in your proof, make sure you clearly write what they represent. b. Carefully prove that f is surjective (onto). Justify your answer
We have shown that the function f : Z × Z → Z × Z defined by f(m, n) = (2 − n, 3 + m) is both injective (one-to-one) and surjective (onto), satisfying the conditions of a bijective function.
a. To prove that the function f : Z × Z → Z × Z defined by f(m, n) = (2 − n, 3 + m) is injective (one-to-one), we need to show that for any two distinct inputs (m1, n1) and (m2, n2) in Z × Z, their corresponding outputs under f are also distinct.
Let (m1, n1) and (m2, n2) be two arbitrary distinct inputs in Z × Z. We assume that f(m1, n1) = f(m2, n2) and aim to prove that (m1, n1) = (m2, n2).
By the definition of f, we have (2 − n1, 3 + m1) = (2 − n2, 3 + m2). From this, we can deduce two separate equations:
1. 2 − n1 = 2 − n2 (equation 1)
2. 3 + m1 = 3 + m2 (equation 2)
From equation 1, we can see that n1 = n2, and from equation 2, we can observe that m1 = m2. Therefore, we conclude that (m1, n1) = (m2, n2), which confirms the injectivity of the function.
b. To prove that the function f : Z × Z → Z × Z defined by f(m, n) = (2 − n, 3 + m) is surjective (onto), we need to show that for every element (a, b) in the codomain Z × Z, there exists an element (m, n) in the domain Z × Z such that f(m, n) = (a, b).
Let (a, b) be an arbitrary element in Z × Z. We need to find values for m and n such that f(m, n) = (2 − n, 3 + m) = (a, b).
From the first component of f(m, n), we have 2 − n = a, which implies n = 2 − a.
From the second component of f(m, n), we have 3 + m = b, which implies m = b − 3.
Therefore, by setting m = b − 3 and n = 2 − a, we have f(m, n) = (2 − n, 3 + m) = (2 − (2 − a), 3 + (b − 3)) = (a, b).
Hence, for every element (a, b) in the codomain Z × Z, we can find an element (m, n) in the domain Z × Z such that f(m, n) = (a, b), demonstrating the surjectivity of the function.
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Solve 4x 2
+24x−5=0 by completing the square. Leave your final answers as exact values in simplified form.
To solve the quadratic equation 4x^2 + 24x - 5 = 0 by completing the square, we follow a series of steps. First, we isolate the quadratic terms and constant term on one side of the equation.
Then, we divide the entire equation by the coefficient of x^2 to make the leading coefficient equal to 1. Next, we complete the square by adding a constant term to both sides of the equation. Finally, we simplify the equation, factor the perfect square trinomial, and solve for x.
Given the quadratic equation 4x^2 + 24x - 5 = 0, we start by moving the constant term to the right side of the equation:
4x^2 + 24x = 5
Next, we divide the entire equation by the coefficient of x^2, which is 4:
x^2 + 6x = 5/4
To complete the square, we add the square of half the coefficient of x to both sides of the equation. In this case, half of 6 is 3, and its square is 9:
x^2 + 6x + 9 = 5/4 + 9
Simplifying the equation, we have:
(x + 3)^2 = 5/4 + 36/4
(x + 3)^2 = 41/4
Taking the square root of both sides, we obtain:
x + 3 = ± √(41/4)
Solving for x, we have two possible solutions:
x = -3 + √(41/4)
x = -3 - √(41/4)
These are the exact values in simplified form for the solutions to the quadratic equation.
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4. Let f : A → B.
(a) Decide if the following statement is true or false, and prove your answer: for all subsets S and T of A, f(S \ T) ⊆ f(S) \ f(T). If the statement is false, decide if the assumption that f is one-to-one, or that f is onto, will make the statement true, and prove your answer.
(b) Repeat part (a) for the reverse containment.
(a) The statement f(S \ T) ⊆ f(S) \ f(T) is false and here is the proof:
Let A = {1, 2, 3}, B = {4, 5}, and f = {(1, 4), (2, 4), (3, 5)}.Then take S = {1, 2}, T = {2, 3}, so S \ T = {1}, then f(S \ T) = f({1}) = {4}.
Moreover, we have f(S) = f({1, 2}) = {4} and f(T) = f({2, 3}) = {4, 5},thus f(S) \ f(T) = { } ≠ f(S \ T), which implies that the statement is false.
Then to show that the assumption that f is one-to-one, or that f is onto, will make the statement true, we can consider the following two cases. Case 1: If f is one-to-one, the statement will be true.We will prove this statement by showing that f(S \ T) ⊆ f(S) \ f(T) and f(S) \ f(T) ⊆ f(S \ T).
For f(S \ T) ⊆ f(S) \ f(T), take any x ∈ f(S \ T), then there exists y ∈ S \ T such that f(y) = x. Since y ∈ S, it follows that x ∈ f(S).
Suppose that x ∈ f(T), then there exists z ∈ T such that f(z) = x.
But since y ∉ T, we get y ∈ S and y ∉ T,
which implies that z ∉ S.
Thus, we have f(y) = x ∈ f(S) \ f(T).
Therefore, f(S \ T) ⊆ f(S) \ f(T).For f(S) \ f(T) ⊆ f(S \ T),
take any x ∈ f(S) \ f(T), then there exists y ∈ S such that f(y) = x, and y ∉ T. Thus, y ∈ S \ T, and it follows that x = f(y) ∈ f(S \ T).
Therefore, f(S) \ f(T) ⊆ f(S \ T).
Thus, we have shown that f(S \ T) ⊆ f(S) \ f(T) and f(S) \ f(T) ⊆ f(S \ T), which implies that f(S \ T) = f(S) \ f(T) for all subsets S and T of A,
when f is one-to-one.
Case 2: If f is onto, the statement will be true.
We will prove this statement by showing that f(S \ T) ⊆ f(S) \ f(T) and f(S) \ f(T) ⊆ f(S \ T).For f(S \ T) ⊆ f(S) \ f(T),
take any x ∈ f(S \ T), then there exists y ∈ S \ T such that f(y) = x.
Suppose that x ∈ f(T), then there exists z ∈ T such that f(z) = x.
But since y ∉ T, it follows that z ∈ S, which implies that x = f(z) ∈ f(S). Therefore, x ∈ f(S) \ f(T).For f(S) \ f(T) ⊆ f(S \ T), take any x ∈ f(S) \ f(T),
then there exists y ∈ S such that f(y) = x, and y ∉ T. Since f is onto, there exists z ∈ A such that f(z) = y.
Thus, z ∈ S \ T, and it follows that f(z) = x ∈ f(S \ T).
Therefore, x ∈ f(S) \ f(T).Thus, we have shown that f(S \ T) ⊆ f(S) \ f(T) and f(S) \ f(T) ⊆ f(S \ T), which implies that f(S \ T) = f(S) \ f(T) for all subsets S and T of A, when f is onto.
The statement f(S \ T) ⊆ f(S) \ f(T) is false. The assumption that f is one-to-one or f is onto makes the statement true.(b) Repeat part (a) for the reverse containment.Since the conclusion of part (a) is that f(S \ T) = f(S) \ f(T) for all subsets S and T of A, when f is one-to-one or f is onto, then the reverse containment f(S) \ f(T) ⊆ f(S \ T) will also hold, and the proof will be the same.
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Solve the system of equation by the method of your choice if the the system has a unique solution, type in that answer as an ordered triple. If the system is inconsistebt or dependent type in "no solutio"
-4x-6z=-12
-6x-4y-2z = 6
−x + 2y + z = 9
The solution is given as (-4 + z, (-46z + 240)/56, z), where z can take any real value.
To solve the system of equations:
-4x - 6z = -12 ...(1)
-6x - 4y - 2z = 6 ...(2)
-x + 2y + z = 9 ...(3)
We can solve this system by using the method of Gaussian elimination.
First, let's multiply equation (1) by -3 and equation (2) by -2 to create opposite coefficients for x in equations (1) and (2):
12x + 18z = 36 ...(4) [Multiplying equation (1) by -3]
12x + 8y + 4z = -12 ...(5) [Multiplying equation (2) by -2]
-x + 2y + z = 9 ...(3)
Now, let's add equations (4) and (5) to eliminate x:
(12x + 18z) + (12x + 8y + 4z) = 36 + (-12)
24x + 8y + 22z = 24 ...(6)
Next, let's multiply equation (3) by 24 to create opposite coefficients for x in equations (3) and (6):
-24x + 48y + 24z = 216 ...(7) [Multiplying equation (3) by 24]
24x + 8y + 22z = 24 ...(6)
Now, let's add equations (7) and (6) to eliminate x:
(-24x + 48y + 24z) + (24x + 8y + 22z) = 216 + 24
56y + 46z = 240 ...(8)
We are left with two equations:
56y + 46z = 240 ...(8)
-x + 2y + z = 9 ...(3)
We can solve this system of equations using various methods, such as substitution or elimination. Here, we'll use elimination to eliminate y:
Multiplying equation (3) by 56:
-56x + 112y + 56z = 504 ...(9) [Multiplying equation (3) by 56]
56y + 46z = 240 ...(8)
Now, let's subtract equation (8) from equation (9) to eliminate y:
(-56x + 112y + 56z) - (56y + 46z) = 504 - 240
-56x + 112y - 56y + 56z - 46z = 264
-56x + 56z = 264
Dividing both sides by -56:
x - z = -4 ...(10)
Now, we have two equations:
x - z = -4 ...(10)
56y + 46z = 240 ...(8)
We can solve this system by substitution or another method of choice. Let's solve it by substitution:
From equation (10), we have:
x = -4 + z
Substituting this into equation (8):
56y + 46z = 240
Simplifying:
56y = -46z + 240
y = (-46z + 240)/56
Now, we can express the solution as an ordered triple (x, y, z):
x = -4 + z
y = (-46z + 240)/56
z = z
Therefore, the solution is given as (-4 + z, (-46z + 240)/56, z), where z can take any real value
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The population of a pod of bottlenose dolphins is modeled by the function A(t)=15(1.2)t, where t is given in years. To the nearest whole number, what will the pod population be after 5 years? The pod population will be dolphins.
Rounding to the nearest whole number, the pod population after 5 years will be approximately 37 dolphins.
To find the pod population after 5 years, we can substitute t = 5 into the given function [tex]A(t) = 15(1.2)^t[/tex] and evaluate it.
[tex]A(t) = 15(1.2)^t\\A(5) = 15(1.2)^5[/tex]
Calculating the expression:
[tex]A(5) = 15(1.2)^5[/tex]
≈ 15(2.48832)
≈ 37.3248
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Suppose that an arithmetic sequence has \( a_{12}=60 \) and \( a_{20}=84 \). Find \( a_{1} \).
Find \( a_{1} \) if \( S_{14}=168 \) and \( a_{14}=25 \)
Suppose that an arithmetic sequence has [tex]\( a_{12}=60 \) and \( a_{20}=84 \)[/tex] Find [tex]\( a_{1} \)[/tex] Also, find [tex]\( a_{1} \) if \( S_{14}=168 \) and \( a_{14}=25 \).[/tex]
Given, an arithmetic sequence has [tex]\( a_{12}=60 \) and \( a_{20}=84 \)[/tex] .We need to find [tex]\( a_{1} \)[/tex]
Formula of arithmetic sequence is: [tex]$$a_n=a_1+(n-1)d$$$$a_{20}=a_1+(20-1)d$$$$84=a_1+19d$$ $$a_{12}=a_1+(12-1)d$$$$60=a_1+11d$$[/tex]
Subtracting above two equations, we get
[tex]$$24=8d$$ $$d=3$$[/tex]
Put this value of d in equation [tex]\(84=a_1+19d\)[/tex], we get
[tex]$$84=a_1+19×3$$ $$84=a_1+57$$ $$a_1=27$$[/tex]
Therefore, [tex]\( a_{1}=27 \)[/tex]
Given, [tex]\(S_{14}=168\) and \(a_{14}=25\).[/tex] We need to find[tex]\(a_{1}\)[/tex].We know that,
[tex]$$S_n=\frac{n}{2}(a_1+a_n)$$ $$S_{14}=\frac{14}{2}(a_1+a_{14})$$ $$168=7(a_1+25)$$ $$24= a_1+25$$ $$a_1=-1$$[/tex]
Therefore, [tex]\( a_{1}=-1 \).[/tex]
Therefore, the first term of the arithmetic sequence is -1.
The first term of the arithmetic sequence is 27 and -1 for the two problems given respectively.
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In a town whose poputation is 3300 , a disease creaces an 4 ? a) How many are insaly indected with the dasease (t = O)? Round to the nearest whole number os needed.) b) Find the number indected affer 2 doys, 5 days, 8 day, 12 dpys, and 16 daya. The rumber infected after 2 days a (Found to the nearett whole namber at needed) The number infecied afler 5 days is . Feound to the rearest whole numbers as needed.) The number intected ater 8 days is (Alound fo the nearest whoie numbers as needed.) The namber zeected atter 12 days is (Found fo the nearest mhole mambere as needed.). The number infected after 16 days is. (Round to the nearest whole numben as needed ) A As (→6,N(1)−3300,103300 be00le wit be infeched after days.
a) The number of people that are initially infected with the disease are 145 people.
b) The number infected after 2 days is 719 people.
The number infected after 5 days is 2659 people.
The number infected after 8 days is 3247 people.
The number infected after 12 days is 3299 people.
The number infected after 16 days is 3300 people.
c) As t → e, N(t) → 3300, so 3300 people will be infected after 16 days.
How many are initially infected with the disease?Based on the information provided above, the number of people N infected t days after the disease has begun can be modeled by the following exponential function;
[tex]N(t)=\frac{3300}{1\;+\;21.7e^{-0.9t}}[/tex]
When t = 0, the number of people N(0) infected can be calculated as follows;
[tex]N(0)=\frac{3300}{1\;+\;21.7e^{-0.9(0)}}[/tex]
N(0) = 145 people.
Part b.
When t = 2, the number of people N(2) infected can be calculated as follows;
[tex]N(2)=\frac{3300}{1\;+\;21.7e^{-0.9(2)}}[/tex]
N(2) = 719 people.
When t = 5, the number of people N(5) infected can be calculated as follows;
[tex]N(5)=\frac{3300}{1\;+\;21.7e^{-0.9(5)}}[/tex]
N(5) = 2659 people.
When t = 8, the number of people N(8) infected can be calculated as follows;
[tex]N(8)=\frac{3300}{1\;+\;21.7e^{-0.9(8)}}[/tex]
N(8) = 3247 people.
When t = 12, the number of people N(12) infected can be calculated as follows;
[tex]N(12)=\frac{3300}{1\;+\;21.7e^{-0.9(12)}}[/tex]
N(12) = 3299 people.
When t = 16, the number of people N(16) infected can be calculated as follows;
[tex]N(16)=\frac{3300}{1\;+\;21.7e^{-0.9(16)}}[/tex]
N(16) = 3300 people.
Part c.
Based on this model, we can logically deduce that 3300 people will be infected after 16 days because as t tends towards e, N(t) tends towards 3300.
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Missing information:
The question is incomplete and the complete question is shown in the attached picture.
A person sitting on a Ferris wheel rises and falls as the wheel turns. Suppose that the person's height above ground is described by the following function. h(t)=18.3+16.6cos1.6r In this equation, h(t) is the height above ground in meters, and f is the time in minutes. Find the following. If necessary, round to the nearest hundredth. An object moves in simple harmonic motion with amplitude 8 m and period 4 minutes. At time t = 0 minutes, its displacement d from rest is 0 m, and initially it moves in a positive direction. Give the equation modeling the displacement d as a function of time f.
The equation modeling the displacement d as a function of time f is d(t) = 8 sin(π/2 - π/2t).
motion:
Amplitude = 8m
Period = 4 minutes
Displacement from rest = 0m
Initially moves in a positive direction
We need to find the equation that models the displacement d of the object as a function of time f.Therefore, the equation that models the displacement d of the object as a function of time f is given by the formula:
d(t) = 8 sin(π/2 - π/2t)
To verify that the displacement is 0 at time t = 0, we substitute t = 0 into the equation:
d(0) = 8 sin(π/2 - π/2 × 0)= 8 sin(π/2)= 8 × 1= 8 m
Therefore, the displacement of the object from its rest position is zero at time t = 0, as required.
:Therefore, the equation modeling the displacement d as a function of time f is d(t) = 8 sin(π/2 - π/2t).
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Solve for v. ²-3v-28=0 If there is more than one solution, separate them with commas. If there is no solution, click on "No solution." v =
The equation ²-3v-28=0 has two solutions, v = 7, -4.
Given quadratic equation is:
²-3v-28=0
To solve for v, we have to use the quadratic formula, which is given as: [tex]x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$[/tex]
Where a, b and c are the coefficients of the quadratic equation ax² + bx + c = 0.
We need to solve the given quadratic equation,
²-3v-28=0
For that, we can see that a=1,
b=-3 and
c=-28.
Putting these values in the above formula, we get:
[tex]v=\frac{-(-3)\pm\sqrt{(-3)^2-4(1)(-28)}}{2(1)}$$[/tex]
On simplifying, we get:
[tex]v=\frac{3\pm\sqrt{9+112}}{2}$$[/tex]
[tex]v=\frac{3\pm\sqrt{121}}{2}$$[/tex]
[tex]v=\frac{3\pm11}{2}$$[/tex]
Therefore v_1 = {3+11}/{2}
=7
or
v_2 = {3-11}/{2}
=-4
Hence, the values of v are 7 and -4. So, the solution of the given quadratic equation is v = 7, -4. Thus, we can conclude that ²-3v-28=0 has two solutions, v = 7, -4.
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The solutions to the equation ²-3v-28=0 are v = 7 and v = -4.
To solve the quadratic equation ²-3v-28=0, we can use the quadratic formula:
v = (-b ± √(b² - 4ac)) / (2a)
In this equation, a, b, and c are the coefficients of the quadratic equation in the form ax² + bx + c = 0.
For the given equation ²-3v-28=0, we have:
a = 1
b = -3
c = -28
Substituting these values into the quadratic formula, we get:
v = (-(-3) ± √((-3)² - 4(1)(-28))) / (2(1))
= (3 ± √(9 + 112)) / 2
= (3 ± √121) / 2
= (3 ± 11) / 2
Now we can calculate the two possible solutions:
v₁ = (3 + 11) / 2 = 14 / 2 = 7
v₂ = (3 - 11) / 2 = -8 / 2 = -4
Therefore, the solutions to the equation ²-3v-28=0 are v = 7 and v = -4.
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Determine the degree of each of the following polynomial functions. a. f(x) = 1 + x + x² + x³ Degree of f: b. g(x)=x82x² - 7 Degree of g: c. h(x) = x³ + 2x³ + 1 Degree of h: d. j(x) = x² - 16 De
The degree of a polynomial function is the highest power of the variable that occurs in the polynomial.
a.[tex]f(x) = 1 + x + x^2 + x^3[/tex], degree of f: 3
b. [tex]g(x)=x82x^2 - 7[/tex], degree of g: 8
c. [tex]h(x) = x^3 + 2x^3 + 1[/tex], degree of h: 3
d. [tex]j(x) = x^2 - 16[/tex], degree of j: 2.
a. [tex]f(x) = 1 + x + x^2 + x^3[/tex]
The degree of a polynomial function is the highest power of the variable that occurs in the polynomial. The polynomial function given is [tex]f(x) = 1 + x + x^2 + x^3[/tex].
The degree of the polynomial is the highest power of the variable in the polynomial. The highest power of x in the polynomial is x³.Therefore, the degree of f(x) is 3.
b. [tex]g(x)=x82x^2 - 7[/tex]
The degree of a polynomial function is the highest power of the variable that occurs in the polynomial. The polynomial function given is [tex]g(x)=x82x^2 - 7[/tex].
Rearranging the polynomial expression, we obtain;
[tex]g(x) = x^8 + 2x^2 - 7[/tex]
The degree of the polynomial is the highest power of the variable in the polynomial. The highest power of x in the polynomial is x^8.
Therefore, the degree of g(x) is 8.
c. [tex]h(x) = x^3 + 2x^3 + 1[/tex]
The degree of a polynomial function is the highest power of the variable that occurs in the polynomial. The polynomial function given is [tex]h(x) = x^3 + 2x^3 + 1[/tex].
Collecting like terms, we have; [tex]h(x) = 3x^3+ 1[/tex]
The degree of the polynomial is the highest power of the variable in the polynomial. The highest power of x in the polynomial is x^3.Therefore, the degree of h(x) is 3.
d. [tex]j(x) = x^2 - 16[/tex]
The degree of a polynomial function is the highest power of the variable that occurs in the polynomial. The polynomial function given is [tex]j(x) = x^2 - 16[/tex].
The degree of the polynomial is the highest power of the variable in the polynomial. The highest power of x in the polynomial is x².Therefore, the degree of j(x) is 2.
In conclusion;
a.[tex]f(x) = 1 + x + x^2 + x^3[/tex], degree of f: 3
b. [tex]g(x)=x82x^2 - 7[/tex], degree of g: 8
c. [tex]h(x) = x^3 + 2x^3 + 1[/tex], degree of h: 3
d. [tex]j(x) = x^2 - 16[/tex], degree of j: 2.
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Using flat rate depreciation, the value of another machine after 5 years will be \( \$ 2695 \) and after a further 7 years it will become worthless. The value \( T_{n} \) of this machine after \( n \)
Answer: The value Tₙ of the machine after n years using flat rate depreciation is Tₙ = $4620 - $385n.
Step-by-step explanation:
To determine the value of the machine after a given number of years using flat rate depreciation, we need to find the common difference in value per year.
Let's denote the initial value of the machine as V₀ and the common difference in value per year as D. We are given the following information:
After 5 years, the value of the machine is $2695.
After a further 7 years, the value becomes $0.
Using this information, we can set up two equations:
V₀ - 5D = $2695 ... (Equation 1)
V₀ - 12D = $0 ... (Equation 2)
To solve this system of equations, we can subtract Equation 2 from Equation 1:
(V₀ - 5D) - (V₀ - 12D) = $2695 - $0
Simplifying, we get:
7D = $2695
Dividing both sides by 7, we find:
D = $2695 / 7 = $385
Now, we can substitute this value of D back into Equation 1 to find V₀:
V₀ - 5($385) = $2695
V₀ - $1925 = $2695
Adding $1925 to both sides, we get:
V₀ = $2695 + $1925 = $4620
Therefore, the initial value of the machine is $4620, and the common difference in value per year is $385.
To find the value Tₙ of the machine after n years, we can use the formula:
Tₙ = V₀ - nD
Substituting the values we found, we have:
Tₙ = $4620 - n($385)
So, To determine the value of the machine after a given number of years using flat rate depreciation, we need to find the common difference in value per year.
Let's denote the initial value of the machine as V₀ and the common difference in value per year as D. We are given the following information:
After 5 years, the value of the machine is $2695.
After a further 7 years, the value becomes $0.
Using this information, we can set up two equations:
V₀ - 5D = $2695 ... (Equation 1)
V₀ - 12D = $0 ... (Equation 2)
To solve this system of equations, we can subtract Equation 2 from Equation 1:
(V₀ - 5D) - (V₀ - 12D) = $2695 - $0
Simplifying, we get:
7D = $2695
Dividing both sides by 7, we find:
D = $2695 / 7 = $385
Now, we can substitute this value of D back into Equation 1 to find V₀:
V₀ - 5($385) = $2695
V₀ - $1925 = $2695
Adding $1925 to both sides, we get:
V₀ = $2695 + $1925 = $4620
Therefore, the initial value of the machine is $4620, and the common difference in value per year is $385.
To find the value Tₙ of the machine after n years, we can use the formula:
Tₙ = V₀ - nD
Substituting the values we found, we have:
Tₙ = $4620 - n($385)
So, the value Tₙ of the machine after n years using flat rate depreciation is Tₙ = $4620 - $385n.
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