Is the gfc of a pair of numbers ever greater than both numbers

Answer: C

Step-by-step explanation:

Answer:

C.

Step-by-step explanation:

Trust

Answer:

33.33

Step-by-step explanation:

60% of 33.33 is 20. 100% of 33.33 is 33.33, therefore 60 percent of 33.33 equals 20.

Answer:

\( \huge{ \boxed{ \tt{33.33}}}\)

Step-by-step explanation:

\( \star{ \: \sf{ \: Suppose, \: the \: number \: be \: x}}\)

\( \star \: \sf \: Create \: an \: equation \: and \: solve \: for \: x\)

\( \underline{ \underline{ \mathfrak{lets \: solve}}} : \)

\( \sf{ \: 60\% \: \: of \: \: x \: = 20}\)

\( \text{Step \: 1} : \sf{ \: Convert \: percentage \: ( \%) \: into \: fraction}\)

\( \sf{To \: do \: so, \: Divide \: it \: by \: 100 \: and \: remove \: the \: \% \: symbol. \: }\)

\( \sf{Then, \: Simplify \: the \: fraction \: to \: its \: lowest \: term.}\)

\( \hookrightarrow{ \sf{ \frac{60}{100} x = 20}}\)

\( \hookrightarrow{ \sf{ \frac{ \cancel{60} ^{ \: \: 3} }{ \cancel{100} \: ^{ \: 5} } x = 20}}\)

\( \hookrightarrow{ \sf{ \frac{3}{5}x = 20}} \)

\( \hookrightarrow{ \sf{0.6x = 20}}\)

\( \text{Step \: 2} : \sf{Divide \: both \: sides \: by \: 0.6}\)

\( \hookrightarrow{ \sf{ \frac{0.6x}{0.6} = \frac{20}{0.6}}} \)

\( \text{Step \: 3} : \sf{Calculate}\)

\( \hookrightarrow{ \sf{x = 33.33}}\)

The required number is 33.33

Hope I helped!

Best regards! :D

~\( \sf{TheAnimeGirl} \)

Thus, the general solution of the given differential equation `2y′ y = 5t²` is `y = Ke^(5t³/6)` where `K` is a constant of integration.

The general solution of the given differential equation `2y′ y = 5t²` and using it to determine how solutions behave as t is discussed below:Solving the differential equation:

Separating the variables of the differential equation `2y′ y = 5t²` we get:dy/y = (5/2)t² dtIntegrating both sides, we have:ln|y| = (5/6) t³ + C1

Taking the exponential of both sides, we get:y = Ke^(5t³/6) , where K = ± e^(C1) is a constant of integration.

The general solution of the given differential equation is given by `y = Ke^(5t³/6)` where `K` is a constant of integration.

How solutions behave as `t`:When `t → ∞` (i.e., as t grows large), `e^(5t³/6) → ∞`. So solutions of the given differential equation `y′ y = 5t²` grow exponentially as `t → ∞`.

When `t → -∞` (i.e., as t gets very negative), `e^(5t³/6) → 0`. So solutions of the given differential equation `y′ y = 5t²` approach `y = 0` as `t → -∞`.

Thus, the general solution of the given differential equation `2y′ y = 5t²` is `y = Ke^(5t³/6)` where `K` is a constant of integration.

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The value of a is 37°, the value of b is 34° and the value of c is 143° in the given triangle.

According to the question,

We have the following information:

We have a triangle with angles given.

We know that angle made on a straight line is 180° and the sum of all three angles of a triangle is 180°.

We also know that vertically opposite angles are equal.

So, a is vertically opposite to 37°.

So, the value of a is 37°.

Now, a and c make a straight line.

c = 180-37

c = 143°

a + b + 109° = 180°

37° + b + 109° = 180°

b+146° = 180°

b = 180-146

b = 34°

Hence, the value of a is 37°, the value of b is 34° and the value of c is 143° in the given triangle.

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The value for x is 7π/6 and 11π/6.

sin2xsecx+2cosx=0

(2sinxcosx)(1/cosx)+2cosx=0

sinxcosx/cosx+2cosx=0

So,

2sinx+2cosx=0

2(sinx+cosx)=0

(2(sinx+cosx))²=0

4(sin²x+cos²x+2sinxcosx)=0

hence,

(sin²x+cos²x+2sinxcosx)/4=0/4

sin2x+cos2x+2sinxcosx=0

1+sin2x=0

sinx=−1/2

=7π/6 and 11π/6

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The linear relationship between time and distance can be considered proportional. The constant of proportionality in this case is 4.

In a proportional relationship, the ratio between the two quantities remains constant. Here, as time increases by 2 hours, the distance also increases by 8 miles. Let's calculate the ratio of distance to time for each pair of values:

For the first pair (2 hours, 8 miles):

Ratio = distance / time = 8 miles / 2 hours = 4 miles/hour

For the second pair (4 hours, 16 miles):

Ratio = distance / time = 16 miles / 4 hours = 4 miles/hour

For the third pair (6 hours, 24 miles):

Ratio = distance / time = 24 miles / 6 hours = 4 miles/hour

For the fourth pair (8 hours, 32 miles):

Ratio = distance / time = 32 miles / 8 hours = 4 miles/hour

As we can see, the ratio of distance to time remains constant at 4 miles per hour for all the pairs. This indicates a proportional relationship between time and distance.

Therefore, the linear relationship is also proportional, and the constant of proportionality is 4.

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The power supplied to the pump is 260.79 kW. Thus, option B is correct.

(a) Given that,The water is to be delivered at a rate of 0.030 m³/s.

The pressure in the tank where the water is discharged is 95.0 kPa.

The pipe from the pump to the tank is 100 m long (including all vertical and horizontal lengths) and has an inside diameter of 0.100 m.

The water has a density of 1000 kg/m³ and a viscosity of 1.10 mPa s.

We are to determine the pressure where the water leaves the pump. Now, using Bernoulli's principle, we have:

P1 + 1/2ρv1² + ρgh1 = P2 + 1/2ρv2² + ρgh2

The height difference (h2 - h1) is 10 m.

Therefore, the equation becomes:

P1 + 1/2ρv1² = P2 + 1/2ρv2² + ρgΔh

where; Δh = h2 - h1 = 10 mρ = 1000 kg/m³g = 9.81 m/s²

v1 = Q/A1 = (0.030 m³/s) / (π/4 (0.100 m)²) = 0.95 m/s

A1 = A2 = (π/4) (0.100 m)² = 0.00785 m²

Then, v2 can be determined from: P1 - P2 = 1/2

ρ(v2² - v1²) + ρgΔh95 kPa = P2 + 1/2(1000 kg/m³) (0.95 m/s)² + (1000 kg/m³) (9.81 m/s²) (10 m)1 Pa = 1 N/m²

Thus, 95 × 10³ Pa = P2 + 436.725 Pa + 98100 PaP2 = 94709.275 Pa

Therefore, the pressure where the water leaves the pump is 94.7093 kPa.

Hence, option A is correct. (b)

The power supplied to the pump is given by:

P = QΔP/η

where; η is the efficiency of the pump, Q is the volume flow rate, ΔP is the pressure difference,

P = (0.030 m³/s) (95.0 × 10³ Pa - 1 atm) / (1.10 × 10⁻³ Pa s)P = 260790.91 Watt

Hence, the power supplied to the pump is 260.79 kW. Thus, option B is correct.

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so hmmm tripling $380 we end up with $1140, now, since 7/8 is 0.875, that means that Layla's rate is 5.875, whilst Brandon's is 6.125.

\(~~~~~~ \stackrel{ \textit{\LARGE Layla}}{\textit{Compound Interest Earned Amount}} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\dotfill & \$ 1140\\ P=\textit{original amount deposited}\dotfill &\$380\\ r=rate\to 5.875\%\to \frac{5.875}{100}\dotfill &0.05875\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{quarterly, thus four} \end{array}\dotfill &4\\ t=years \end{cases}\)

\(1140 = 380\left(1+\frac{0.05875}{4}\right)^{4\cdot t} \implies \cfrac{1140}{380}=1.0146875^{4t} \\\\\\ 3=1.0146875^{4t}\implies \log(3)=\log(1.0146875^{4t}) \\\\\\ \log(3)=t\log(1.0146875^{4})\implies \cfrac{\log(3)}{\log(1.0146875^{4})}=t\implies \boxed{18.84\approx t} \\\\[-0.35em] ~\dotfill\)

\(~~~~~~ \stackrel{ \textit{\LARGE Brandon}}{\textit{Compound Interest Earned Amount}} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\dotfill & \$ 1140\\ P=\textit{original amount deposited}\dotfill &\$380\\ r=rate\to 6.125\%\to \frac{6.125}{100}\dotfill &0.06125\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{monthly, thus twelve} \end{array}\dotfill &12\\ t=years \end{cases}\)

\(1140 = 380\left(1+\frac{0.06125}{12}\right)^{12\cdot t} \implies \cfrac{1140}{380}=\left( \cfrac{9649}{9600} \right)^{12t}\implies 3=\left( \cfrac{9649}{9600} \right)^{12t} \\\\\\ \log(3)=\log\left[ \left( \cfrac{9649}{9600} \right)^{12t} \right]\implies \log(3)=t\log\left[ \left( \cfrac{9649}{9600} \right)^{12} \right]\)

\(\cfrac{\log(3)}{ ~~ \log\left[ \left( \frac{9649}{9600} \right)^{12} \right] ~~ }=t\implies \boxed{17.98\approx t} \\\\[-0.35em] ~\dotfill\\\\ 18.84~~ - ~~17.98 ~~ \approx ~~ \text{\LARGE 0.86}\)

\(a_n = a_(n-1) + a_(n-2)\)

where a_n is the number of bit strings of length n that contain a pair of consecutive 0s, a_(n-1) is the number of bit strings of length (n-1) that contain a pair of consecutive 0s, and a_(n-2) is the number of bit strings of length (n-2) that contain a pair of consecutive 0s.

To find a recurrence relation for the number of bit strings of length n that contain a pair of consecutive 0s, we can start by noting that the number of bit strings of length n that contain a pair of consecutive 0s (a_n) is dependent upon the number of bit strings of length (n-1) that contain a pair of consecutive 0s (a_(n-1)) and the number of bit strings of length (n-2) that contain a pair of consecutive 0s (a_(n-2)). This is because the last two bits of a bit string of length n can be either 00, 01, or 10. If the last two bits are 00, then the bit string of length n contains a pair of consecutive 0s and the remaining bit string must be of length (n-2). If the last two bits are 01 or 10, then the bit string of length n does not contain a pair of consecutive 0s and the remaining bit string must be of length (n-1). Therefore, we can conclude that the recurrence relation for the number of bit strings of length n that contain a pair of consecutive 0s is a_n = a_(n-1) + a_(n-2).

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Answer:

12, 1/2 Kg in 6 Kg

Step-by-step explanation:

6 divided by 1/2= 12

Answer:

I'm pretty sure the answer is C

Step-by-step explanation:

What question do you need help with

Answer:

The slope of the line perpendicular to the line segment having slope \(-\dfrac{3}{4}\) is \(\dfrac{4}{3}\)

Step-by-step explanation:

Given the slope of the line perpendicular to the line segment is \(-\dfrac{3}{4}\)

The expressions for the slope of line which are perpendicular to each other is given as \(m_{1} m_{2} =-1\)

So the slope of the perpendicular line will be

\(m_{2} =-\frac{1}{m_{1} }\)

\(m_{2} =-\frac{1}{-\frac{3}{4} }\)

\(m_{2} =\dfrac{4}{3}\)

Answer:

The slope of line perpendicular to the line segment that has a slope of \(\dfrac{-3}{4}\) is \(\dfrac{4}{3}\)

Step-by-step explanation:

Given,

The line segment has a slope of \(\dfrac{-3}{4}\).

To find:

The slope of the line that is perpendicular to the given line segment.

Now,

Let suppose a line \(L_1\) has a slope \(m_1\) and line \(L_2\) has a slope \(m_2\).

Condition for perpendicularity:

If the product of the slopes of these lines equals -1, then the lines are perpendicular to each other.

Therefore, apply the above condition to our question.

So,

\(\begin{aligned}\dfrac{-3}{4}\times{\rm{slope\;of\;perpendicular\;line}}&=-1\\{\rm{slope\;of\;perpendicular\;line}}&=\dfrac{4}{3}\end{aligned}\)

Answer:

hope this helps

Step-by-step explanation:

When a small group of farmers in eastern China looked at their watermelon fields one spring morning in 2011, they had quite a shock. The fields were not full of a bumper crop like that shown in this picture. Instead, the fields were full of shattered watermelons. The melons looked like they had been blown apart.

A row of watermelons in a field.

Field of ripe watermelons

The problem was limited to about 45 acres (18 hectare) of melons near Danyang. Experts concluded a chemical called forchlorfenuron was at fault. This chemical makes fruits and vegetables grow faster and larger than normal. In fact, they can grow 20 percent bigger!

China’s Watermelon Crop

China grows more watermelons than any other country. All parts of the watermelon are used in traditional Chinese medicine. Farmers see benefits in increasing the size of each melon. Bigger melons bring bigger profits. Melons that grow faster also get to market more quickly. As a result, several farmers had treated their watermelon crop. But the melons grew much faster than anyone expected—in fact, so fast that they burst apart.

Scientists who studied the situation stated that several important things went wrong. First, the type of watermelon these farmers grow is fairly thin-skinned. Under the best conditions, it would be more likely than other types of melon to split. Adding the growth stimulant made that result even more likely.

Watermelons are also more likely to split when the amount of water they receive fluctuates. Watermelons are 92 percent water. During drier weather, their skin tightens. If a heavy rain follows, the plant must absorb a lot of water very fast. The result is cracking. In this case, farmers applied the chemical before a period of heavy rain. The pressure from the water plus the growth stimulant caused the melons to burst.

Chemical Use—Good and Bad

Farmers use many chemicals on crops. That is especially true in a country like China where many people make a living as farmers, and where agriculture is an important part of the economy. The chemical used on the watermelons was a growth stimulant. It makes each acre more productive. Farmers also use pesticides. These chemicals kill pests such as insects or rodents that destroy crops. They use other chemicals to control weeds or kill organisms that spread disease.

Much of China’s farmland has been used for a long time. Fertilizers are needed to add nutrients to soil that years of use have pulled out. China’s farmers also raise livestock and poultry under conditions where many animals are crowded together. Antibiotics are widely used to control the spread of disease.

The goal of heavy chemical use is to boost farm productivity. In fact, China is the world’s largest user of agricultural chemicals. It uses 30 percent of the world total on just 9 percent of the world’s farmland. They do help China produce more food. But chemicals can also cause problems—and not just exploding watermelons.

Some chemicals can be harmful to people or the environment. Most of China’s farms are small—fewer than 5 acres (2 hectares). Studies show that small farms are more likely to use chemicals inefficiently. They may use them in way that can be harmful to the environment. Small farms are also less likely to use innovations in technology that reduce the need for chemicals.

Still, China knows the harm these chemicals are doing to the land. It has set a goal of no further growth in the use of these chemicals by 2020. That means farmers will have to cut back on the chemicals they spray onto their crops. And that should also mean no more exploding watermelons.

In general, the origin of the new object is the same because the scalation is made by using the origin of the previous object as a fixed point.

The dimensions of the new object are found simply by multiplying each coordinate of special points of the object by the scale factor.

For example, if the points (x1,y1), (x2,y2) and (x3,y3) are vertices of an object with triangle shape, the new object will have the coordinates

(k*x1,k*y1), (k*x2,k*y2) and (k*x3,k*y3) where the scale factor is k.

No, enough information is not given to prove that triangles are congruent using the SSS congruence theorem.

The SSS congruence theorem  expresses that assuming in two triangles, three sides of one are consistent to three sides of the other, then, at that point, the two triangles are said to be congruent.

Now, we are given that triangle ACB and triangle DEB that share a common vertex b. So, both triangles share a common side. So, one side is same for both of triangles.

From the diagram, we can see that both triangles opposite side are equal. So, second side is equal for both of triangles.

Now, from the diagram, we can see that both triangles vertically opposite angles are equal. So, we have two sides equal and one angle is equal.

So, given triangle is proved to congruent by SAS congruency not by SSS congruency.

Hence, enough information is not given.

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According to the information we can infer that the greater area is the striped region.

To calculate what is the greater area we have to calculate the area of each segment of the target. Fist we can calculate the area of the striped region.

Then we have to calculate the area of the shaded region with the following procedure:

According to the above, the striped area has the greater area.

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Answer:

Step-by-step explanation:

All of the angles will equal up to 180

10x-10=180

180+10 is 190

190=10x     190/10=x     x=19 or R

<Q: 19*2=38-5=33    <Q:33

<P:133-5=128    <P:128

128+33+19=180,

Double checking to fact check the asnwers.

The correct option is (D). Approximately 77.3% of new machine set ups are completed in less than 25 minutes.

Given a local manufacturing plant, employees must complete new machine set ups within 30 minutes.

The new machine set-up times can be described by a normal model with a mean of 22 minutes and a standard deviation of four minutes. The typical worker needs five minutes to adjust to their surroundings before beginning their duties.

To find the percentage of new machine set ups completed in less than 25 minutes, we need to calculate the z-score. For this, we will use the formula:

z = (X - μ) / σ

where X = 25 minutes, μ = 22 minutes, and σ = 4 minutes

z = (25 - 22) / 4z = 0.75

We can now look up the percentage of the area under the normal distribution curve that corresponds to z = 0.75. Using a standard normal distribution table, we find that the area to the left of z = 0.75 is approximately 0.7734.

So, the percentage of new machine set ups completed in less than 25 minutes is approximately 77.34%.

Therefore, the correct option is (D).Approximately 77.3% of new machine set ups are completed in less than 25 minutes.

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For a marketing firm which is considering making up to three new hires.

a) The probability that the firm will make at least one hire is equals to the 0.90.

b) The expected value and the standard deviation of the number of hires are 1.57 and 0.78 respectively.

We have a marketing firm is considering making up to three new hires. Let's consider the a random variable X that represents the hiring of candidates. So, possible values of X = 0, 1, 2,3.. Now, The chances of probability of hiring at least two candidates, P( X ≥ 2) = 60% = 0.60

The chances of probability that hiring of none of them = P( X = 0) = 10% = 0.10

The chances of probability that hiring of all of them = P( X = 3) = 5% = 0.05

We have to determine probability that the firm will make at least one hire, P(X ≥ 1).

By probability law, sum of any possible probability values = 1

=> P( X≥ 1 ) = 1 - P( X = 0)

= 1 - 0.10 = 0.90

b) The expected value, E(X), or we say mean μ of a discrete random variable X, is equals to the sum of resultant of multiply each value of the random variable by its probability. The formula is, E ( X ) = μ = ∑ x P ( x )

So, Probability that hiring of exactly two candidates, P( X= 2). As we know, P( X ≥ 2) = 0.60

=> P ( 3) + P (2) = 0.60

=> P(2) = 0.60 - 0.05 = 0.55

Probability that hiring of exactly one candidates, P( X= 1). From, P( X≥ 1 ) = 0.90

=> P( 1) + P ( 3) + P (2) = 0.90

=> P(1) = 0.30

Hence, excepted value, E(X) = ∑ x P ( x )

= 0 × 0.10 + 1× 0.30 + 2× 0.55 + 3× 0.05

= 1.57

Now, standard deviations, σ =√E(X²) - (E(X))²

E(X²) = ∑ x² P ( x )

= 0² ×0.10 + 1² ×0.30 + 2²× 0.55 + 3²× 0.05

= 3.07

so, the standard deviation of the number of hires = √3.07² - 1.57² = 0.78

Hence, required value is 0.78.

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Hi there! See an attachment below! (May not be clear for an explanation so I will explain it to you.)

We can find real roots of polynomial function by counting their x-intercepts of graph. See in the attachment that the graph intersects 3 points on x-axis/plane.

x-axis represents the solutions of the graph or equation of the graph. For example, a graph of (x-1)² would be a parabola that has x-intercept at (1,0). If we solve the equation for (x-1)² = 0, our solution would be x = 1 like the x-intercepts.

Conclusions

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the area of the dartboard is approximately 254.34 square inches.

The area of a dartboard can be calculated using the formula for the area of a circle, which is given by:

Area = π * \(radius^2\)

Given that the radius of the dartboard is 9 inches, we can substitute this value into the formula:

Area = π * \(9^2\)

Area = π * 81

To calculate the area, we need to use the value of π (pi). Pi is an irrational number and is commonly approximated as 3.14. Using this approximation, we can calculate the area:

Area ≈ 3.14 * 81

Area ≈ 254.34 square inches

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A recurrence relation for the number of ternary strings of length n that do not contain two consecutive 0s : T(n) = 2T(n-1) + T(n-2) and the initial conditions are : T(1) = 3 and T(2) = 9 .

Let T(n) be the number of ternary strings of length n that do not contain two consecutive 0s.

Consider a ternary string of length n. The last digit can be either 0, 1, or 2. If the last digit is 1 or 2, then we can append either 0, 1, or 2 to any string of length n-1 that does not end in 0 to get a string of length n that does not contain two consecutive 0s. Therefore, the number of such strings of length n is 2T(n-1).

If the last digit is 0, then the second-to-last digit must be 1 or 2, and the rest of the string of length n-2 can be any string of length n-2 that does not end in 0. Therefore, the number of such strings of length n is T(n-2).

Putting these cases together, we get the recurrence relation:

T(n) = 2T(n-1) + T(n-2)

with initial conditions:

T(1) = 3 (0, 1, or 2)

T(2) = 9 (00 is not allowed, but each of the other two digits can be followed by any of the three digits)

Therefore, the number of ternary strings of length n that do not contain two consecutive 0s can be calculated using the recurrence relation T(n) = 2T(n-1) + T(n-2), with initial conditions T(1) = 3 and T(2) = 9.

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Answer:

V = 66 cm³

Step-by-step explanation:

The volume (V) of the prism is calculated as

V = Ah ( A is the area of triangular end and h is perpendicular height )

A = \(\frac{1}{2}\) × 3 × 4 = 6 cm² and h = 11

V = 6 × 11 = 66 cm³

Answer:

ok

Step-by-step explanation:

The first derivative of the function f(x) = x³ - 3x + 1 at the point x0 = 2 using a five-point formula with h = 5 is -31.17 (approx) and the differentiation error is 0.

Given function is f(x) = x³ - 3x + 1. We have to compute the first derivative of the function at the point x0 = 2 using a five-point formula with h = 5.

Here, we use the five-point formula for differentiation. The five-point formula is given by

\(f′(x) = (-f(x+2h) + 8f(x+h) − 8f(x−h) + f(x−2h))/(12h)\)

Using h = 5 and x0 = 2, we have, h = 5, x = 2

Therefore, x−2h = −8, x−h = −3, x+h = 7, x+2h = 12

Now substitute these values in the above formula:

\(f′(2) = (-f(12) + 8f(7) − 8f(−3) + f(−8))/(12×5)f′(2) = [-1993 - 8(359) + 8(13) + 1349]/60f′(2) = −31.17\) (approx)

Now, we need to find the differentiation error. To find differentiation errors, we use the error formula

|E| = K * h⁴,

where K is a constant, h is the step size and |E| is the maximum error|E| = K * h⁴. Since h = 5, we get|E| = K * 5⁴. To find K, we need to find the maximum value of

\(|f⁽⁵⁾(x)|.f(x) = x³ - 3x + 1\)

∴ \(f′(x) = 3x² - 3\)

∴ \(f′′(x) = 6x\)

∴ \(f′′′(x) = 6\)

∴ \(f⁽⁴⁾(x) = 0\)

∴ \(f⁽⁵⁾(x) = 0\)

So, the maximum value of \(|f⁽⁵⁾(x)| = 0\).

∴ K = 0|E| = K * h⁴ = 0.

f′(2) = -31.17 (approx). The differentiation error is 0.

We are given a function f(x) = x³ - 3x + 1 and we have to find the first derivative of the function at the point x0 = 2 using the five-point formula with h = 5.

The five-point formula is

\(f′(x) = (-f(x+2h) + 8f(x+h) − 8f(x−h) + f(x−2h))/(12h)\)

Using h = 5 and x0 = 2, we have h = 5 and x = 2.

Substituting these values in the formula for the five-point formula for differentiation, we get

\(f′(2) = (-f(12) + 8f(7) − 8f(−3) + f(−8))/(12×5)\)

Substituting the values of f(12), f(7), f(-3) and f(-8) from the given function, we get

\(f′(2) = [-1993 - 8(359) + 8(13) + 1349]/60= −31.17\)(approx)

Now, we need to find the differentiation error. To find the differentiation error, we use the error formula

|E| = K * h⁴,

where K is a constant, h is the step size and |E| is the maximum error.

|E| = K * h⁴

Since h = 5, we get|E| = K * 5⁴. To find K, we need to find the maximum value of |f⁽⁵⁾(x)|.

Differentiating f(x) five times, we get

\(f′(x) = 3x² - 3f′′(x) = 6xf′′′(x) = 6f⁽⁴⁾(x) = 0f⁽⁵⁾(x) = 0\)

The maximum value of |f⁽⁵⁾(x)| = 0. Therefore, K = 0|E| = K * h⁴ = 0. Therefore, the differentiation error is 0. Hence, the first derivative of the function f(x) = x³ - 3x + 1 at the point x0 = 2 using a five-point formula with h = 5 is -31.17 (approx) and the differentiation error is 0.

The first derivative of the function f(x) = x³ - 3x + 1 at the point x0 = 2 using a five-point formula with h = 5 is -31.17 (approx) and the differentiation error is 0.

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The equation that shows the relationship between the angles in the triangle is m∠1 + m∠2 + m∠3 = 180.

The equation that shows the relationship between the angles in the triangle is determined as follows;

The sum of angles in a triangle is equal to 180 degrees.

So based on this information, the equation that shows the relationship between the angles in the triangle is formulated as;

angle 1 + angle 2 + angle 3 = 180

We can re-write the equation as follows;

m∠1 + m∠2 + m∠3 = 180

Thus, the equation that shows the relationship between the angles in the triangle is the sum of angle 1 plus angle 2 plus angle 3.

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Answer:

First, we need to find the length of the slope connecting the shallow end and the deep end. We can use the Pythagorean theorem to find the length of the hypotenuse of the right triangle formed by the slope and the bottom of the pool:

h = sqrt((12.5 - 9)² + 4.5²) = 6.08 feet

Next, we need to find the difference in depth between the shallow end and the deep end. We can use the tangent of the angle of depression to find this difference:

tan(16.5) = (4.5 - d) / 6.08

where d is the depth of the pool at the deep end. Solving for d, we get:

d = 4.5 - 6.08 * tan(16.5) = 2.24 feet

Therefore, the depth of the pool at the deep end is approximately 2.2 feet.