The perimeter of a square is 28 cm. Find its area and the length of its diagonal.


help me ​

Answer:

90 = π/2

45 = π/4

0 and 360 = 0 and 2π

135 = 3π/4

180 = π

225 = 5π/4

270 = 2π/3

315 = 7π/4

315 =

Step-by-step explanation:

Y(s) = 6/s * 1/(s * (s - 3)) + 1/s * 5 + 1/(s - 3) * 12

The solution to the initial value problem is :

y(t) = 12e³ᵗ + 3.

We have 3t y'' - 9y' + 18y = 6t e

Taking Laplace transform on both sides, we get

3L(ty'') - 9L(y') + 18L(y) = 6L(te)

Using Laplace transform formulas, we get:

3[s²Y(s) - sy(0) - y'(0)] - 9[sY(s) - y(0)] + 18Y(s) = 6/s²L(e)

⇒ 3s²Y(s) - 3s(5) + 6 - 9sY(s) + 45 + 18Y(s) = 6/s² * 1/sY(s)[3s² - 9s + 18] = 6/s² * 1/s - 3s + 12Y(s) = 6/s * 1/(s * (s - 3)) + 1/s * 5 + 1/(s - 3) * 12

Now, we need to find inverse Laplace transform of Y(s) to obtain the solution y(t).

Let's solve for the first term by Partial Fraction Expansion.

6/s * 1/(s * (s - 3))= A/s + B/(s - 3)6 = A(s - 3) + Bs

Therefore, A = -2 and B = 2y(t) = L⁻¹[Y(s)] = L⁻¹[6/s * 1/(s * (s - 3)) + 1/s * 5 + 1/(s - 3) * 12]= -2L⁻¹[1/s] + 2L⁻¹[1/(s - 3)] + 5L⁻¹[1/s] + 12L⁻¹[1/(s - 3)]= -2 + 2e³ᵗ + 5 + 12e³ᵗ= 12e³ᵗ + 3

Therefore, Y(s) = 6/s * 1/(s * (s - 3)) + 1/s * 5 + 1/(s - 3) * 12 and the solution to the initial value problem is y(t) = 12e³ᵗ + 3.

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Answer:

(Thermometer B reading - Thermometer A reading)

Step-by-step explanation:

The thermometer reading aren't given in the question.

However, hypothetically.

The difference between two temperature values (morning and evening values) would be :

Temperature in the evening - morning temperature

Therefore,

If ;

Thermometer A reading = morning temperature

Thermometer B reading = evening temperature

Difference in the temperature :

(Thermometer B reading - Thermometer A reading)

Answer:

45/49

decimal form:

0.91836734

Step-by-step explanation:

The exact value of A in the general solution is 1 and B is 7

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

y = Ax² + Bx + C

The differential equation is given as

y' = 2x + 7

When y = Ax² + Bx + C is differentiated, we have

y' = 2Ax + B

So, we have

2x + 7 = 2Ax + B

By comparing both sides of the equation, we have

2Ax = 2x

B = 7

So, we have

2A = 2

B = 7

Divide both sides of 2A = 2 by 2

A = 1

B = 7

Hence, the value of A in the general solution is 1 and B is 7

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Answer:

between Jupiter and mars

Answer:

Choice A

Step-by-step explanation:

The Asteroid Belt in our Solar System is in-between the planets Jupiter and Mars.

The asteroid belt is a torus-shaped region in the Solar System, located roughly between the orbits of the planets Jupiter and Mars, that is occupied by a great many solid, irregularly shaped bodies, of many sizes but much smaller than planets, called asteroids or minor planets.

Step-by-step explanation:

megan thee stallion songs

cardi b's songs

space cadet-gunna

astronaut in the ocean

Flo milli

and my personal favorite

knock knock- sofaygo

The solution to the initial-value problem using the Laplace transform is y(t) = (1/25)(1 - [tex]e^{(-5t)[/tex]) - (1/25)t + (1/125)[tex]e^{(-5t)[/tex].

To solve the given initial-value problem using Laplace transform, we will first take the Laplace transform of the given differential equation and apply the initial condition.

Take the Laplace transform of the differential equation:

Applying the Laplace transform to the equation y' + 5y = f(t), we get:

sY(s) - y(0) + 5Y(s) = F(s),

where Y(s) represents the Laplace transform of y(t) and F(s) represents the Laplace transform of f(t).

Apply the initial condition:

Using the initial condition y(0) = 0, we substitute the value into the transformed equation:

sY(s) - 0 + 5Y(s) = F(s).

Substitute the given function f(t):

The given function f(t) is defined as:

f(t) = t, 0 ≤ t < 1

f(t) = 0, t ≥ 1

Taking the Laplace transform of f(t), we have:

F(s) = L{t} = 1/s²,

Solve for Y(s):

Substituting F(s) and solving for Y(s) in the transformed equation:

sY(s) + 5Y(s) = 1/s²,

(Y(s)(s + 5) = 1/s²,

Y(s) = 1/(s²(s + 5)).

Inverse Laplace transform:

To find y(t), we need to take the inverse Laplace transform of Y(s). Using partial fraction decomposition, we can write Y(s) as:

Y(s) = A/s + B/s² + C/(s + 5),

Multiplying both sides by s(s + 5), we have:

1 = A(s + 5) + Bs + Cs².

Expanding and comparing coefficients, we get:

A = 1/25, B = -1/25, C = 1/125.

Therefore, the inverse Laplace transform of Y(s) is:

y(t) = (1/25)(1 - [tex]e^{(-5t)[/tex]) - (1/25)t + (1/125)[tex]e^{(-5t)[/tex].

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Answer:

$247.50

Step-by-step explanation:

Answer:

45°

Step-by-step explanation:

x should be the equivalent angle as the 45° given, as this is a perfect circle, so the distance from the center shouldn't affect the angle

Answer:

9.8 units

Step-by-step explanation:

distance = sqrt (x2 - x1)^2 + ( y2 - y1)^2

sqrt (-3 - (-6))^2 + (5 - 1)^2

sqrt (9)^2 + (4)^2

sqrt 81 + 16

sqrt 97

9.848857802

Answer:

27/86

Step-by-step explanation:

the difference is multiplying by 3. next number is

27/86

Answer:

y intercept (0;4)

Step-by-step explanation:

let x = 0 because the graph will intersect the y-axis at the value of 0 for the x-axis

The number of labeled bonsai forests with n vertices and k trees is given by (3)^(kn-k).

The number of labeled bonsai forests with n vertices and k trees is (3)^(kn-k).

To understand why this is the case, let's break it down step by step.

First, let's consider a single bonsai tree with a root and n-1 other vertices connected to the root.

Each of these n-1 vertices can have one of three choices: either it is connected to the root, it is not connected to the root, or it is the root itself. Therefore, for a single bonsai tree, we have 3^(n-1) possibilities.

Now, if we have k bonsai trees, we can treat each tree as an independent entity. Therefore, the total number of labeled bonsai forests with k trees would be the product of the number of possibilities for each individual tree.

Hence, the total number of labeled bonsai forests with n vertices and k trees is (3)^(n-1) * (3)^(n-1) * ... * (3)^(n-1) (k times), which can be written as (3)^(kn-k).

In simpler terms, for each vertex in the bonsai forest, there are three possible choices: being connected to the root, not connected to the root, or being the root itself. As each vertex is independent and has the same three choices, the total number of possibilities for the entire forest is calculated by multiplying the number of possibilities for each vertex (3) by itself (n-1) times, for a total of kn-k times.

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Answer:

i need the rest of the problem to figure it out sorry

Step-by-step explanation:

Answer:

60 ounces

Step-by-step explanation:

got i t on edmentum

Step-by-step explanation:

1/2(7x+48) = 7x ÷2 +48÷2 = 7x÷2 + 24

and

-(1/2x-3)+4(x+5) = 7x ÷2 + 46÷2 = 7x÷2 +23

Answer: there is no picture

Answer:

g=4

Step-by-step explanation:

this is a 30 60 90 triangle. the hypotenuse is 2x while the shortest side is x. if 8=2x then x must be 4.

The distinction between parameters and statistics is crucial for inferential statistics, the correct is option D.

The population of interest in the scenario,

1."For all named storms that have made landfall in the United States since 2000, of interest is to determine the mean sustained wind speed of the storms at the time they made landfall," is:

all named storms that have made landfall in the United States since 2000.

5.The correct answer is D. Statistic.

A parameter is a numerical or other measurable factor that characterizes a given population, while a statistic is a numerical value calculated from a sample of data.

Parameters are used to describe a population, while statistics are used to describe a sample from a population.

The distinction between parameters and statistics is crucial for inferential statistics.

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The area of the surface formed by the part of the sphere [tex]x^2 + y^2 + z^2 = 4z[/tex] that lies inside the paraboloid [tex]z = x^2 + y^2[/tex] is π/6 square units.

To find the area of the surface, we need to calculate the double integral over the region that lies inside both the sphere and the paraboloid.

The given sphere equation can be rewritten as [tex]x^2 + y^2 + (z - 2)^2 = 4[/tex]. This represents a sphere centered at (0, 0, 2) with a radius of 2.

The paraboloid equation [tex]z = x^2 + y^2[/tex] represents an upward-opening paraboloid centered at the origin.

To find the region of intersection, we set the sphere equation equal to the paraboloid equation:

[tex]x^2 + y^2 + (x^2 + y^2 - 2)^2 = 4[/tex]

Simplifying, we get [tex]x^4 + y^4 - 4x^2 - 4y^2 + 4 = 0[/tex].

This equation represents the boundary curve of the region of intersection.

By evaluating the double integral over this region, we find the area of the surface to be π/6 square units.

Therefore, the area of the surface formed by the given part of the sphere lying inside the paraboloid is π/6 square units.

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Answer:

(-2, -4)

Step-by-step explanation:

If a population proportion p = 0.68, the expected value and the standard error of p' with n = 30 is 0.68 and 0.090 respectively.

To calculate the expected value and standard error of the sample proportion p' with a known population proportion p = 0.68 and a sample size n = 30, we use the formulas:

Expected value of p' (E[p']) = p

Standard error of p' (SE[p']) = √((p * (1 - p)) / n)

Given that the population proportion p = 0.68 and the sample size n = 30, we can substitute these values into the formulas:

E[p'] = p = 0.68

SE[p'] = √((p * (1 - p)) / n) = √((0.68 * (1 - 0.68)) / 30) = √(0.2176 / 30) ≈ 0.090

Therefore, the expected value of the sample proportion p' is 0.68, indicating that, on average, we expect the sample proportion to be equal to the population proportion.

The standard error of the sample proportion is approximately 0.090, representing the estimated standard deviation of the sampling distribution of p' and indicating the variability in the estimates of p'.

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The given metric is as follows: $$ ds^2 = e^{2A(r)} dt^2 - e^{2B(r)} dr^2 - 2(r^2 +\sin^2\theta) (d\phi^2 + \sin^2\theta d\phi^2) $$

The Riemann tensor is given as: $$ R^a_{bcd} = \partial_c \Gamma^a_{bd} - \partial_d \Gamma^a_{bc} + \Gamma^a_{ce}\Gamma^e_{bd} - \Gamma^a_{de}\Gamma^e_{bc} $$

Here, $\Gamma^a_{bc}$ is the Christoffel symbol of the second kind defined as:

$$ \Gamma^a_{bc} = \frac{1}{2} g^{ad}(\partial_b g_{cd} + \partial_c g_{bd} - \partial_d g_{bc}) $$

In this problem, we need to calculate the 20 independent components of the Riemann tensor. First, let's calculate the Christoffel symbols of the second kind.

Here, $g_ {00} = e^{2A(r)}$, $g_ {11} = -e^{2B(r)} $, $g_ {22} = -(r^2 + \sin^2\theta) $, and $g_{33} = -(r^2 + \sin^2\theta) \sin^2\theta$.

Using these, we get:$$ \Gamma^0_{00} = A'(r)e^{2A(r)}$$$$ \Gamma^0_{11} = B'(r)e^{2B(r)}$$$$ \Gamma^1_{01} = A'(r)e^{2A(r)}$$$$ \Gamma^1_{11} = -B'(r)e^{2B(r)}$$$$ \Gamma^2_{22} = -r(r^2 + \sin^2\theta)^{-1}$$$$ \Gamma^3_{33} = -\sin^2\theta(r^2 + \sin^2\theta)^{-1}$$$$ \Gamma^2_{33} = \cos\theta\sin\theta(r^2 + \sin^2\theta)^{-1}$$$$ \Gamma^3_{32} = \Gamma^3_{23} = \cot\theta $$

Using these Christoffel symbols, we can now calculate the components of the Riemann tensor. There are a total of $4^4 = 256$ components of the Riemann tensor, but due to symmetry, only 20 of these are independent. Using the formula for the Riemann tensor, we get the following non-zero components:

$$ R^0_{101} = -A''(r)e^{2A(r)}$$$$ R^0_{202} = R^0_{303} = (r^2 + \sin^2\theta)(\sin^2\theta A'(r) + rA'(r))e^{2(A-B)}$$$$ R^1_{010} = -A''(r)e^{2A(r)}$$$$ R^1_{121} = -B''(r)e^{2B(r)}$$$$ R^2_{232} = r(r^2 + \sin^2\theta)^{-1}$$$$ R^3_{323} = \sin^2\theta(r^2 + \sin^2\theta)^{-1}$$$$ R^2_{323} = -\cos\theta\sin\theta(r^2 + \sin^2\theta)^{-1}$$$$ R^3_{322} = -\cos\theta\sin\theta(r^2 + \sin^2\theta)^{-1}$$$$ R^0_{121} = A'(r)B'(r)e^{2(A-B)}$$$$ R^1_{020} = A'(r)B'(r)e^{2(A-B)}$$$$ R^2_{303} = -\sin^2\theta A'(r)e^{2(A-B)}$$$$ R^3_{202} = -rA'(r)e^{2(A-B)}$$$$ R^0_{202} = (r^2 + \sin^2\theta)\sin^2\theta A'(r)e^{2(A-B)}$$$$ R^0_{303} = (r^2 + \sin^2\theta)A'(r)e^{2(A-B)}$$$$ R^1_{010} = A''(r)e^{2(A-B)}$$$$ R^1_{121} = B''(r)e^{2(A-B)}$$$$ R^2_{232} = r(r^2 + \sin^2\theta)^{-1}$$$$ R^3_{323} = \sin^2\theta(r^2 + \sin^2\theta)^{-1}$$

Therefore, these are the 20 independent components of the Riemann tensor.

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The expression 10/3 can be expressed as a percent by multiplying it by 100. The result is approximately 333.33%.

To express a fraction as a percent, we need to convert it into a decimal and then multiply by 100 to get the percentage representation. In this case, we have 10/3 as the fraction.

To convert the fraction 10/3 to a decimal, we divide 10 by 3, which gives us approximately 3.3333. To express this decimal as a percentage, we multiply it by 100. Thus, 3.3333 * 100 = 333.33%.

Therefore, the expression 10/3 can be expressed as approximately 333.33% when converted to a percentage.

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Step-by-step explanation:

This is an odd question  (do we have all of the info??)....I had to make an assumption...

Well..... you will not get a numerical answer...it is a quadratic equation

area = (x+5) ft  (3x+6) ft         (I assumed one was length and one was width)

area =   (3x^2 +21x + 30)     ft^2

It is important to carefully review the calculations and ensure the data has been entered correctly. Double-checking the formulas and verifying the input values will help identify any mistakes and provide an accurate interpretation of the F statistic.

If you calculate an F statistic and find that it is negative, it is highly likely that a calculation error has occurred. The F statistic is a measure of the ratio of variances, specifically the ratio of the between-groups variance to the within-groups variance. The F statistic is always expected to be positive, as it represents the difference among group means relative to the variation within the groups.

A negative F statistic contradicts the fundamental nature of the statistic, as it implies that the between-groups variance is smaller than the within-groups variance, suggesting that the difference among group means is less than what would have occurred by chance. This scenario is highly unlikely and indicates that an error has been made during the calculation or data entry process.

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Answer:

(5,354 + x)

or

536.4*x

Step-by-step explanation:

We know that x = 10.

Now we want to write an expression (in terms of x) for the number 5,364.

This could be really trivial, remember that x = 10.

Then:  (x - 10) = 0

And if we add zero to a number, the result is the same number, then if we add this to 5,364 the number does not change.

5,364 = 5,364 + (x - 10) = 5,364 + x - 10

5,364 = 5,354 + x

So (5,354 + x) is a expression for the number 5,364 in terms of x.

Of course, this is a really simple example, we could do a more complex case if we know that:

x/10 = 1

And the product between any real number and 1 is the same number.

Then:

(5,364)*(x/10) = 5,364

(5,364/10)*x = 5,364

536.4*x = 5,364

So we just found another expression for the number 5,364 in terms of x.

The Wronskian formula is given by:$$W(y_1,y_2)=\begin {vmatrix} y_1 & y_2 \\ y_1' & y_2' \end{vmatrix}$$To prove the Wronskian formula of two functions, let $y_1$ and $y_2$ be two non-zero solutions of the differential equation $y'' + p(x)y' + q(x)y = 0$.

Then the Wronskian of these two functions is given by: $W(y_1,y_2)=\begin{vmatrix} y_1 & y_2 \\ y_1' & y_2' \end{vmatrix}=Ce^{-\int p(x)dx}$ where $C$ is a constant that depends on $y_1$ and $y_2$ but not on $x$.

Part (a) of the given Wronskian formulas is: $$W(2\sin v(x), J_v(x))=\begin{vmatrix} 2\sin v(x) & J_v(x) \\ 2v\cos v(x) & J_v'(x) \end{vmatrix}=2\sin v(x)J_v'(x)-2v\cos v(x)J_v(x)$$

Note that this formula is almost the same as the standard Wronskian formula, but with the constant $C$ replaced by $2\sin v(x)$.

We can verify that this is indeed a valid Wronskian by taking the derivative with respect to $x$:$$\frac{d}{dx}[2\sin v(x)J_v'(x)-2v\cos v(x)J_v(x)]=2\cos v(x)J_v'(x)-2\sin v(x)[vJ_v(x)+J_v'(x)]=0$$

The last step follows from the differential equation satisfied by the Bessel functions: $x^2y''+xy'+(x^2-v^2)y=0$

Part (b) of the given Wronskian formulas is: $$W(Y_\nu(x),Y_{\nu+1}(x))=\begin{vmatrix} Y_\nu(x) & Y_{\nu+1}(x) \\ Y_\nu'(x) & Y_{\nu+1}'(x) \end{vmatrix}=W_0Y_{\nu+1}(x)-W_1Y_\nu(x)$$where $W_0$ and $W_1$ are constants that depend on $\nu$ but not on $x$. This formula is also a valid Wronskian, since we can verify that its derivative with respect to $x$ is zero:

$$\frac{d}{dx}[W_0Y_{\nu+1}(x)-W_1Y_\nu(x)]=W_0Y_{\nu+1}'(x)-W_1Y_\nu'(x)=0$$

This follows from the recurrence relations satisfied by the Bessel functions:$Y_{\nu-1}'(x)-\frac{\nu}{x}Y_{\nu-1}(x)+\frac{\nu+1}{x}Y_{\nu+1}(x)=0$ $Y_{\nu+1}'(x)-\frac{\nu+1}{x}Y_{\nu+1}(x)+\frac{\nu+2}{x}Y_{\nu+2}(x)=0$

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