Show that Φ (Phi/fa1) uppercase Φ, lowercase φ or ϕ) is onto and one-to-one

Answer:

If there was a flood, you would calulate how high the water is.

Answer:

-n-2

Step-by-step explanation:

the focus of all points Y that occur as midpoints of segments PX, where X lies on the given circle, is a circle with radius half of the original circle (R/2) and its center located at the midpoint M of segment OP.

Given a circle centered at point O and an arbitrary point P, let's consider the focus of all points Y that occur as midpoints of segments PX, where X lies on the given circle.
To show that this focus is a circle with radius half of the original circle, let's first analyze the properties of the midpoint Y. By definition, Y is the midpoint of segment PX, so PY = YX.
Let the radius of the original circle be R, and let OP = d. Using the law of cosines, we can derive the distance between X and P as follows:

PX² = OP² + OX² - 2(OP)(OX)cos∠POX.

Since OX = R, OP = d, and ∠POX = θ, we have:

PX² = d² + R² - 2dRcosθ.

Now, let's consider the distance between Y and P. Since Y is the midpoint of segment PX, we have:

PY² = (1/2PX)² = (1/4)(d² + R² - 2dRcosθ).

Now, let's find the locus of Y. To do so, we will fix the distance PY and let the angle θ vary. The locus of Y will be the set of all points equidistant from P with distance (1/2)PX. This forms a circle centered at a point M with radius half of the original circle (R/2).

To locate the center M of the locus, let's consider the midpoint of segment OP. Since Y lies on the line segment PX, and Y is the midpoint of PX, the locus of Y will be the set of all points equidistant from the midpoint of segment OP. This forms a circle centered at point M, where M is the midpoint of segment OP.

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Answer: 640801

Step-by-step explanation: Simply add both together, 637,982 plus 2819.

Answer:

637, 982 + 2,819 = 640,801

1) The solution to the initial-value problem is y(t) = 2 cos(4t) - (1/2) sin(4t).

2) The solution to the initial-value problem is y(t) = 0.

Let's solve the initial-value problems given.

Initial-Value Problem: y'' + 16y = 0, y(0) = 2, y'(0) = -2

We have a second-order linear homogeneous differential equation. To solve this, we assume a solution of the form y(t) = \(e^{rt\), where r is a constant to be determined.

Substituting this assumed solution into the differential equation, we get the characteristic equation:

\(r^2\) + 16 = 0

Solving the characteristic equation for r, we have:

r = ±4i

Since the roots are complex (r = 4i and r = -4i), the general solution will involve complex exponentials.

The general solution for the differential equation is:

y(t) = \(c_1\) cos(4t) + \(c_2\) sin(4t)

To find the particular solution that satisfies the initial conditions, we substitute the initial conditions y(0) = 2 and y'(0) = -2 into the general solution:

y(0) = \(c_1\) cos(0) + \(c_2\) sin(0) = \(c_1\) = 2

y'(0) = -4\(c_1\) sin(0) + 4\(c_2\) cos(0) = 4\(c_2\) = -2

Solving these equations, we find \(c_1\) = 2 and \(c_2\) = -1/2.

Therefore, the solution to the initial-value problem is:

y(t) = 2 cos(4t) - (1/2) sin(4t)

2) Initial-Value Problem: y'' + y' + 2y = 0, y(0) = y'(0) = 0

We have a second-order linear homogeneous differential equation. To solve this, we assume a solution of the form y(t) = \(e^{rt\), where r is a constant to be determined.

Substituting this assumed solution into the differential equation, we get the characteristic equation:

\(r^2\) + r + 2 = 0

Solving the characteristic equation for r, we have:

r = (-1 ± √(1 - 4(1)(2)))/2

= (-1 ± √(-7))/2

= (-1 ± i√7)/2

Since the roots are complex, the general solution will involve complex exponentials.

The general solution for the differential equation is:

y(t) = \(c_1\) \(e^{-t/2\) cos((√7/2)t) + \(c_2\) \(e^{-t/2\) sin((√7/2)t)

To find the particular solution that satisfies the initial conditions, we substitute the initial conditions y(0) = 0 and y'(0) = 0 into the general solution:

y(0) = \(c_1\) = 0

y'(0) = -(1/2)\(c_1\) + (√7/2)\(c_2\) = 0

Solving this equation, we find \(c_1\) = 0 and \(c_2\) = 0.

Therefore, the solution to the initial-value problem is:

y(t) = 0

In this case, the zero solution satisfies the initial conditions.

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Answer:

The solution of the system of equations are;

(-2, -6) and (4, 6)

Step-by-step explanation:

-2·x + y = -2...............(1)

\(y = -\dfrac{1}{2} \cdot x^2 + 3 \cdot x + 2\)........(2)

Equation (1) gives;

y = 2·x - 2

From which we have;

\(2 \cdot x - 2 = -\dfrac{1}{2} \cdot x^2 + 3 \cdot x + 2\)

\(0= -\dfrac{1}{2} \cdot x^2 + x + 4\)

x² -2·x -8 = 0

(x - 4)·(x + 2) = 0

x = 4 or x = -2

The y-coordinate values are;

y = 2×(-2) - 2 = -6 and y = 2×(4) - 2 = 6

The solution points are;

(-2, -6) and (4, 6).

The points where the equation, -2·x + y = -2 and the equation \(y = -\dfrac{1}{2} \cdot x^2 + 3 \cdot x + 2\) intersect are  (-2, -6) and (4, 6).

By evaluating the integral \(\int_1^5 xg'(x) dx\) we will get -27.

Given that g(1) = 2, g(5) = -6, and \(\int_1^5 g(x) dx = -5\)

Given integral \(\int_1^5 g(x) dx\) = -5⇒ \(\int_1^5 g(x) dx\) = -5------(1)

∫x g'(x) dx = x g(x) - ∫g(x) dx

As function g(x) is not given, we need to use the above formula to find the \(\int_1^5 xg'(x) dx\).

\(\int_1^5 xg'(x) dx = [x g(x)]_1^5 - \int_1^5 g(x) dx\)

We know that for function g(x), g(1) = 2 and g(5) = -6, so:

\([xg(x)]_1^5 = 5g(5) - 1g(1) = 5(-6) - 2 = -32\)

And we also know that \(\int_1^5 g(x) dx = -5\), so:

\(\int_1^5 xg'(x) dx = [xg(x)]_1^5 - \int_1^5 g(x) dx\)

= -32 - (-5)

= -27

Therefore, \(\int_1^5 xg'(x) dx\) is equal to -27.

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In analysis of variance, sample variance is measured by an MS value.

The variability in a particular sample is determined using sample variance. A sample is a collection of observations taken from a population that can accurately reflect the entire population. The sample variance is calculated in relation to the data set mean. Additionally called the estimated variance.

Due to the fact that data can be grouped or ungrouped, there are two formulas available to determine the sample variance. The sample standard deviation is also produced by taking the square root of the sample variance. The spread of the data points in a particular data set around the mean is measured using sample variance.

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Answer:

The correct answer:

C. 7

3(2) + 1 = 6 + 1 = 7

Step-by-step explanation:

So the question is 3x+1. We are given the information that x=2. Therefor we would plug in 2 to where x is. 3(2)+1. Now we would multiply first for 6+1. Add to get 7. 7 would be our answer giving us C.

Hamid is going to take one of his T-shirts at random.

The probability of selecting a light blue T-shirt is \(\frac{1}{4}\) or \(0.25\).

a) To find the probability that the T-shirt will be grey, we need to determine the ratio of grey T-shirts to the total number of T-shirts.

From the given information, we know that Hamid has 2 grey T-shirts out of a total of 20 T-shirts.

Therefore, the probability of selecting a grey T-shirt at random is \(\frac{2}{20}\), which simplifies to \(\frac{1}{10}\) or \(0.1\).

b) To find the probability that the T-shirt will not be red, we need to determine the ratio of non-red T-shirts to the total number of T-shirts.

Hamid has 4 red T-shirts, so the number of non-red T-shirts is

\(20 - 4 = 16\).

Therefore, the probability of selecting a T-shirt that is not red is \(\frac{16}{20}\), which simplifies to \(\frac{4}{5}\) or \(0.8\).

c) If Hamid takes one of his blue T-shirts at random, we need to consider the ratio of blue T-shirts to the total number of T-shirts.

From the information provided, we know that Hamid has a total of 4 blue T-shirts.

Therefore, the probability of selecting a blue T-shirt at random is \(\frac{4}{20}\), which simplifies to \(\frac{1}{5}\) or \(0.2\).

d) If Hamid takes one of his blue T-shirts at random, the probability that the T-shirt is light blue depends on the ratio of light blue T-shirts to the total number of blue T-shirts.

From the information provided, there is only 1 light blue T-shirt and a total of 4 blue T-shirts.

Hence, the probability of selecting a light blue T-shirt is \(\frac{1}{4}\) or \(0.25\).

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Answer:

the answer is D i believe

Answer:

Approximately 9.211 times

Step-by-step explanation:

Given

\(x = 7.6 * 10^{-22}s\) --- half life of helium-5

\(y = 7 * 10^{-21}s\) --- half life of helium-9

Required

How many times greater is y than x

To do this, we simply divide y by x

\(\frac{y}{x} = \frac{7 * 10^{-21}s}{7.6 * 10^{-22}s}\)

\(\frac{y}{x} = \frac{7 * 10^{-21}}{7.6 * 10^{-22}}\)

Apply law of indices

\(\frac{y}{x} = \frac{7 * 10^{-21--22}}{7.6}\)

\(\frac{y}{x} = \frac{7 * 10^{-21+22}}{7.6}\)

\(\frac{y}{x} = \frac{7 * 10}{7.6}\)

\(\frac{y}{x} = \frac{70}{7.6}\)

\(\frac{y}{x} = 9.211\)

Since the distance from a point to point C is 1 and the distance from that same point to point B is 4, the point must be: C. point D.

In Mathematics and Geometry, a number line simply refers to a type of graph with a graduated straight line which comprises both positive and negative numbers that are placed at equal intervals along its length.

This ultimately implies that, a number line primarily increases in numerical value towards the right from zero (0) and decreases in numerical value towards the left from zero (0).

From the number line shown above, we have:

Distance = 4 + (-1)

Distance = 4 - 1

Distance = 3 (point D).

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The results indicate that encountering sales clerks who smile could potentially increase customer purchases.

From the information given, we have that;

The four stores in the smile condition were 36, 40, 36, and 44;

To determine the t-value, we need to first calculate the mean for the control condition, we have;

Mean = 36 + 40 + 36+ 44/4

Mean =126/4

Mean = 32

Mean for the smile condition;

Mean = 39

One can employ a t-test to compare the average values of the two situations in order to scrutinize the outcomes. With a significance level of 0. 05, a two-tailed test with six degrees of freedom would require a critical t-value of around 2. 447

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Answer:

12 over -8

Step-by-step explanation:

We can write the given partial equilibrium model on the form Ax = d, and then use Cramer's rule to solve for the values of Qs* and P*.

To write the model on the form Ax = d, we need to express the equations in a matrix form.

The given equations are:

Qd = a1 + b1P

Qs = a2 + b2(P - t)

We can rewrite these equations as:

-Qd + 0P + Qs = a1

0Qd - b2P + Qs = a2 - b2t

Now, we can represent the coefficients of the variables and the constants in matrix form:

| -1 0 1 | | Qd | | a1 |

| 0 -b2 1 | * | P | = | a2 - b2t |

| 0 1 0 | | Qs | | 0 |

Let's denote the coefficient matrix as A, the variable matrix as x, and the constant matrix as d. So, we have:

A * x = d

Using Cramer's rule, we can solve for the variables Qs* and P*:

Qs* = | A_qs* | / | A |

P* = | A_p* | / | A |

where A_qs* is the matrix obtained by replacing the Qs column in A with d, and A_p* is the matrix obtained by replacing the P column in A with d.

By calculating the determinants, we can find the values of Qs* and P*.

It's important to note that Cramer's rule allows us to solve for the variables in this system of equations. However, the applicability of Cramer's rule depends on the properties of the coefficient matrix A, specifically its determinant. If the determinant is zero, Cramer's rule cannot be used. In such cases, alternative methods like substitution or elimination may be required to solve the equations.

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Answer:

B)

Step-by-step explanation:

given the 2 angles of 60°, that makes the 3rd angles also 60° (remember, the sum of all angles in a triangle is always 180°).

and that makes the triangle an equilateral triangle (all 3 sides are equally long : 20).

BD = sin(60)×20

CD = cos(60)×20

tan(60) = sin(60)/cos(60) = sin(60)×20/(cos(60)×20)

CD = 20/2 = 10

for BD we need Pythagoras

20² = BD² + 10²

400 = BD² + 100

300 = BD²

BD = sqrt(300) = sqrt(4×75) = sqrt(4×25×3) =

= 2×5×sqrt(3) = 10×sqrt(3)

so,

tan(60) = BD/CD = 10×sqrt(3) / 10

which is sqrt(3), by the way.

Answer:

48/5in2=9.6in2 ~ 10in2

Step-by-step explanation:

216p³ - 125

p³ = 125 / 216

p = 5/6

»216p³ - 125

»216p³ = 125

»p³ = 125/216

»p = ³√125/³√216

»p = 5/6✅

Let the number of hours = h

Multiply the rate per hour by hours and add to the rental fee.

This at most can equal the amount you have.

Equation: 6h+ 13.50 <= 37.50

Solve for h:

6h + 13.50 <= 37.50

Subtract 13.50 from both sides:

6h <= 24.00

Divide both sides by 6:

H <= 4

You can rent the canoe for 4 hours maximum.

A good way to get a small standard error is to use a Large sample. the correct answer is option 1.

The standard error is a measure of the variability of the sampling distribution of a statistic. A smaller standard error indicates that the statistic is more precise and is likely closer to the true population value.

One way to obtain a smaller standard error is to use a larger sample size. This is because a larger sample size tends to produce a more accurate estimate of the population parameter, with less variability. Therefore, option 1, "Large sample," is the correct answer.

The other options, such as a large population, repeated sampling, small sample, or small population, are not necessarily related to obtaining a small standard error.

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Step-by-step explanation:

The program I used tells me the lengths, though I could use the law of cosines or Pythagoras's Therom for this.

AB: 2

BC: 3

AC: About 3.61

Also, point A is (-2, 4), point B is (-2, 2), and point C is (1, 2)

Hope this helps!

A total number of 4 x 10⁶ DVD's were sold in the span of next three years.

A function is a relation between a dependent and independent variable. We can write the examples of functions as -

y = f(x) = ax + b

y = f(x, y, z) = ax + by + cz

Given is that in 1996, 1 million DVDs were sold.

Since no further information is given, assume that the increase in the sales of the DVD's took place at the same rate every year. So, initially, 1 million DVD's have been sold. So, for the next {x} years, we can write the total number of DVD's sold as -

y = mx + c

y = 10⁶x + 10⁶

y = 10⁶(x + 1)

For {x} = 3, we can write -

y = 4 x 10⁶

Therefore, a total number of 4 x 10⁶ DVD's were sold in the span of next three years.

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The length of the rectangle is 10 inches and the rectangle is 10 inches long and 2 inches wide.

The total distance around the outside of the rectangle is termed as Perimeter of a rectangle. It is equal to the sum of the lengths of all four sides of the rectangle.

Let's assume that the width as "w" inches.

As per the problem statement, the length of the rectangle is 5 times its width, which means that the length is 5w inches.

The perimeter of a rectangle is :

P = 2(l + w)

Substituting the values, we get:

24 = 2(5w + w)

Simplifying the above equation, we get:

24 = 12w

Dividing both sides by 12, we get:

w = 2

So, the width of the rectangle is 2 inches.

The length is 5 times of width, that is:-

l = 5w = 5(2) = 10

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Answer: 9

Step-by-step explanation:

Formula : Mean = \(\dfrac{\text{Sum of observations}}{\text{Number of observaions}}\)

Given: 5 students participated in a push-up competition.

The amount of push-ups each student completed is listed: 12, 7, 10, 11,5.

Mean = \(\dfrac{12+7+10+77+5}{5}\)

\(=\dfrac{45}{5}=9\)

Hence, the mean number of push-ups = 9

The option that correctly describes scatter plots is:

Option 2: are used to see two numeric variables plotted in comparison with each other.

A scatterplot is defined as a type of graph or mathematical graph that uses Cartesian coordinates to show the values ​​of usually two variables in a data set. Additional variables can be displayed if the points are coded

Scatterplots are used to identify possible relationships between changes observed in two different sets of variables. It provides a visual and statistical way to test the strength of the relationship between two variables.

Looking at the given options and comparing with the definition of scatter plots above, it is very clear that only option 2 is correct because they are used to see two numeric variables plotted in comparison with each other.

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Complete question is:

Group of answer choices

1. are not used for revealing patterns.

2. are used to see two numeric variables plotted in comparison with each other.

3. must lie on straight lines.

4. are used to see two categorical variables plotted in comparison with each other.

Answer:

a, h and k have values 1, -5 and -10 respectively

The area of the square is 144 \(cm^{2}\).

Let x be the side of the square. Then the length of the triangle is (x+4). Perimeter is the length of all sides of a geometric figure combined. For an equilateral triangle, it's equal to thrice the length of one side. For a square, it's four times the length of one side. The Perimeter of the Triangle is 3(x+4) & the Perimeter of the square is 4x.

We know, both these perimeters are equal. Hence,

4x = 3(x+4)

To further simplify the above equation.

4x = 3x + 12

x = 12

Hence, the length of one side of the square is 12 cm. The area of the square can be calculated as follows:

Area = \((side)^{2}\)

Area = 12 * 12

Area = 144 \(cm^{2}\)

Hence, the Area of the Square is 144 \(cm^{2}\)

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Answer:

13)69°{corresponding angles}

14)52°{exterior alternate angles}

15)71°{vertically opposite angle}

16)110°{supplementary angles}

17)126°{interior alternate angles}

18)109°{co interior}

13. 69*

14. 52*

15. 71*

16. 110*

17. 126*

18. 109*

The results for the variable t in each linear equation are listed below:

In this question we have linear equations with a single variable that must be cleared by using algebra properties. Now we proceed to solve each equation for the variable t:

Equation 1

16 · t + 4 · (5 - t) = 8

16 · t + 20 - 4 · t = 8

12 · t = - 12

t = - 1

Equation 2

18 = 2 · t - 3 · (t + 2)

18 = 2 · t - 3 · t - 6

24 = - t

t = - 24

Equation 3

- 10 · t - (t + 5) = 28

- 10 · t - t - 5 = 28

- 11 · t = 33

t = - 3

Equation 4

- 6 · (3 · t + 5) = 6

- 18 · t - 30 = 6

- 18 · t = 36

t = - 2

Equation 5

12 = 3 · (t + 14)

12 = 3 · t + 42

3 · t = - 30

t = - 10

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