Classify the following non-identity isometries of R². If the isometry is not unique, justify all possibilities. (a) Let f be an isometry, without fixed points, given by a reflection followed by a glide reflection. (b) Let g be an isometry that fixes two points, g(P) = P and g(Q) = Q. (c) Let h be the composition of three reflections, h = Fc Fy Fa. Suppose that the distinct lines a, b, c are concurrent (i.e., have a common point). (d) Now, suppose a || b and cla. Classify the isometry h. Justify.

Answer: 1/2

Step-by-step explanation:

From the question, we are informed that Eva uses 10 tiles to make a mosaic and that five of the tiles are blue.

The fraction that represents the tiles that are blue will be:

= Number of blue tiles / Total number of tiles

= 5/10

= 1/2

Answer:

13.) 237 millimeters

14.) 87.5 feet (87 feet 6 inches)

15.) 28

16.) 121 degrees

Step-by-step explanation:

16.) The "arc length" formula is s = rФ, where Ф represents the central angle in radians (not degrees).

Here r = 18 ft and s = 38 ft, and so:

                                     38 ft

s = rФ becomes Ф = ------------ = 2.111 radians

                                     18 ft

which, in degrees, is:

2.111 rad      180 deg

------------- * --------------- = 121 degrees, to the nearest degree

  1            3.142

Answer:

Hi

Step-by-step explanation:

Complete question

A square pyramid. The square base has side lengths of 6.5 meters and the height of 6 meters.

What is the volume of the pyramid?

Answer:

84.5m³

Step-by-step explanation:

The volume of a square. Pyramid is given as :

V = 1/3 * base * height

Base = l² ; l = side length

Base = 6.5² = 42.25 m²

V = 1/3 * 42.25 * 6

V = 42.25 * 2

V = 84.5

The volume of square pyramid = 84.5m³

Step-by-step explanation:

hey, hope it helps.. :))

Answer:

10.10 15.15

Step-by-step explanation:

25.30 15.12 25.13

The mean absolute deviation of last week's temperatures is 7.2°F.

Given below is the table of last week's and this week's low temperatures:Last week's temperatures: 29°F, 35°F, 42°F, 46°F, 52°FThis week's temperatures: 27°F, 31°F, 35°F, 38°F, 42°F, 46°F, 50°F

The range of this week's temperatures is found by subtracting the smallest value from the largest value. Therefore, the range of this week's temperatures is 50°F - 27°F = 23°F.

Mean absolute deviation is a measure of how much the data deviates from the mean of the data. To find the mean absolute deviation of last week's temperatures, we first need to find the mean of the data set.Using the formula for mean, we have:

Mean = (29 + 35 + 42 + 46 + 52)/5 = 40.8°F

To find the absolute deviations of each temperature from the mean, we need to subtract each temperature from the mean and take the absolute value.Absolute deviations:

|29 - 40.8| = 11.8|35 - 40.8|

= 5.8|42 - 40.8|

= 1.2|46 - 40.8|

= 5.2|52 - 40.8|

= 11.2

Next, we need to find the mean of these absolute deviations. Using the formula for mean again, we have:Mean absolute deviation = (11.8 + 5.8 + 1.2 + 5.2 + 11.2)/5 = 7.2°F

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Objective function is Z = 9 XC,RB + 11 XCB,GM, and the constraint for minimum vitamin C for Red Berry is75XC,RB + 350XCB,RB + 200XA,RB >= 0.

Objective function for the given statement is Z = 9 XC,RB + 11 XCB,GM,

where, XC, RB is the number of gallons of Cherry juice used in Red Berry, XCB, GM is the number of gallons of Cranberry juice used in Green Mush and also, XA, RB is the number of gallons of Avocado juice used in Red Berry, XC, GM is the number of gallons of Cherry juice used in Green Mush, XCB, GM is the number of gallons of Cranberry juice used in Green Mush, XA, GM is the number of gallons of Avocado juice used in Green Mush.

Hence, the objective function is Z = 9 XC,RB + 11 XCB,GM.

Minimum vitamin C for Red Berry will be given by the equation,

350XCB,RB + 400XC,RB + 200XA,RB >= 325XC,RB

=> 75XC,RB + 350XCB,RB + 200XA,RB >= 0

So, the constraint for minimum vitamin C for Red Berry is75XC,RB + 350XCB,RB + 200XA,RB >= 0.

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The probability that an individual reaches age 100 is 0.000001.

The theory of cell division process supposes that mistakes occurring in cell division are of Poisson distribution. The given Poisson parameter is 2.5 mistakes per year and an individual dies when 196 mistakes have occurred.

Let X denote the number of mistakes before an individual dies.

(a) The mean lifetime of an individual. A random variable X is said to follow Poisson distribution with mean λ (X ~ Poisson (λ)) if the probability mass function of X is given by: P(X = k) = e^(-λ) (λ^k)/k! Here, rate = 2.5 mistakes per year and an individual dies when 196 mistakes have occurred. Therefore, λ = rate x time = 2.5 mistakes/year × T years = 196 mistakes. T = 196/2.5 = 78.4 years. The mean lifetime of an individual is given by: μ = E(X) = λ = 78.4 years.  

(b) The variance of the lifetime of an individual. The variance of a Poisson distribution is given by: Var(X) = λ. Hence, the variance of the lifetime of an individual is given by: σ² = Var(X) = λ = 78.4 years

(c) .The probability that an individual dies before age 67.2Let Y denote the lifetime of an individual. The number of mistakes before an individual dies is given by X. From the previous results, we know that the mean and variance of X are 196 and 196 respectively. Let y = 67.2 be the age of the individual. We have to find the probability that the individual dies before y. In other words, we need to find P(Y < y). P(Y < y) = P(X < 196/y) = P(X < 196/67.2) = P(X < 2.9137) = 0.9868 approximately

(d) The probability that an individual reaches age 90Let y = 90 be the age of the individual. We have to find the probability that the individual reaches 90 years. In other words, we need to find P(Y ≥ y). P(Y ≥ y) = P(X ≥ 2.5 × 90) = P(X ≥ 225) = 1 - P(X < 225) = 1 - P(X ≤ 224). From Poisson distribution tables, we get:P(X ≤ 224) = 0.9993 approximately. Therefore, P(X ≥ 225) = 1 - P(X ≤ 224) = 1 - 0.9993 = 0.0007 approximately.

(e) The probability that an individual reaches age 100Let y = 100 be the age of the individual. We have to find the probability that the individual reaches 100 years. In other words, we need to find P(Y ≥ y). P(Y ≥ y) = P(X ≥ 2.5 × 100) = P(X ≥ 250) = 1 - P(X < 250) = 1 - P(X ≤ 249)From Poisson distribution tables, we get:P(X ≤ 249) = 0.999999 approximately.

Therefore, P(X ≥ 250) = 1 - P(X ≤ 249) = 1 - 0.999999 = 0.000001 approximately

Therefore, the probability that an individual reaches age 100 is 0.000001.

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Answer:

please stop being lazy by using OCR! next time type the question!

there's nothing such as  = -x2 3x + 1

it's probablhy f(x) = -x² + 3x + 1

Answer:

2 distinct irrational roots

Step-by-step explanation:

[tex]4000(1.003)^{(x-1)}[/tex]

Step-by-step explanation:

Answer:

abd

Step-by-step explanation:

im smart

Answer:

The answer is A B D F

Step-by-step explanation:

I took the test sorry I don't have a better explanation.

Answer:

this means there is no answer to the problem

Step-by-step explanation:

Answer:

9/15 = 3/5 (simplified)

Step-by-step explanation:

In a right triangle, the cosine is the ratio of the adjacent side to the hypotenuse:

[tex]\cos\theta = \frac{\text{adjacent}}{\text{hypotenuse}}[/tex]

The adjacent side is the side that makes up the angle but is not the hypotenuse.

I'm assuming the answer box should ask for cos(theta) rather than cos(x).

To find the arc length, we can use the formula:

[tex]\[ L = \frac{\theta}{360} \times 2\pi r \]\\Given:\( r = \frac{d}{2} = \frac{30}{2} = 15 \) ft\( \theta = 42 \) degrees\( \pi = 3.14 \)\\Substituting the given values into the formula:\[ L = \frac{42}{360} \times 2 \times 3.14 \times 15 \]\[ L = \frac{42}{360} \times 94.2 \]\[ L = \frac{3956.4}{360} \]\[ L \approx 10.99 \] ftTherefore, the arc length traveled by the car, to the nearest hundredth, is 10.99 feet. The correct option is c.[/tex]

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Given piecewise functions are: g(x) = {x² – 5 , 2€ (-[infinity], -7)9x - 17, 9 x € (-7,2](x + 1)(x - 5) , 2 € (2,00)

We are supposed to evaluate g(x) at x = 7. As per the given conditions,

we have the following; g(x) = {x² – 5 , 2€ (-[infinity], -7)9x - 17, 9 x € (-7,2](x + 1)(x - 5) , 2 € (2,00)

Now, g(7) represents the value of function g(x) at x = 7. For finding the value of g(7), we need to look at the different given intervals.

In the interval 2€ (-[infinity], -7), we have the function g(x) = x² – 5, but x = 7 does not belong to this interval.

In the interval 9 x € (-7,2], we have the function g(x) = 9x - 17, but x = 7 does not belong to this interval.

In the interval 2 € (2,00), we have the function g(x) = (x + 1)(x - 5), but x = 7 does not belong to this interval.

As x = 7 does not belong to any of the given intervals, g(7) is not defined.

Hence, the correct option is "Not defined".

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Answer:

whats ur question?

Step-by-step explanation:

The particular solution to the differential equation with the initial condition y(0) = 1 is:

[tex]y = 1 + cos(t) + 1 - e^{(cos(t))[/tex]

To solve the given differential equation:

dy/dt + sin(t) = 1

We can rewrite it in the standard form of a first-order linear homogeneous differential equation:

dy/dt = 1 - sin(t)

The integrating factor for this equation is [tex]e^{(\int(-sin(t))dt)} = e^{(-cos(t)).[/tex]

Now, multiply both sides of the equation by the integrating factor:

[tex]e^{(-cos(t)) \times dy/dt} = (1 - sin(t)) \times e^{(-cos(t))[/tex]

The left-hand side can be rewritten using the chain rule:

[tex]d/dt [e^{(-cos(t))} \times y] = (1 - sin(t)) \times e^{(-cos(t))[/tex]

Integrate both sides with respect to t:

[tex]\int d/dt [e^{(-cos(t))} \times y] dt = \int (1 - sin(t)) \times e^{(-cos(t)) dt[/tex]

[tex]e^{(-cos(t))} \times y = \int (1 - sin(t)) \times e^{(-cos(t)) dt[/tex]

To evaluate the integral on the right-hand side, let u = -cos(t), du = sin(t) dt:

[tex]e^{(-cos(t))} \times y = \int (1 - sin(t)) \times e^{(-cos(t)) dt[/tex]

[tex]= \int (1 - sin(t)) \times e^u du[/tex]

[tex]= \int (e^u - sin(t) \times e^u) du[/tex]

[tex]= e^u - \int sin(t) \times e^u du[/tex]

Now, integrate the second term by parts:

[tex]\int sin(t) \times e^u du = -e^u \times cos(t) + \int e^u \times cos(t) dt[/tex]

Substituting the expression back into the equation:

[tex]e^{(-cos(t))} \times y = e^u - (-e^u \times cos(t) + \int e^u \times cos(t) dt)[/tex]

Simplifying:

[tex]e^{(-cos(t))} \times y = e^{(-cos(t))} + e^{(-cos(t))} \times cos(t) - \int e^{(-cos(t))} \times cos(t) dt[/tex]

To solve the remaining integral, let v = -cos(t), dv = sin(t) dt:

[tex]\int e^{(-cos(t))} \times cos(t) dt = \int e^v \times (-dv)[/tex]

[tex]= -\int e^v dv[/tex]

[tex]= -e^v + C[/tex]

[tex]= -e^{(-cos(t))} + C[/tex]

Substituting back into the equation:

[tex]e^{(-cos(t))} \times y = e^{(-cos(t))} + e^{(-cos(t))} \times cos(t) - (-e^{(-cos(t))} + C)[/tex]

Divide both sides by [tex]e^{(-cos(t))[/tex]:

[tex]y = 1 + cos(t) + 1 + C \times e^{(cos(t))[/tex]

Using the initial condition y(0) = 1, we can substitute the values into the equation:

[tex]1 = 1 + cos(0) + 1 + C \times e^{(cos(0))[/tex]

[tex]1 = 1 + 1 + C \times e^1[/tex]

[tex]C \times e = -1[/tex]

Therefore, the particular solution to the differential equation with the initial condition y(0) = 1 is:

[tex]y = 1 + cos(t) + 1 - e^{(cos(t))[/tex]

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Complete question =

Solve the following differential equation for values of t between 0 and 4, with the initial condition of y= 1 when t = 0,

dy/dt + sin(t) = 1

Answer:0.3 m

Step-by-step explanation:

Given

Ken can Jump [tex]1\ \frac{1}{5}\ m[/tex]

Steve can Jump [tex]1.5\ m[/tex]

Converting mixed fraction into a fraction

[tex]\Rightarrow 1\ \dfrac{1}{5}=\dfrac{1\times 5+1}{5}\\\\\Rightarrow \dfrac{6}{5}\ m=1.2\ m[/tex]

The difference between their Jumps is

[tex]\Rightarrow 1.5-1.2=0.3\ m[/tex]

Steve Jumps 0.3 m more than ken

To find the 95% confidence interval when n=100 and p=0.72, we can use the following formula:

$$\left(\hat{p}-z_{\alpha/2} \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}, \hat{p}+z_{\alpha/2} \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\right)$$

Where, $\hat{p}$ is the point estimate of the population proportion, $n$ is the sample size, $z_{\alpha/2}$ is the critical value of the standard normal distribution at a significance level of $\alpha$, which can be obtained from a table. For a 95% confidence level, $\alpha$ is equal to 0.05/2 = 0.025 on each tail.

The corresponding z-value is 1.96 (approximately).Hence, plugging in the values, we get

$$\begin{aligned}\left(0.72-1.96 \sqrt{\frac{0.72(0.28)}{100}}, 0.72+1.96 \sqrt{\frac{0.72(0.28)}{100}}\right) \\ \left(0.631, 0.809\right)\end{aligned}$$

Therefore, the 95% confidence interval is (0.631, 0.809) rounded to three decimal places.

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Answer:

c

Step-by-step explanation:

The specific enthalpy at the exit of the pipe is 2804 kJ/kg.

The expression of specific enthalpy at the exit of the pipe can be found by the following method:

Here, m = mass flow rate of the steam = 3 kg/s

q1 = specific enthalpy at inlet of the pipe = 2744 kJ/kg

W = work done on the system = 0 (steady-state)

Q = rate of heat transfer = 696 kW

z1 = elevation of the inlet = 0 m

z2 = elevation of the outlet = +71 m

Since the rate of heat transfer is given as

Q = -m(h2 - h1) + W + Q_hot

So, h2 = [Q - Q_hot + m(h1) - W]/m

Where W = 0 and Q_hot = 0

h2 = (696 - 0 + 3(2744))/3

h2 = 2804 kJ/kg

Therefore, the specific enthalpy at the exit of the pipe is 2804 kJ/kg.

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The proportion of values between 152 and 124 cm is 0.8996 or 89.96%.

Proportion between 152 and 124 cm according to a normal distribution with mu = 138cm and sigma = 7cm is 0.8996.

According to the given question, we know that the mean is μ = 138 cm, standard deviation is σ = 7 cm.

We have to find the proportion that is between 152 and 124 cm.

To solve the problem, first we have to find the z-scores for 152 and 124 cm.

We can calculate the z-scores as follows:Z-score for 152 cm is given by:z152​=(152−138)7=2z_{152}=\frac{(152-138)}{7}=2z152​=(152−138)7=2Z-score for 124 cm is given by:z124​=(124−138)7=−2z_{124}=\frac{(124-138)}{7}=-2z124​=(124−138)7=−2

We can use a z-table or calculator to find the proportion of values between these two z-scores.

The area under the standard normal curve between z = -2 and z = 2 is approximately 0.8996.

Therefore, the proportion of values between 152 and 124 cm is 0.8996 or 89.96%.

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Answer:

12 students.

Step-by-step explanation:

If you had 4 sheets, you would want to start by dividing them all in 3's, since that's easiest. (open the file i just for a better example of how the sheet should look).

Now, let's assume these sheets are regular office sized paper.

Start by handing 1 third out to A student.

Now, since there are 4 sheets, divided in 3 sections, we multiply to see the result of how many sections there are.

4 x 3 = 12.

Therefore, 12 students will receive a third of the tissue paper.

Answer:

2(11*9) + 2(11*h) + 2(9*h) = 558

h = 9 in.

Step-by-step explanation:

2(11*9) + 2(11*h) + 2(9*h) = 558

198 + 22h + 18h = 558

40h = 558-198

40h = 360

h = 9 in.