Consider the following third-order IVP: Ty''(t) + y"(t) – (1 – 2y (t) 2)y'(t) + y(t) =0 y(0)=1, y'(0)=1, y''(0)=1, where T=-1. Use the midpoint method with a step size of h=0.1 to estimate the value of y(0.1) + 2y'(0.1) + 3y" (0.1), writing your answer to three decimal places.

Answer: 14.6

Step-by-step explanation:

I made a square around the triangle which I then counted the squares, found the Pythagorean theorem, and then added the missing sides together

Answer:

Yes. See explanation below.

Step-by-step explanation:

The central angle and the degree arc measure of a sector of a circle are  equal. Doubling the arc measure, doubles the central angle measure and vice versa.

Area of the the original sector:

[tex] A_{sector} = \dfrac{n}{360^\circ}\pi r^2 [/tex]

where n = measure of the central angle of the sector

Since the central angle and the arc measure of the sector are equal, changing the arc measure has the same effect as changing the central angle measure.

Let's double the central angle to 2n which is the same as doubling the arc measure.

Area of the sector with a doubled central angle or a doubled arc measure:

[tex] A_{sector} = \dfrac{2n}{360^\circ}\pi r^2 [/tex]

Now we divide the area of the doubled sector by the area of the original sector.

[tex]\dfrac{\frac{2n}{360^\circ}\pi r^2}{\frac{n}{360^\circ}\pi r^2} =[/tex]

Simplify:

[tex]= \dfrac{2n}{n} \times \dfrac{360^\circ \pi r^2}{360^\circ \pi r^2}[/tex]

[tex] = \dfrac{2n}{n} [/tex]

[tex] = 2 [/tex]

The ratio of the areas is 2, so the area of the sector is indeed doubled.

Answer: Yes.

Answer:

x = -4 and 2

Step-by-step explanation:

When x = -4 and 2, y = 0 so -4 and 2 are the roots

Answer:

Question A)

[tex]=\sqrt{6}x[/tex]

Question B)

[tex]=3a\sqrt{a}[/tex]

Question C)

[tex]=5\sqrt{2}b^2[/tex]

Step-by-step explanation:

A)

We are given:

[tex]\sqrt{6x^2}\, \text{ where } x\geq 0[/tex]

We can rewrite the expression:

[tex]=\sqrt{6}\cdot \sqrt{x^2}[/tex]

The square root and square will cancel each other out. Thus:

[tex]=\sqrt{6}x[/tex]

B)

We are given:

[tex]\sqrt{9a^3}[/tex]

Rewrite:

[tex]=\sqrt{9}\cdot \sqrt{a^3}[/tex]

Note that the square root of 9 is simply 3. We can also factor the second part:

[tex]=3\cdot \sqrt{a^2\cdot a}[/tex]

Rewriting:

[tex]=3\cdot\sqrt{a^2}\cdot\sqrt{a}[/tex]

Simplify:

[tex]=3a\sqrt{a}[/tex]

C)

We are given:

[tex]\sqrt{50b^4}[/tex]

Rewrite. Note that 50 = 25(2):

[tex]=\sqrt{25}\cdot \sqrt{2}\cdot \sqrt{b^4}[/tex]

Simplify. We can rewrite the factor as:

[tex]=5\cdot \sqrt{2}\cdot \sqrt{(b^2)^2}[/tex]

The square and square root will cancel out. Thus:

[tex]=5\sqrt{2}b^2[/tex]

Answer:

$11.25

Step-by-step explanation:

55.50 divided by 6 = 9.25

9.25 + 2 = 11.25

Answer:

y=13 degrees

Step-by-step explanation:

This is an isosceles triangle, we know this because NO and NM are equal.

In an isosceles triangle, the base angles are congruent. In this case, they are angle NOM and angle NMO.

We also know that the sum of the interior angles of a triangle are equal to 180.

With this information, we can make an equation by gathering all the interior angles:

8y+2(3y-1)=180

Solve for y.

8y+6y-2=180

14y-2=180

14y=182

y=13

To solve the given initial value problem using the method of Laplace transforms, we'll take the Laplace transform of both sides of the differential equation. Let's denote the Laplace transform of the function y(t) as Y(s).

The Laplace transform of the second derivative y" is s²Y(s) - sy(0) - y'(0), where y(0) and y'(0) are the initial conditions given.

The Laplace transform of the first derivative y' is sY(s) - y(0).

The Laplace transform of the term 6y is 6Y(s).

The Laplace transform of the term -24e^t can be found using the table of Laplace transforms.

Applying the Laplace transform to the entire differential equation, we get:

s²Y(s) - sy(0) - y'(0) + 5(sY(s) - y(0)) + 6Y(s) - 24/(s-1) = 0

Substituting the initial conditions y(0) = -5 and y'(0) = -19, we have:

s²Y(s) + 5sY(s) + 6Y(s) - 5s + 19 - 24/(s-1) = 0

Now, we can solve this equation for Y(s). Once we find Y(s), we can take the inverse Laplace transform to obtain y(t), the solution to the initial value problem.

Since the given question doesn't specify a particular form for Y(s), I'm unable to provide the exact solution y(t) in terms of e.

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Answer:

A

Step-by-step explanation:

Answer:

A.

0.5

Step-by-step explanation:

Answer:

1m²

Step-by-step explanation:

Answer:

A = 157.3 units²

Step-by-step explanation:

A = 1/2(6.9)(8 x 5.7) = 157.3 units²

Answer:

To understand their poems better

Step-by-step explanation:

First you have a general knowledge on why the author made the poem

Second you now have more info to use in order to understand the poems

Third is the same reason why there are people in history books.

Using cylindrical coordinates, the volume of the solid region bounded on the top by the paraboloid z = 12 − x^2 − y^2 and bounded on the bottom by the cone z = x^2 y^2 can be found. The explanation below provides the step-by-step process.

In cylindrical coordinates, we can express the paraboloid and the cone equations as follows:

Paraboloid: z = 12 -[tex]r^2[/tex]

Cone: z = [tex]r^2 cos^2(θ) sin^2(θ)[/tex]

To find the volume of the solid region, we need to determine the limits of integration. The region is bounded by the paraboloid on top and the cone on the bottom. The paraboloid and the cone intersect when 12 - [tex]r^2[/tex] = [tex]r^2 cos^2(θ) sin^2(θ)[/tex]. Simplifying this equation, we get 12 = [tex]r^2[/tex](1 - [tex]cos^2(θ)[/tex] [tex]sin^2(θ[/tex])). Since r is always non-negative, we can rewrite the equation as 12 =[tex]r^2[/tex][tex]sin^2(θ) (1 - sin^2(θ)[/tex]). This equation defines the boundary curve in the polar coordinate plane (r, θ).

To determine the limits of integration for r, we need to find the values of r that satisfy the equation above for each θ. For a fixed value of θ, the equation becomes 12 = [tex]r^2 sin^2(θ) (1 - sin^2(θ))[/tex]. This equation represents a circle with radius [tex]\sqrt(12 sin^2(θ) (1 - sin^2(θ)))[/tex]. Thus, the limits for r are 0 and [tex]\sqrt(12 sin^2(θ) (1 - sin^2(θ)))[/tex].

For the limits of integration for θ, we need to consider the range in which the paraboloid and the cone intersect. The cone is defined in the range 0 ≤ θ ≤ π, and the paraboloid intersects the cone when θ satisfies 12 = [tex]r^2 sin^2(θ) (1 - sin^2(θ))[/tex]. By solving this equation, we find that 0 ≤ θ ≤ π/2.

To calculate the volume, we integrate over the cylindrical coordinates as follows:

V = ∫∫∫ dV

= ∫[0,π/2]∫[0,√[tex](12sin^2(θ)(1-sin^2(θ)))]∫[r^2cos^2(θ)sin^2(θ),12-r^2][/tex] r dz dr dθ

Evaluating this triple integral will yield the volume of the solid region bounded by the given surfaces.

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Answer:

c for both

Step-by-step explanation:

9514 1404 393

Answer:

  AB = [[-6, -1][-4, 6][-15, 10]]

Step-by-step explanation:

Any of a number of on-line, spreadsheet, or calculator tools will find the matrix product for you.

The input and output of one such tool is shown below.

__

As you know, each term in the product matrix is the sum of products of a row in the left matrix and a column in the right matrix. The coordinates of that row and column are the coordinates of the result in the product matrix.

For example, row 2, column 1 of the product matrix is the sum of products ...

  (4)(-3) +(-2)(-4) = -12 +8 = -4 . . . . row 2, column 1 of the result

Answer:

i just got it right.

Step-by-step explanation:

Answer:

the unit rate of pesos to dollars is 1 MXN = 0.04960 USD

Step-by-step explanation:

Quick Conversions from Mexican Peso to United States Dollar : 1 MXN = 0.04960 USD

$ or MEX$ 10 $, US$ 0.50

$ or MEX$ 50 $, US$ 2.48

$ or MEX$ 100 $, US$ 4.96

$ or MEX$ 250 $, US$ 12.40

Answer:

61 ft

Step-by-step explanation:

since it's equal

u look cute in that pfp

Answer:

y=3x+6

Step-by-step explanation:

Just test this on like a graph or something and you will see it works. Let me know if there is a fault in my answer. Thanks! Have a good day.

Answer:

The inequality that can be used to determine the number of guitar lessons Gale could take using her birthday money is:

[tex]25x + 50 \leqslant 200[/tex]

The solution to the inequality is:

[tex]x \leqslant 6[/tex]

She can take at most, 6 guitar lessons

Step-by-step explanation:

Total amount Gale received is $200

For rental equipment, she paid $50

Each guitar lesson will cost $25

Let x represent the number of lessons she could take, then the total amount she will spend must bot exceed $200

The guitar lesson will cost a total of 25x

So, her total spending will be:

[tex]25x + 50[/tex]

This must be at most $200

[tex]25x + 50 \leqslant 20[/tex]

Solving the above inequality:

[tex]25x \leqslant 200 - 50 = 150[/tex]

[tex]x \leqslant \frac{150}{25} = 6[/tex]

Answer:

a = √39 (exact)

a = 6.24 (dec.)

Step-by-step explanation:

a^2 + b^2 = c^2

a^2 + 5^2 = 8^2

a^2 + 25 = 64

a^2 = 39

a = √39 (exact)

a = 6.24 (dec.)

To solve the equation 7 sin(2x) = 6 for the two smallest positive solutions A and B, we can use algebraic techniques and trigonometric properties.

The solutions A and B are approximately equal to A ≈ 0.287 and B ≈ 1.569, respectively.

To explain the solution, let's begin by rearranging the equation: sin(2x) = 6/7. Since the range of the sine function is between -1 and 1, the equation has solutions only if 6/7 is within this range. We can find the corresponding angles by taking the inverse sine (arcsin) of 6/7. Using a calculator, we find that the arcsin(6/7) is approximately 0.942.

However, this gives us only one of the solutions. To find the other solution, we can use the periodicity of the sine function. We know that sin(θ) = sin(π - θ), where θ is the angle in radians. Therefore, the second solution is π - 0.942, which is approximately 2.199. However, since we're looking for the smallest positive solutions, we need to consider only the values between 0 and 2π. Thus, the two smallest positive solutions are A ≈ 0.287 and B ≈ 1.569.

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Answer:

I think the answers would be :

a . 4/3

b. 1/5

c . - 5/4

hope it helps u ^^

Answer:

A. 4x¹² + 9x⁷ + 3x³ -x

Step-by-step explanation:

Hi!

==================================================================

To write a polynomial in descending order, we write the terms with higher degrees, or exponents, first.

3x³ + 9x⁷ -x + 4x¹²

4x¹² has the highest degree, so it is written as the first term.

⇒4x¹²

9x⁷ has the next highest degree, so it is written next.

⇒4x¹² + 9x⁷

3x³ has the next highest degree, so it is written next.

⇒4x¹² + 9x⁷ + 3x³

-x has the lowest degree, so it is written last.

⇒4x¹² + 9x⁷ + 3x³ -x

4x¹² + 9x⁷ + 3x³ -x

==================================================================

Hope I Helped, Feel free to ask any questions to clarify :)

Have a great day!

       -Aadi x

Step-by-step explanation:

the difference is the previous term multiplied by 4 to get the next term

2,8,32,128,512,2048,8192

Hope that helps :)

Answer:

a2 + 2al (or) a2 + √a24+h2 a 2 4 + h 2

Step-by-step explanation:

Answer:

SA=L²+2L√(L²/4+H²)

Step-by-step explanation:

Where H is the height and L is the length

we have shown that if [tex]$a_1, a_2, \ldots, a_n$[/tex] are n distinct real numbers, exactly [tex]$n-1$[/tex] multiplications are used to compute their product, regardless of how parentheses are inserted into the product.

What is the principle of mathematical induction?

The principle of mathematical induction is a powerful proof technique used to establish the validity of an infinite sequence of statements.

To prove that exactly [tex]$n-1$[/tex] multiplications are used to compute the product of n distinct real numbers, regardless of how parentheses are inserted into their product, we will use the principle of mathematical induction.

[tex]\textbf{Base Case:}[/tex]

For [tex]$n=2$[/tex], we have two distinct real numbers [tex]a_1$ and $a_2$.[/tex] The product is simply [tex]a_1 \cdot a_2$,[/tex] which requires only one multiplication. Thus, the base case holds true.

[tex]\textbf{Inductive Step:}[/tex]

Assume the statement holds true for [tex]$n=k$[/tex], where [tex]k \geq 2$.[/tex] That is, when multiplying k distinct real numbers, exactly [tex]$k-1$[/tex] multiplications are used.

Now, consider the case for [tex]$n=k+1$[/tex], where we have [tex]$k+1$[/tex] distinct real numbers [tex]$a_1, a_2, \ldots, a_{k+1}$[/tex]. The product can be computed by multiplying [tex]$a_1$[/tex] with the product of the remaining k numbers, which can be denoted as [tex]$(a_2 \cdot a_3 \cdot \ldots \cdot a_{k+1})$[/tex].

By our induction hypothesis, computing the product of k distinct real numbers requires [tex]$k-1$[/tex] multiplications. Therefore, multiplying[tex]$a_1$[/tex] with the product of the remaining [tex]$k$[/tex] numbers requires an additional multiplication, resulting in a total of k multiplications.

Hence, we have shown that if [tex]a_1, a_2, \ldots, a_n$ are $n$[/tex] distinct real numbers, exactly [tex]$n-1$[/tex] multiplications are used to compute their product, regardless of how parentheses are inserted into the product.

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The integral for the area becomes:

A = ∫₀ᵃʳᶜᶜᵒˢ(ᵃ/₂) ∫ₐ(₁+ᶜᵒˢ(θ))³ᵃᶜᵒˢ(θ) r dr dθ

To find the area inside the circle r = 3acos(θ) and outside the cardioid r = a(1 + cos(θ)), we can set up a double integral in polar coordinates.

First, let's find the points of intersection between the two curves. The circle r = 3acos(θ) and the cardioid r = a(1 + cos(θ)) intersect when:

3acos(θ) = a(1 + cos(θ))

Simplifying, we get:

3acos(θ) - a(1 + cos(θ)) = 0

2acos(θ) - a = 0

acos(θ) = a/2

θ = arccos(a/2)

Now, let's set up the integral. We want to find the area inside the circle and outside the cardioid, so the region of integration is defined by:

0 ≤ θ ≤ arccos(a/2)

a(1 + cos(θ)) ≤ r ≤ 3acos(θ)

Evaluating this double integral will give us the desired area inside the circle and outside the cardioid.

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Answer:

Vince is not correct because he should have interpreted the solution as having to save a minimum of $103.55.

Step-by-step explanation:my

my sister which is in college helped me with one.

Answer: its D

Step-by-step explanation:

To find a function that satisfies the given criteria, we can start by determining the requirements for its derivative, f'(x).

Let's break down the given properties and find the corresponding requirements for f'(x): f(x) is decreasing at x = -6: This means that the slope of the function should be negative at x = -6. Therefore, f'(-6) < 0. f(x) has a local minimum at x = -3: At a local minimum, the slope changes from negative to positive. Thus, f'(-3) = 0. f(x) has a local maximum at x = 3: At a local maximum, the slope changes from positive to negative. Hence, f'(3) = 0.

Now, let's integrate f'(x) to obtain f(x): Integrating f'(x) = -6 < x < -3 will give us a decreasing function on that interval. Integrating f'(x) = -3 < x < 3 will give us an increasing function on that interval. Integrating f'(x) = 3 < x < 6 will give us a decreasing function on that interval. To simplify the process, let's assume that f'(x) is a quadratic function with roots at -6, -3, and 3. We can represent it as: f'(x) = k(x + 6)(x + 3)(x - 3), where k is a constant that affects the steepness of the curve. By setting f'(-3) = 0, we find that k = -1/18.

Therefore, f'(x) = -1/18(x + 6)(x + 3)(x - 3). Integrating f'(x) will give us f(x): f(x) = ∫[-6,x] -1/18(t + 6)(t + 3)(t - 3) dt. Evaluating this integral is a bit complicated. Let's denote F(x) as the antiderivative of f(x): F(x) = ∫[-6,x] -1/18(t + 6)(t + 3)(t - 3) dt. Now, we can find f(x) by differentiating F(x): f(x) = d/dx[F(x)]. To get an explicit equation for f(x), we need to calculate the integral and differentiate the resulting antiderivative. Once you have the equation for f(x), you can plot it on the provided graphing option to verify that it matches the criteria mentioned in the question.

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