1. Write a function of × that performs the following operations: Raise x to the ninth
power, multiply by 6, and then add 4.
y = f(x) = _____
2. Find the inverse to the function you found in
part (a).
x = g (y) =

The average rate of change of g(x) from x = 1 to x = 6 is 7

The average rate of change of a function over an interval is given by the difference in the values of the function at the endpoints of the interval, divided by the length of the interval.

In this case, we want to find the average rate of change of g(x) = 2x² - 7x from x = 1 to x = 6.

The value of g(x) at x = 1 is:

g(1) = 2(1)² - 7(1) = -5

The value of g(x) at x = 6 is:

g(6) = 2(6)² - 7(6) = 30

So the difference in the values of g(x) is:

g(6) - g(1) = 30 - (-5) = 35

The length of the interval is:

6 - 1 = 5

Therefore, the average rate of change of g(x) from x = 1 to x = 6 is:

Rate = 35/5

Evaluate

Rate = 7

So, the rate is 7

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The average rate of change is 7.

We know that,

The average rate of change = (final value - initial value)/change in the value of x.

Now,

The given function is,

g(x)=2x²-7x

The initial value of x is 1 (given)

∴ The initial value of the function g(x), at x=1,

g(1)=2(1)²-7(1)=2-7

or, g(1)= -5

Now, the final value of x is 6,

∴ Finding the final value of the function g(x) at x=6,

i.e, g(6)=2(6)²-7(6)

or,  g(6)=72-42 = 30

∴ The change in the value of function g(x), from x= to x=6,

= g(6)-g(1)

= 30-(-5)

= 35

Now, change in the value of x = 6-1=5

∴ The average rate of change = 35/5 = 7

Hence the average rate of change is 7.

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The amount of milk left is given by A = 1.150 Liters of milk = 1150 mL

Given data ,

Let the total amount of milk be = 2 Liters = 2000 mL

Now , Tom gave Kevin 350 ml and Sold 500mL to Bob

So , the remaining amount of milk is given by A

where A = total amount of milk - 350mL - 500mL

On simplifying the equation , we get

A = 2000 - 350 - 500

A = 1150 mL

Hence , the amount of milk left is 1150 mL

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Answer:

Annual % Rate: 26.43%

Amount of time it takes to double: 2.62yr

Step-by-step explanation:

Solving for r(Annual%Rate)

A=P⋅e^(r⋅t)

1405=100⋅^(r⋅10)

1405=100⋅e(^10r)

1405/100=100/100⋅e^(10r)

14.05 = e^(10r)

log(14.05) = log(e^10r)

1.1476 = 10r⋅log(e)

1.1476/log(e) = 10r

1.1476/10⋅log(e) = r

r = 0.26426

=26.43%

Now we have r=26.43%. We can use this information to solve for t, the time period to double the initial investment:

2P = Pe^(rt)

2 = e^(0.2643t)

ln(2) = 0.2643t

t = ln(2)/0.2643

t = 2.62yr

Your answer is D. 1/18  is the probability that she will choose a chocolate bar and then a piece of gum

First, let's determine the total number of snacks in the bag:
4 bags of chips + 5 fruit snacks + 6 chocolate bars + 3 pieces of bubble gum = 18 snacks

Next, let's find the probability of choosing a chocolate bar:
There are 6 chocolate bars and 18 snacks total, so the probability is 6/18, which simplifies to 1/3.

Since she puts the chocolate bar back, the total number of snacks remains the same. Now, let's find the probability of choosing a piece of gum:
There are 3 pieces of gum and 18 snacks total, so the probability is 3/18, which simplifies to 1/6.

Finally, to find the probability of both events happening, multiply the probabilities together:
(1/3) * (1/6) = 1/18

So, the probability that Debra will choose a chocolate bar and then a piece of gum is 1/18. Your answer is D. 1/18.

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The line passes through the same point on figure A and D.

From the figure, we have the original shape to be figure E

The center point of figure E is the origin

This means that a line drawn from the origin that passes through point A must pass through the same point on the shape

From the figure, the line passes through the same point on figure A and D.

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The surface area of pyramid B to the nearest hundredth is 58.67 cm²

Similar figures are two figures having the same shape. The ratio of the corresponding sides of similar shapes are equal.

The scale ratio of the height of the pyramid A to B is

9/6 = 3/2

Area factor = (3/2)² = 9/4

9/4 = 132/x

9x = 132×4

9x = 528

divide both sides by 9

x = 528/9

x = 58.67cm²

Therefore the surface area of pyramid B is 58.67cm².

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The solutions to each trigonometric equation are:

Case 6: x = 50.768° (0.282π) or x = 129.232° (0.718π) or x = 230.768° (1.282π) or x = 309.232° (1.718π)

Case 9: x = 41.810° (0.232π) or x = 138.190° (0.768π) or x = 199.471° (1.108π) or x = 340.529° (1.892π)

Case 11: x = 208.164° (1.156π) or x = 331.836° (1.844π)

In this problem we find four cases of quadratic-like trigonometric equations, whose solutions must be found by means of algebra properties and trigonometric formulas. Now we proceed to solve for each case:

Case 6

cos² x + 3 / 5 = 1

cos² x = 2 / 5

cos x = ± √(2 / 5)

x = 50.768° (0.282π) or x = 129.232° (0.718π) or x = 230.768° (1.282π) or x = 309.232° (1.718π)

Case 9

9 · sin² x - 3 · sin x - 2 = 0

sin x = 2 / 3 or sin x = - 1 / 3

x = 41.810° (0.232π) or x = 138.190° (0.768π) or x = 199.471° (1.108π) or x = 340.529° (1.892π)

Case 11

sin² x - 8 · sin x - 4 = 0

sin x = - 0.472

x = 208.164° (1.156π) or x = 331.836° (1.844π)

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The number of staff members under the age of 40 are 28 and the number of staff members 40 and older are 22.

From the given histogram,

A. Between the age 20-29 there are 10 members.

B. Under the age of 40 = 10+18

= 28

40 and older = 12+7+2+1

= 22

C) Total staff members = 28+22=50

D) Number of staff members are 50 years or older

= 7+2+1

= 10

Therefore, the number of staff members under the age of 40 are 28 and the number of staff members 40 and older are 22.

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We can solve this problem by using some properties of centroids of the triangle and the fact that the centroid divides each median in a 2:1 ratio.

The centroid of a triangle is the point intersection of the three medians of the triangle.

First find the value of MR. The centroid divides each median in a 2:1 ratio, so we have:

MR = 2/3 * R + 1/3 * M

R is the centroid, so R = (P + V + M)/3.

Substituting, we get: MR = 2/3 * [(P + V + M)/3] + 1/3 * M

= 2/9 * P + 2/9 * V + 5/9 * M

Now, substitute the given values of PV and M to find MR:

MR = 2/9 * (3w+7) + 2/9 * (12y-9) + 5/9 * (5x-9) = (2w/3 + 8y/9 + 25x/9) - 1

Simplifying the expression: MR = (2w + 24y + 25x - 27)/9

Next, let's find the value of RP using the centroid. Since R is the midpoint of PV:

RP = 2/3 * R + 1/3 * P

Substituting the values of R and P:

RP = 2/3 * [(3w+7)/3 + (12y-9)/3 + (5x-9)/3] + 1/3 * (3w+7)

= (2w/9 + 8y/9 + 5x/3 + 7/3) + (w+7)/3

= (5w/3 + 8y/9 + 5x/3 + 10)/3

Simplifying this:

RP = (5w + 8y + 5x + 30)/9

Next, find the value of RV using the centroid. R is the midpoint of PV:

So, RV = 2/3 * R + 1/3 * V

Substituting R and V values:

RV = 2/3 * [(3w+7)/3 + (12y-9)/3 + (5x-9)/3] + 1/3 * (12y-9) = (2w/9 + 8y/9 + 5x/3 + 7/3) + 4y/3 - 3

Simplifying: RV = (5w + 20y + 5x - 18)/9

Find the value of RW using the centroid. R is the midpoint of VW, so: RW = 2/3 * R + 1/3 * W

Substituting the values of R and W:

RW = 2/3 * [(3w+7)/3 + (12y-9)/3 + (5x-9)/3] + 1/3 * 1.75x = (2w/9 + 8y

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To evaluate the indefinite integral ∫x tan^2 x dx, we can use integration by parts.

Let u = x and dv = tan^2 x dx. Then, du/dx = 1 and v = ∫tan^2 x dx = tan x - x.

Using the formula for integration by parts, we have:

∫x tan^2 x dx = uv - ∫v du/dx dx
= x(tan x - x) - ∫(tan x - x) dx
= x(tan x - x) + ln|cos x| + C

Therefore, the indefinite integral of x tan^2 x dx is x(tan x - x) + ln|cos x| + C, where C is the constant of integration.
To evaluate the indefinite integral ∫x tan^2(x) dx, we can use integration by parts, which is defined as ∫udv = uv - ∫vdu.

First, let's choose our u and dv:
u = x, so du = dx
dv = tan^2(x) dx

To find v, we need to integrate dv. Since tan^2(x) = sec^2(x) - 1, we get:
v = ∫(sec^2(x) - 1) dx = tan(x) - x

Now, using integration by parts:
∫x tan^2(x) dx = x(tan(x) - x) - ∫(tan(x) - x) dx

Let's evaluate the remaining integral:
∫(tan(x) - x) dx = ∫tan(x) dx - ∫x dx
= ln|sec(x)| - (1/2)x^2 + C

So, the final answer is:
∫x tan^2(x) dx = x(tan(x) - x) - [ln|sec(x)| - (1/2)x^2] + C

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For the given data set of miles driven to see a space shuttle launch, the mean is 28.14, the median is 28, and the mode is 28.

To analyze this data, let's find the mean (average), median, and mode.

1. Mean (average): Add all the miles together and divide by the total number of data points.
(19 + 27 + 14 + 28 + 30 + 51 + 28) / 7 = 197 / 7 = 28.14

The mean miles driven to see a space shuttle launch is 28.14.

2. Median: Arrange the data points in ascending order and find the middle value.
14, 19, 27, 28, 28, 30, 51
Since there are 7 data points, the median is the 4th value, which is 28.

The median miles driven to see a space shuttle launch is 28.

3. Mode: Identify the most frequently occurring value in the data set.
14, 19, 27, 28, 28, 30, 51
The number 28 appears twice, which is more than any other value.

The mode for miles driven to see a space shuttle launch is 28.

In summary, for the given data set of miles driven to see a space shuttle launch, the mean is 28.14, the median is 28, and the mode is 28.

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The profit per pig is $4 and the profit per goose is $8.

Let's start by defining the variables:

a = number of pigs Farmer Jones raises

y = number of geese Farmer Jones raises

The problem tells us that he wants to raise at most 16 animals, so we can write:

a + y ≤ 16

It also tells us that he wants no more than 12 geese, so we can write:

y ≤ 12

We know that it costs $5 to raise a pig and $2 to raise a goose, and he has $550 available to spend. So the cost of raising the animals can be expressed as:

5a + 2y ≤ 550

Finally, we want to maximize his profits. The profit per pig is $4 and the profit per goose is $8. So the objective function for Farmer Jones' profits is:

Profit = 4a + 8y

To summarize, the linear programming model for this problem is:

Maximize: Profit = 4a + 8y

Subject to:

a + y ≤ 16

y ≤ 12

5a + 2y ≤ 550

where a and y are non-negative integers.

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The vocabulary for the segment a is the apothem

The area of the polygon is about 41.6 yd²

The area of a regular figure is the extent of the planer space the figure occupies.

The length of each side of the regular polygon, s = 4 yd

The length of the segment a = 2·√3

The vocabulary term for the segment a drawn from from the center of the polygon and perpendicular to one of its sides is the apothem

Therefore, the vocabulary term for segment a is the apothem

The polygon is a hexagon.

The area of a hexagon is; A = ((3·√3)/2) × s²

Therefore, the area of the polygon is; A = ((3·√3)/2) × (4)² = 24·√3 ≈ 41.6

The area of the polygon is about 41.6 yd²

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The horizontal distance from the bottom of the ramp to the bottom of the platform is 57.74 feet.

In order to find the horizontal distance between the bottom of the ramp and the bottom of the platform, we need to use the Pythagorean theorem. Let's call this distance "d". We know that the vertical distance from the bottom of the ramp to the bottom of the platform is 50 feet, and the length of the ramp is 70 feet.

Using the Pythagorean theorem, we can solve for the horizontal distance:

[tex]d^2 = 70^2 - 50^2[/tex]

[tex]d^2[/tex] = 4,900 - 2,500

[tex]d^2[/tex]= 2,400

d = √2,400

d = 48.99 (rounded to the nearest hundredth)

Therefore, the horizontal distance from the bottom of the ramp to the bottom of the platform is 48.99 feet (rounded to the nearest hundredth).

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The derivatives of the given functions are:

A) h'(x) = (1/2)√xen(x)[2+In(x)]

B) h'(x) = (1/x) - V2V

C) h'(x) = (2/x) + 2

D) h'(x) = 0

A) To find the derivative of h(x) = √xen(x), we use the product rule of differentiation. Let u = √x and v = en(x).

Then, h(x) = uv, and h'(x) = u'v + uv'.

We have u' = (1/2)x^(-1/2) and v' = en(x)(1/x).

Substituting the values, we get h'(x) = (1/2)√xen(x)[2+In(x)].

B) To find the derivative of h(x) = In(x) + V2V, we use the sum rule of differentiation.

Using the properties of logarithms, we rewrite the function as h(x) = In(x) + (1/2)ln(x).

Taking the derivative, we get h'(x) = (1/x) - V2V.

C) To find the derivative of h(x) = V2 In(x) + 2, we use the sum rule of differentiation.

Taking the derivative, we get h'(x) = (2/x) + 2.

D) To find the derivative of h(x) = In(30) + 2ln(x), we use the sum rule of differentiation.

Taking the derivative, we get h'(x) = 0, since the derivative of a constant is always zero.

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The tomatoes start to get mushy after more than 6 days.

Note: The calculation assumes that the number of tomatoes harvested is large enough for the continuous division process to approach zero.

Let's assume that Marcos initially harvested X cherry tomatoes.

According to the given information, each day Marcos keeps one-third of the tomatoes and brings the remaining amount into the office. This means that after the first day, Marcos would have 2/3 (or 2/3X) of the original harvest left.

On the second day, he would keep one-third of the remaining tomatoes and bring 2/3 × 1/3 (or 2/9) of the original harvest into the office. This would leave him with 2/3 × 2/3 (or 4/9) of the original harvest.

In general, after N days, the amount of tomatoes left can be calculated using the following formula:

Amount of tomatoes left = (2/3)^N × X

The question states that Marcos has less than one eighty-first of the original harvest left when the tomatoes start to get mushy. In other words, when the amount of tomatoes left is less than 1/81 of the original harvest, we can write the following inequality:

[tex](2/3)^N × X < 1/81[/tex]

To find the number of days (N) when the tomatoes start to get mushy, we need to solve for N in the inequality above.

Taking the logarithm (base 2/3) of both sides of the inequality:

[tex]N > log2/3(1/81) / log2/3(2/3)[/tex]

N > 4 / (2/3)

N > 4 × 3/2

N > 12/2

N > 6

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It seems that the question provided is not clear and has some typos or formatting issues, making it difficult to understand the exact problem you need help with. Please rephrase or clarify the question, and I'll be more than happy to help you!

To find W,(1,0), we need to use the formula for the partial derivative of F with respect to u at (6,7) and (6,-7) and plug in the given values:

F_u(6,7) = 6,0(6,7) = -5
F_u(6,-7) = 6,0(6,-7) = -7

Now we can use these values to solve for W,(1,0) using the formula:

W,(1,0) = F(6,t) - F_u(6,7)(1-6) - F_u(6,-7)(1-6)

Plugging in the given values, we get:

W,(1,0) = F(6,t) - (-5)(-5) - (-7)(-5)
W,(1,0) = F(6,t) + 30

We still need to find F(6,t). To do this, we use the formula for the partial derivative of F with respect to s at (1,0) and plug in the given values:

F_s(1,0) = 1,0(6,0) - 7,0(1,0) - 9,0(1,0)
F_s(1,0) = -7

Now we can use F_u(6,7), F_u(6,-7), and F_s(1,0) to solve for F(6,t) using the formula:

F(6,t) = F_u(6,7)(6,t-7) + F_u(6,-7)(6,t+7) + F_s(1,0)(t)

Plugging in the given values, we get:

F(6,t) = (-5)(6,t-7) + (-7)(6,t+7) + (-7)(t)
F(6,t) = -77t - 188

Now we can substitute this value of F(6,t) into our formula for W,(1,0) to get the final answer:

W,(1,0) = -77t - 188 + 30
W,(1,0) = -77t - 158

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measure of arc JG = 160 degrees

Main Concept: Intersecting chords

Chords are line segments with ends points that are both on the edge of the circle.  Intersecting chords are a pair of chords on the same circle that intersect.

In an extreme example, the chords may intersect at one of the end points, making the intersecting chords an inscribed angle.

Because Intersecting chords intersect, if the line segments are extended into lines, the lines form two pairs of vertical angles.  Vertical angles are congruent.  Given one vertical angle pair, they will contain two arcs (in the extreme case, the arc will have a measure of zero).

The measure of each of the vertical angles is the average of the two contained arcs.

This problem

For this problem, FG and HJ are chords of the same circle, and they intersect.

If we call the intersection P, angle GPJ is given with a measure of 100 degrees.

Angle GPJ and Angle FPJ form a vertical angle pair, so they are congruent, because vertical angles are congruent.

The measure of each of the vertical angles is the average of the two contained arcs.

The two arcs that this vertical angle pair contain are the arc JG and arc FH.

The measure of arc FH is given as 40 degrees.

Substitute these known quantities into the equation describing the relationship between one of the vertical angles and the contained arcs.

[tex]m \angle GPJ=\frac{1}{2}(m ~\text{arc}JG + m ~\text{arc}FH)[/tex]

[tex](100^o)=\frac{1}{2}(m ~\text{arc}JG + (40^o))[/tex]

Multiply both sides by 2...

[tex]200^o=m ~\text{arc}JG + 40^o[/tex]

Subtract 40 degrees from both sides...

[tex]160^o=m ~\text{arc}JG[/tex]

Gabby does not have any fabric left, as she used more fabric than she had to start with.

How much fabric does Gabby have left after using 3215 yards, and what is the estimate?

Based on the information given, Gabby started with 578 yards of fabric and used 3215 yards of fabric. To estimate the amount of fabric Gabby has left, we need to subtract the amount of fabric used from the starting amount of fabric:

578 yards - 3215 yards = -2637 yards

Since the result is a negative number, it doesn't make sense in this context. It's possible that there was a mistake in the numbers given, or Gabby used more fabric than she had to start with.

Without further information or clarification, we cannot estimate the amount of fabric Gabby has left.

To estimate the amount of fabric Gabby has left, we can subtract the amount of fabric she used from the amount of fabric she had initially.

So, to find the estimate for the amount of fabric Gabby has left, we can perform the following calculation:

Estimate for the amount of fabric Gabby has left = 578 yards (initial amount) - 3215 yards (amount used)

Estimate for the amount of fabric Gabby has left = -2637 yards

However, the result is negative, which means that Gabby doesn't have any fabric left, and she needs to purchase an additional 2637 yards to make up for the shortfall.

Therefore, the estimate for the amount of fabric Gabby has left is 0 yards (she needs to purchase more fabric to continue her work).

In the next 20 students surveyed, you would expect 5 to own a dog and a cat

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

Dog = 42

Cat = 34

Dog and cat = 15

Neither = 9

This means that

P(Dog and cat) = 15/100

When evaluated, we have

P(Dog and cat) = 5/20

So, when the next 20 students surveyed, we have

Dog and cat = 5/20 * 20

Evaluate

Dog and cat = 5

Hence, the number of dogs and cats is 5

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Cubic inches of popcorn does the jumbo size hold compared to the regular size round to the nearest tenth

To determine how many more cubic inches of popcorn the jumbo size holds compared to the regular size, you would need to:

1. Find the volume (in cubic inches) of both the jumbo and regular size popcorn containers.
2. Subtract the volume of the regular size container from the volume of the jumbo size container.

Unfortunately, without specific dimensions for the jumbo and regular size popcorn containers, I cannot provide a numerical answer. Please provide the dimensions, and I would be happy to help you with the calculations.

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A) The equation relates the number of cups of flour to the number of pound cakes that the bakery makes is f = 2.75p

B) Cups of flour are needed for 14 cakes is 38.5

A) The number of cups of flour, f, that a bakery needs for the number of pound cakes that they make, p is directly proportional to each other which can be written in form ,

f/p = 8.25/3

f = (8.25/3)×p

f = 2.75 p

The equation forms is f = 2.75p

B) Cups of flour are needed for 14 cakes

Here p = 14

by putting the value in the equation we get ,

f = 2.75(14)

f = 38.5

hence , cups of flour are needed for 14 cakes is 38.5

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Tim's truck has a volume of (6.5 feet) x (5 feet) x (7 feet) = 227.5 cubic feet. Each box of papers has a volume of (2 feet) x (2 feet) x (2 feet) = 8 cubic feet. To determine how many boxes of papers Tim can pack into the truck, we need to divide the total volume of the truck by the volume of each box:

227.5 cubic feet ÷ 8 cubic feet per box = 28.44 boxes

Since we can't pack a fraction of a box, Tim can pack a maximum of 28 boxes of papers into his truck. However, this assumes that there is no wasted space due to irregular shapes of the boxes or other items in the truck.

In reality, Tim may be able to pack slightly fewer boxes depending on how he arranges them in the truck.

Hence, Tim's truck has a volume of 227.5 cubic feet. Each box of papers has a volume of 8 cubic feet.

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An appropriate and interesting word for 15,000 is "fifteen thousand," while for 3,000, it is "three thousand."

To solve this question, we need to write the numbers 15,000 and 3,000 as words. To do this, first, we look at the place values of each digit. In 15,000, the "15" is in the thousands place, so we write it as "fifteen thousand." Similarly, in 3,000, the "3" is in the thousands place, so we write it as "three thousand."

By doing this, we have expressed the numbers using their word forms, which helps in understanding and communicating numerical values more effectively, especially in written or spoken contexts. Remember, when writing large numbers as words, we use the place values to guide us in expressing them accurately and understandably.

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Answer:

108y^10x^17

Step-by-step explanation:

Answer:

x^9√x

Step-by-step explanation:

The car depreciated at an annual rate of approximately 45.81%.

In 2016, Dave bought a new car for $15,500, and its current value is $8,400. To find the annual depreciation rate, we'll use the formula A(t) = P(1 ± r)t, where A(t) is the future value, P is the initial value, r is the annual rate, and t is the time in years.

Here, A(t) = $8,400, P = $15,500, and t = 1 (one year). We are solving for r, the annual depreciation rate.

$8,400 = $15,500(1 - r)¹

To isolate r, we'll first divide both sides by $15,500:

$8,400/$15,500 = (1 - r)

0.541935 = 1 - r

Now, subtract 1 from both sides:

-0.458065 = -r

Finally, multiply both sides by -1 to find r:

0.458065 = r

To express r as a percentage, multiply by 100:

0.458065 x 100 = 45.81%

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The car depreciated at an annual rate of 12.2%.

The car depreciated in value over time, so we want to find the rate of decrease. We can use the formula:

A(t) = P(1 - r)t

where A(t) is the current value of the car, P is the original price of the car, r is the annual rate of depreciation, and t is the time elapsed in years.

We can plug in the given values and solve for r:

$8,400 = $15,500(1 - r)⁵

Dividing both sides by $15,500, we get:

0.54 = (1 - r)⁵

Taking the fifth root of both sides, we get:

(1 - r) = 0.878

Subtracting 1 from both sides, we get:

-r = -0.122

Dividing both sides by -1, we get:

r = 0.122

Multiplying by 100 to express as a percentage, we get:

r = 12.2%

Therefore, the car depreciated at an annual rate of 12.2%.

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Answer:

There are 16 unit squares on the geoboard. Each unit square represents 5 meters, so the total area is 16*5 = 80 square meters.

Here is a diagram of the geoboard with the unit squares labeled:

[Image of a geoboard with 16 unit squares labeled]

I hope this helps! Let me know if you have any other questions.

A fraction that is greater than 1/2 and less than 4/5, we need to consider fractions between these two values. A fraction that is greater than 1/2 and less than 4/5 is 6/10.

Compare the two given fractions by finding a common denominator.
The lowest common denominator for 1/2 and 4/5 is 10.
Convert both fractions to equivalent fractions with a denominator of 10.
1/2 = 5/10
4/5 = 8/10
Identify a fraction between 5/10 and 8/10.
One possible fraction between these two is 6/10.
A fraction that is greater than 1/2 and less than 4/5 is 6/10.

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