can someone help pls!!!!!!!!!!!!!

The solution as 75 + 34√5 while solving (5+2√5)(7+4 √5).

To get the product of the given two binomials, (5+2√5) and (7+4√5), use FOIL multiplication method. Here, F stands for First terms, O for Outer terms, I for Inner terms, and L for Last terms. Then simplify the expression. The solution is shown below:

First, multiply the first terms together which give: (5)(7) = 35.

Second, multiply the outer terms together which give: (5)(4 √5) = 20√5.

Third, multiply the inner terms together which give: (2√5)(7) = 14√5.

Finally, multiply the last terms together which give: (2√5)(4√5) = 40.

When all the products are added together, we get; 35 + 20√5 + 14√5 + 40 = 75 + 34√5

Therefore, (5+2√5)(7+4√5) = 75 + 34√5.

Thus, we got the solution as 75 + 34√5 while solving (5+2√5)(7+4 √5).

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The exact area A of the region bounded between the curve f(x) = 1 - ln(x) and the line g(x) = 1 over the interval [1, 5] is given by:

A = -5ln(5) + 5 units²

To find the area A of the region bounded between the curve f(x) = 1 - ln(x) and the line g(x) = 1 over the interval [1, 5], we can integrate the difference between the two functions over that interval.

A = ∫[1, 5] (f(x) - g(x)) dx

First, let's find the difference between the two functions:

f(x) - g(x) = (1 - ln(x)) - 1 = -ln(x)

Now, we can integrate -ln(x) over the interval [1, 5]:

A = ∫[1, 5] -ln(x) dx

To integrate -ln(x), we can use the properties of logarithmic functions:

A = [-xln(x) + x] evaluated from 1 to 5

A = [-5ln(5) + 5] - [-1ln(1) + 1]

Since ln(1) = 0, the second term on the right side becomes 0:

A = -5ln(5) + 5

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The solution to the equation y = - 3tan(½ × x) is 3 sec y' (½ x)²/2

y = - 3tan(½ × x)

Take the derivative

y' = d/dx (- 3tan(½ × x))

Rewrite

y' = d/dx (- 3tan(½ × x))

Use differentiation rules

y' = - 3x × d/dx (tan(½ × x))

Use differentiation rules

y' = - 3 × d/dg (tan(g)) × d/dx (½ × x)

Differentiate

y' = -3 sec (g )² X ½

Substitute back

2 y' = -3sec (½x)² x ½

Calculate

Solution

3 sec y' (½ x)²/2

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Answer:

Step-by-step explanation:

The simplified form of the expression in cube root is 4a^(8/3).

To simplify the radical expression ³√64a⁸¹, we can break it down into its prime factors and simplify each factor separately.

First, let's simplify the number inside the radical, which is 64. We can write it as 2^6, since 2 multiplied by itself 6 times equals 64.

Next, let's simplify the variable inside the radical, which is a^8.

Since we are taking the cube root, we need to find the largest factor of 8 that is a perfect cube. In this case, 2^3 is the largest perfect cube factor of 8.

So, we can rewrite the expression as ³√(2^6 * 2^3 * a).

Using the property of radicals that says ³√(a * b) = ³√a * ³√b, we can simplify further.

³√(2^6 * 2^3 * a) = ²√(2^6) * ³√(2^3) * ³√a

Since ²√(2^6) is 2^3 and ³√(2^3) is 2, we can simplify even more.

2^3 * 2 * ³√a = 8 * 2 * ³√a = 16 * ³√a

Therefore, the simplified radical expression ³√64a⁸¹ is equal to 16 * ³√a.

In summary, to simplify the expression ³√64a⁸¹, we first broke down the number 64 into its prime factors and found the largest perfect cube factor of the exponent 8.

We then used the property of radicals to simplify the expression and arrived at the final answer of 16 * ³√a.

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Answer:

3.(52+81).

Step-by-step explanation:

Hello,

Answer:

[tex]\red{\large{\boxed{156+243 =3(52+81)}}}[/tex]

For a polynomial f(x) with a positive leading coefficient, it can be shown that there exists a value N such that f(x) is always greater than zero for all x greater than N.

Consider the polynomial f(x) = anx^k + ... + a₁x + ao, where an is the leading coefficient and k is the degree of the polynomial. Since an > 0, the polynomial has a positive leading coefficient.

To show that there exists a value N such that f(x) > 0 for all x > N, we need to prove that as x approaches infinity, f(x) also approaches infinity. This can be done by considering the highest degree term in the polynomial, anx^k, as x becomes large.

Since an > 0 and x^k dominates the other terms for large x, the polynomial f(x) becomes dominated by the term anx^k. As x increases, the term anx^k becomes arbitrarily large and positive, ensuring that f(x) also becomes arbitrarily large and positive.

Therefore, by choosing a sufficiently large value N, we can guarantee that f(x) > 0 for all x > N, as the polynomial grows without bound as x approaches infinity.

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The solution to the equation 4x + [3 -7 9] = [-3 11 5 -7] is x = [-3/2 9/2 -1 -7/4].

To solve the equation 4x + [3 -7 9] = [-3 11 5 -7], we need to isolate the variable x.

Given:

4x + [3 -7 9] = [-3 11 5 -7]

First, let's subtract [3 -7 9] from both sides of the equation:

4x + [3 -7 9] - [3 -7 9] = [-3 11 5 -7] - [3 -7 9]

This simplifies to:

4x = [-3 11 5 -7] - [3 -7 9]

Subtracting the corresponding elements, we have:

4x = [-3-3 11-(-7) 5-9 -7]

Simplifying further:

4x = [-6 18 -4 -7]

Now, divide both sides of the equation by 4 to solve for x:

4x/4 = [-6 18 -4 -7]/4

This gives us:

x = [-6/4 18/4 -4/4 -7/4]

Simplifying the fractions:

x = [-3/2 9/2 -1 -7/4]

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The system of equations represented by the given matrix is:

2x + y + z = 1

x + y + z = 1

x - y + z = -1

x - 2y = -2

To interpret the given matrix as a system of equations, we need to organize the elements of the matrix into a coefficient matrix and a constant matrix.

The coefficient matrix is obtained by taking the coefficients of the variables in each equation and arranging them in a matrix form:

2 1 1

1 1 1

1 -1 1

1 -2 0

The constant matrix is obtained by taking the constants on the right-hand side of each equation and arranging them in a matrix form:

1

1

-1

-2

By combining the coefficient matrix and the constant matrix, we can write the system of equations:

2x + y + z = 1

x + y + z = 1

x - y + z = -1

x - 2y + 0z = -2

Here, x, y, and z represent variables, and the numbers on the right-hand side represent the constants in the equations.

The system of equations can be solved using various methods, such as substitution, elimination, or matrix operations.

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Answer:

[tex]\displaystyle{f^{-1}(x)=\dfrac{x}{7}-\dfrac{3}{7}}[/tex]

Step-by-step explanation:

Swap f(x) and x position of the function, thus:

[tex]\displaystyle{x=7f(x)+3}[/tex]

Then solve for f(x), subtract 3 both sides and then divide both by 7:

[tex]\displaystyle{x-3=7f(x)}\\\\\displaystyle{\dfrac{x}{7}-\dfrac{3}{7}=f(x)}[/tex]

Since the function has been inverted, therefore:

[tex]\displaystyle{f^{-1}(x)=\dfrac{x}{7}-\dfrac{3}{7}}[/tex]

And we can prove the answer by substituting x = 1 in f(x) which results in:

[tex]\displaystyle{f(1)=7(1)+3 = 10}[/tex]

The output is 10, now invert the process by substituting x = 10 in [tex]f^{-1}(x)[/tex]:

[tex]\displaystyle{f^{-1}(10)=\dfrac{10}{7}-\dfrac{3}{7}}\\\\\displaystyle{f^{-1}(10)=\dfrac{7}{7}=1}[/tex]

The input is 1. Hence, the solution is true.

The sample space for a roll of a 5-sided die is {1, 2, 3, 4, 5}.

In probability theory, the sample space refers to the set of all possible outcomes of an experiment. In this case, we are rolling a 5-sided die, which means there are 5 possible outcomes. The outcomes are represented by the numbers 1, 2, 3, 4, and 5, as these are the numbers that can appear on the faces of the die. Thus, the sample space for this experiment can be expressed as {1, 2, 3, 4, 5}.

It is important to note that each outcome in the sample space is mutually exclusive, meaning that only one outcome can occur on a single roll of the die. Additionally, the outcomes are collectively exhaustive, as they encompass all the possible results of the experiment. By identifying the sample space, we can analyze and calculate probabilities associated with different events or combinations of outcomes.

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The expected number of absentees on a given day is 3.48

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

Number absent 0 1 2 3 4 5 6

Probability 0.02 0.04 0.15 0.29 0.3 0.13 0.07

The expected number of absentees on a given day is calculated as

E(x) = ∑xP(x)

So, we have

E(x) = 0 * 0.02 + 1 * 0.04 + 2 * 0.15 + 3 * 0.29 + 4 * 0.3 + 5 * 0.13 + 6 * 0.07

Evaluate

E(x) = 3.48

Hence, the expected number is 3.48

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The total amount accumulated at the end of 10 years is approximately $1268.76. Hence, the amount accumulated is $1268.76.

Principal deposited (P): $600

Annual interest rate (r): 5.6%

Number of times interest compounded per year (n): 4

Time in years (t): 10

To find: The total amount accumulated at the end of 10 years.

Solution:

We will use the compound interest formula:

A = P * (1 + r/n)^(nt)

Substituting the given values:

A = 600 * (1 + 0.056/4)^(4 * 10)

Simplifying the expression:

A = 600 * (1.014)^40

Calculating the value:

A ≈ 600 * 2.1146

A ≈ 1268.76

Therefore, , the total money amassed after ten years is around $1268.76.

As a result, the total sum accumulated is $1268.76.

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The tax paid on a grocery bill of $147.56, with a tax rate of 4%, amounts to $5.90.

To calculate this, we multiply the total amount of the bill ($147.56) by the tax rate (4% expressed as 0.04). This gives us the tax amount: $147.56 * 0.04 = $5.90.

Tax amount = Bill amount * Tax rate

In this case, the bill amount is $147.56 and the tax rate is 4% (or 0.04).

Tax amount = $147.56 * 0.04 = $5.90

Therefore, the tax paid on the grocery bill is $5.90.

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a. There are 10 choices for each digit and 26 choices for each letter, so the number of different license plates possible without repetition is 10 * 10 * 10 * 10 * 26 * 26 * 26 * 26 = 456,976,000.

b. With repetition allowed, there are still 10 choices for each digit and 26 choices for each letter. Since repetition is permitted, each digit and letter can be chosen independently, so the total number of different license plates possible is 10^4 * 26^4 = 45,697,600.

In part (a), where repetition is not allowed, we consider each position on the license plate separately. For the four digits, there are 10 choices (0-9) for each position. Similarly, for the four letters, there are 26 choices (A-Z) for each position. Therefore, we multiply the number of choices for each position to find the total number of different license plates possible without repetition.

In part (b), where repetition is permitted, the choices for each position are still the same. However, since repetition is allowed, each position can independently have any of the 10 digits or any of the 26 letters. We raise the number of choices for each position to the power of the number of positions to find the total number of different license plates possible.

It's important to note that the above calculations assume that the order of the digits and letters on the license plate matters. If the order does not matter, such as when considering combinations instead of permutations, the number of possible license plates would be different.

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(t+s)(3) = 16.Given the functions as follows:

s(x)=4x+2     t(x)=x+1

We are to find the expressions for (t⋅s)(x) and (t−s)(x) and evaluate (t+s)(3).

(t.s)(x) = t(x)·s(x)

= (x+1)(4x+2)

= 4x² + 6x + 2

(t-s)(x) = t(x) - s(x)

= (x+1) - (4x+2)

= -3x -1(t+s)(3)

= t(3) + s(3)

= (3+1) + (4(3)+2)

= 16

Therefore, (t.s)(x) = 4x² + 6x + 2,

(t-s)(x) = -3x -1, and (t+s)(3) = 16.

Explanation:

To find (t.s)(x), we need to perform the following operations:

We substitute s(x) = 4x + 2 and t(x) = x + 1 to (t.s)(x) = t(x)·s(x) (x+1)(4x+2) = 4x² + 6x + 2

Therefore, (t.s)(x) = 4x² + 6x + 2

To find (t-s)(x), we need to perform the following operations:

We substitute s(x) = 4x + 2 and t(x) = x + 1 to

(t-s)(x) = t(x) - s(x)(x+1) - (4x+2)

= -3x -1

Therefore, (t-s)(x) = -3x -1

To find (t+s)(3), we need to perform the following operations:

We substitute

s(3) = 4(3) + 2

= 14 and

t(3) = 3 + 1

= 4 in

(t+s)(3) = t(3) + s(3)4 + 14

= 16

Therefore, (t+s)(3) = 16.

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By solving the system of equations and checking the solutions, we can determine if S is linearly independent and if it spans P₂.

a) To determine if the function f(t) = t² - t - 1 is in the span of S = {f₁, f₂, f₃}, we need to check if we can find scalars a, b, and c such that f(t) = af₁(t) + bf₂(t) + cf₃(t).

Let's set up the equation:

f(t) = a(f₁(t)) + b(f₂(t)) + c(f₃(t))

f(t) = a(t² - 2t - 3) + b(t² - 4t - 2) + c(t² + 2t - 5)

f(t) = (a + b + c)t² + (-2a - 4b + 2c)t + (-3a - 2b - 5c)

For f(t) to be in the span of S, the coefficients of t², t, and the constant term in the above equation should match the coefficients of t², t, and the constant term in f(t).

Comparing the coefficients, we get the following system of equations:

a + b + c = 1

-2a - 4b + 2c = -1

-3a - 2b - 5c = -1

By solving this system of equations, we can find the values of a, b, and c. If a solution exists, then f(t) is in the span of S.

b) To determine if S = {f₁, f₂, f₃} is a set of linearly independent functions, we need to check if the only solution to the equation a₁f₁(t) + a₂f₂(t) + a₃f₃(t) = 0 is when a₁ = a₂ = a₃ = 0.

Let's set up the equation:

a₁f₁(t) + a₂f₂(t) + a₃f₃(t) = 0

a₁(t² - 2t - 3) + a₂(t² - 4t - 2) + a₃(t² + 2t - 5) = 0

(a₁ + a₂ + a₃)t² + (-2a₁ - 4a₂ + 2a₃)t + (-3a₁ - 2a₂ - 5a₃) = 0

For S to be linearly independent, the only solution to the above equation should be a₁ = a₂ = a₃ = 0.

To check if S spans P₂, we need to see if every polynomial of degree 2 can be expressed as a linear combination of the functions in S. If the only solution to the equation a₁f₁(t) + a₂f₂(t) + a₃f₃(t) = p(t) is when a₁ = a₂ = a₃ = 0, then S spans P₂.

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The set S is a vector space.

To determine if S is a vector space, we need to check if it satisfies the ten properties of a vector space.

1. The zero vector exists: In this case, the zero vector would be the function y(x) = 0, which satisfies the differential equation y'' + 2y' - y = 0, since the derivative of the zero function is also zero.

2. Closure under addition: If f(x) and g(x) are both functions satisfying the differential equation y'' + 2y' - y = sin(x), then their sum h(x) = f(x) + g(x) also satisfies the same differential equation. This can be verified by taking the second derivative of h(x), multiplying by 2, and subtracting h(x) to check if it equals sin(x).

3. Closure under scalar multiplication: If f(x) is a function satisfying the differential equation y'' + 2y' - y = sin(x), and c is a scalar, then the function g(x) = c * f(x) also satisfies the same differential equation. This can be verified by taking the second derivative of g(x), multiplying by 2, and subtracting g(x) to check if it equals sin(x).

4. Associativity of addition: (f(x) + g(x)) + h(x) = f(x) + (g(x) + h(x))

5. Commutativity of addition: f(x) + g(x) = g(x) + f(x)

6. Additive identity: There exists a function 0(x) such that f(x) + 0(x) = f(x) for all functions f(x) satisfying the differential equation.

7. Additive inverse: For every function f(x) satisfying the differential equation, there exists a function -f(x) such that f(x) + (-f(x)) = 0(x).

8. Distributivity of scalar multiplication over vector addition: c * (f(x) + g(x)) = c * f(x) + c * g(x)

9. Distributivity of scalar multiplication over scalar addition: (c + d) * f(x) = c * f(x) + d * f(x)

10. Scalar multiplication identity: 1 * f(x) = f(x)

By verifying that all these properties hold, we can conclude that the set S is indeed a vector space.

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The resulting matrix T² represents the probabilities of individuals moving between intersections after two time steps.

To calculate T²,  to first understand what the matrix T represents. Let's define the matrix T:

T = | t11 t12 |

| t21 t22 |

In this context, T is a transition matrix that describes the movement of individuals between the four intersections: Owens Bridge on Parkway East (OE), Bay Bridge on Parkway East (BE), Bay Bridge on Parkway West (BW), and Owens Bridge on Parkway West (OW).

Each entry tij of the matrix T represents the probability of an individual moving from intersection i to intersection j. For example, t11 represents the probability of someone moving from Owens Bridge on Parkway East (OE) back to Owens Bridge on Parkway East (OE), t12 represents the probability of someone moving from Owens Bridge on Parkway East (OE) to Bay Bridge on Parkway East (BE), and so on.

The transition matrix T should be constructed based on the given information about the movement of individuals between these intersections. The entries should be probabilities, meaning they should be between 0 and 1, and the sum of each row should be equal to 1 since a person must move to one of the four intersections.

Once the matrix T is defined, calculating T² means multiplying T by itself:

T² = T × T

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Answer:

-24

Step-by-step explanation:

6 divided by -1/4

You can view this as a multiplication problem where you flip the second value.

6 * -4 = -24. This works for other examples as well.

For example, you can do 6 divided by -2/3, and when you flip the second value, you get 6 * -3/2, which gets you -18/2. which is -9.

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The given answer of i 0.53 is incorrect. The correct value is 2.

W(y₁, y₂)(3), we can use the Wronskian determinant formula.

W(y₁, y₂) = y₁y₂' - y₂y₁'

Let's first calculate the derivative of y₂:

y₂' = (d/dt)(y₂)

Next, we can substitute the given values into the formula to find

W(y₁, y₂)W(y₁, y₂)(1) = y₁(1)y₂'(1) - y₂(1)y₁'(1)

Since W(y₁, y₂)(1) is given as 2, we can set up the equation:

2 = y₁(1)y₂'(1) - y₂(1)y₁'(1)

Now, we need to find W(y₁, y₂)(3). To do this, we can use the fact that the Wronskian determinant is constant for linear homogeneous differential equations. Therefore, W(y₁, y₂)(3) will also be equal to 2.

So, W(y₁, y₂)(3) = 2.

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1.10, 2.2821 7.13 31 is an absolute pseudoprime.

1.) Fermat's little theorem states that for a prime number p and any integer a, a^(p-1) ≡ 1 (mod p). If we use p = 23, we get a^(22) ≡ 1 (mod 23).Now, we know that (10^k) ≡ (-1)^(k+1) (mod 11).

Therefore, we can split 1599999999 into 1500000000 + 99999999 = 15 * 10^8 + 99999999.Using the formula, 10^22 ≡ (-1)^23 (mod 23) => 10^22 ≡ -1 (mod 23) => 10^44 ≡ 1 (mod 23) => (10^22)^2 ≡ 1 (mod 23)

Also, 10^8 ≡ 1 (mod 23).

Therefore, we have 15 * (10^22)^8 * 10^8 + 99999999 ≡ 15 * 1 * 1 + 99999999 ≡ 10 (mod 23).

Hence, the remainder when 15999,999,999 is divided by 23 is 10.

2.)A positive integer n is an absolute pseudoprime to the base a if it is composite but satisfies the congruence a^(n-1) ≡ 1 (mod n).2821 7.13 31 => 2821 * 7 * 13 * 31.

Let's verify if 2821 is an absolute pseudoprime.2820 = 2^2 * 3 * 5 * 47

Let a = 2, then we need to verify that 2^2820 ≡ 1 (mod 2821)

Using the binary exponentiation method,

2^2 = 4, 2^4 = 16, 2^8 ≡ 256 (mod 2821), 2^16 ≡ 2323 (mod 2821), 2^32 ≡ 2223 (mod 2821), 2^64 ≡ 1 (mod 2821), 2^128 ≡ 1 (mod 2821), 2^256 ≡ 1 (mod 2821), 2^2816 ≡ 1 (mod 2821)

Therefore, 2^2820 ≡ (2^2816 * 2^4) ≡ (1 * 16) ≡ 1 (mod 2821)

Hence, 2821 is an absolute pseudoprime. Similarly, we can verify for 7, 13 and 31.

Therefore, 2821 7.13 31 is an absolute pseudoprime.

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The solution to the initial value problem is y(t)  = [tex]2e^(-5t) + 4e^(-17t)[/tex].

To solve the given initial value problem using the Laplace transform, we'll follow these steps:

Take the Laplace transform of both sides of the differential equation using the linearity property and the derivatives property of the Laplace transform.

Solve for the Laplace transform of the unknown function, denoted as Y(s).

Apply the initial conditions to find the values of the Laplace transform at s=0.

Inverse Laplace transform Y(s) to obtain the solution y(t).

Let's solve the initial value problem:

Step 1:

Taking the Laplace transform of the differential equation, we have:

s²Y(s) - sy(0) - y'(0) - 12(sY(s) - y(0)) + 85Y(s) = 0

Step 2:

Simplifying the equation and isolating Y(s), we get:

(s² + 12s + 85)Y(s) = s(6) + 58 + 12(6)

Y(s) = (6s + 130) / (s² + 12s + 85)

Step 3:

Applying the initial conditions, we have:

Y(0) = (6(0) + 130) / (0² + 12(0) + 85) = 130 / 85

Step 4:

Inverse Laplace transforming Y(s), we can use partial fraction decomposition or the table of Laplace transforms to find the inverse Laplace transform. In this case, we'll use partial fraction decomposition:

Y(s) = (6s + 130) / (s² + 12s + 85)

= (6s + 130) / [(s + 5)(s + 17)]

Using partial fraction decomposition, we can write:

Y(s) = A / (s + 5) + B / (s + 17)

Multiplying both sides by (s + 5)(s + 17), we get:

6s + 130 = A(s + 17) + B(s + 5)

Expanding and equating coefficients, we have:

6 = 17A + 5B

130 = 5A + 17B

Solving these equations simultaneously, we find A = 2 and B = 4.

Therefore, Y(s) = 2 / (s + 5) + 4 / (s + 17)

Taking the inverse Laplace transform

y(t) = [tex]2e^(-5t) + 4e^(-17t)[/tex].

So the solution to the initial value problem is y(t)  = [tex]2e^(-5t) + 4e^(-17t)[/tex].

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a) The graph originates at the origin( 0, 0) and spirals in exterior as θ increases. b) The graph have two loops centered at the origin. c) The graph is a cardioid. d) The  graph has bigger loop at origin and the innner loop inside it.. e) The graph is helical that starts at the point( 1, 0) and moves in inward direction towards the origin.

a) The function with polar equals is given by dy = θ/( 4π) with 0< θ< 4.

We've to find the crossroad points with θ = 0 and θ = π/ 2,

When θ = 0

dy = 0/( 4π) = 0

therefore, when θ = 0, the function intersects the origin( 0, 0).

Now, θ = π/ 2

dy = ( π/ 2)/( 4π) = 1/( 8)

thus, when θ = π/ 2, the polar function intersects the y- axis at( 0,1/8).

b) The polar function is given by r = sin( 2θ).

We've to find the corners with θ = 0 and θ = π/ 2,

When θ = 0

r = sin( 2 * 0) = sin( 0) = 0

thus, when θ = 0, the polar function intersects the origin( 0, 0).

Now, θ = π/ 2

r = sin( 2 *( π/ 2)) = sin( π) = 0

thus, when θ = π/ 2, the polar function also intersects the origin( 0, 0).

c) The polar function is given by r = 1 cos( θ).

To find the corners with θ = 0 and θ = π/ 2,

At θ = 0

r = 1 cos( 0) = 1 1 = 2

thus, when θ = 0, the polar function intersects thex-axis at( 2, 0).

At θ = π/ 2

r = 1 cos( π/ 2) = 1 0 = 1

thus, when θ = π/ 2, the polar function intersects the circle centered at( 0, 0) with compass 1 at( 1, π/ 2).

d) The polar function is given by r = 1- cos( 2θ).

To find the corners with θ = 0 and θ = π/ 2

At θ = 0

r = 1- cos( 2 * 0) = 1- cos( 0) = 0

thus, when θ = 0, the polar function intersects the origin( 0, 0).

At θ = π/ 2

r = 1- cos( 2 *( π/ 2)) = 1- cos( π) = 2

therefore, when θ = π/ 2, the polar function intersects the loop centered at( 0, 0) with compass 2 at( 2, π/ 2).

e) The polar function is given by r = 1- 2sin( θ).

To find the point of intersection with θ = 0 and θ = π/ 2,

When θ = 0

r = 1- 2sin( 0) = 1- 2( 0) = 1

thus, when θ = 0, the polar function intersects the circle centered at( 0, 0) with compass 1 at( 1, 0).

When θ = π/ 2

r = 1- 2sin( π/ 2) = 1- 2( 1) = -1

thus, when θ = π/ 2, the polar function intersects the negative y-axis at( 0,-1).

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The correct question is given below-

Sketch graphs of the following polar functions. Give the coordinates of intersections with theta = 0 and theta = π/2. a.dy = theta/4pi. with 0 < 0 < 4. b.r =sin(2theta) c.r=1+costheta d) r = 1- cos(2theta) e) r = 1- 2 sin(theta)

The solution of the given differential equation when y(1) = 2e²-e³ and y(0)=1 is given by   y = (1/6)e² + (2/3)e-³

Differential equation is y" + y' - 6y = 0

To show that y = Ae²+ Be-³x is the general solution of the given differential equation, first, we need to find the derivatives of y.

Now,y = Ae²+ Be-³x

Differentiating w.r.t 'x' , we get y' = 2Ae² - 3Be-³x

Differentiating again w.r.t 'x', we get y" = 4Ae² + 9Be-³x

On substituting the derivatives of y in the given differential equation, we get4Ae² + 9Be-³x + (2Ae² - 3Be-³x) - 6(Ae²+ Be-³x) = 0

Simplifying this expression, we getA(6e² - 1)e² + B(3e³ - 2)e-³x = 0

Since this equation should hold for all values of x, we have two possibilities either

A(6 e² - 1) = 0 and

B(3 e³ - 2) = 0or

6 e² - 1 = 0 and

3 e³ - 2 = 0i.e.,

either A = 0 and B = 0 or A = 1/6 and B = 2/3

So, the general solution of the given differential equation is given by

y = A e²+ B e-³x

where A and B are constants, A = 1/6 and B = 2/3

On substituting the given initial conditions, we get

y(1) = 2e²-e³

Ae²+ B e-³y(0) = 1

= Ae²+ Be-³x

Putting A = 1/6 and B = 2/3, we get

2e²-e³ = (1/6)e² + (2/3)e-³And,

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

Therefore, the solution of the given differential equation when y(1) = 2e²-e³ and y(0)=1 is given by   y = (1/6)e² + (2/3)e-³

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The ratio for Pete is 5:6 which is equivalent to 1000 shirts, therefore we will pay $10,000 to Ana, and he will pay $9600 to June.

Ana:

Ana is offering a promotion, which is to buy 5 and get 1 free. Based on this, the ratio would be 5:6 (pay 5 but get 6). Using this ratio, let's calculate the number of shirts that Pete would pay:

1200 / 6 =  200 x 5 = 1000 shirts

1000  shirts x $10 = $10,000

Jun:

The price with Jun is fixed as he will need to pay $8 for each shirt:

1200 shirts x $8 = $9600

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The loan amount for this extension is approximately $32,631. The correct option is (A) $32,631.

To find the loan amount for the extension Liam built onto his home, we can use the loan formula:

Loan formula:

PV = PMT * [{1 - (1 / (1 + r)^n)} / r]

Where,

PV = Present value (Loan amount)

PMT = Monthly payment

r = rate per month

n = total number of months

PMT = $750

r = 4.9% per annum / 12 months = 0.407% per month

n = 48 months

Putting the given values in the loan formula, we get:

PV = $750 * [{1 - (1 / (1 + 0.00407)^48)} / 0.00407]

PV ≈ $32,631 (rounded off to the nearest dollar)

Therefore, This extension's loan amount is roughly $32,631. The correct answer is option (A) $32,631.

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The result of the recurrence: an=9a n−1 for n≥2 with the initial condition a1=−6. an=  -6 × (-9)n-1

There is the recurrence relation: an = 9an - 1 with the initial condition a1 = -6. The task is to find the solution to the recurrence relation. Let's use the backward substitution method to solve the recurrence relation. In the backward substitution method, we start from the value of an and use the relation an = 9an - 1 to calculate an - 1, then use an - 1 = 9an - 2 to calculate an - 2, and so on until we reach the given initial value.

Here, a1 = -6, so we can start with a2. Using the relation an = 9an - 1, we get:

a2 = 9a1 = 9(-6) = -54

Using the relation an = 9 an - 1, we get:

a3 = 9a2 = 9(-54) = -486

Using the relation an = 9an - 1, we get:

a4 = 9a3 = 9(-486) = -4374

Similarly, we can calculate a5:

a5 = 9a4 = 9(-4374 ) = -39366

So, the result of the recurrence relation with the initial condition a1 = -6 is:

an = -6 × (-9)n-1

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1. Objective function: U(x₁, x₂), Constraint function: g(x₁, x₂) = m.

2. Lagrangian equation: L(x₁, x₂, λ) = U(x₁, x₂) - λ(g(x₁, x₂) - m).

3. First derivative with respect to x₁: ∂L/∂x₁ = ∂U/∂x₁ - λ∂g/∂x₁ = 0, First derivative with respect to x₂: ∂L/∂x₂ = ∂U/∂x₂ - λ∂g/∂x₂ = 0.

4. First derivative with respect to λ: ∂L/∂λ = g(x₁, x₂) - m = 0.

1. The objective function can be written as: U(x₁, x₂).

The constraint function can be written as: g(x₁, x₂) = m, where m represents the amount of money.

2. To set up the Lagrangian equation, we multiply the Lagrange multiplier λ to the constraint function and subtract it from the objective function. Therefore, the Lagrangian equation is given as: L(x₁, x₂, λ) = U(x₁, x₂) - λ(g(x₁, x₂) - m).

3. To find the first derivative of L with respect to x₁, we differentiate the Lagrangian equation with respect to x₁ and set it to zero as shown below: ∂L/∂x₁ = ∂U/∂x₁ - λ∂g/∂x₁ = 0.

Similarly, to find the first derivative of L with respect to x₂, we differentiate the Lagrangian equation with respect to x₂ and set it to zero as shown below: ∂L/∂x₂ = ∂U/∂x₂ - λ∂g/∂x₂ = 0.

4. Finally, we find the first derivative of L with respect to λ and set it equal to the constraint function as shown below: ∂L/∂λ = g(x₁, x₂) - m = 0.

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We have shown that the ellipse and hyperbola intersect at right angles.

To show that the ellipse and hyperbola intersect at right angles, we need to prove that their tangent lines at the point of intersection are perpendicular.

Let's first find the equations of the ellipse and hyperbola:

Ellipse: x^2/a^2 + 2y^2 = 1   ...(1)

Hyperbola: x^2/a^2 - 2y^2 = 1   ...(2)

To find the point(s) of intersection, we can solve the system of equations formed by (1) and (2). Subtracting equation (2) from equation (1), we have:

2y^2 - (-2y^2) = 0

4y^2 = 0

y^2 = 0

y = 0

Substituting y = 0 into equation (1), we can solve for x:

x^2/a^2 = 1

x^2 = a^2

x = ± a

So, the points of intersection are (a, 0) and (-a, 0).

To find the tangent lines at these points, we need to differentiate the equations of the ellipse and hyperbola with respect to x:

Differentiating equation (1) implicitly:

2x/a^2 + 4y * (dy/dx) = 0

dy/dx = -x / (2y)

Differentiating equation (2) implicitly:

2x/a^2 - 4y * (dy/dx) = 0

dy/dx = x / (2y)

Now, let's evaluate the slopes of the tangent lines at the points (a, 0) and (-a, 0) by substituting these values into the derivatives we found:

At (a, 0):

dy/dx = -a / (2 * 0) = undefined (vertical tangent)

At (-a, 0):

dy/dx = -(-a) / (2 * 0) = undefined (vertical tangent)

Since the slopes of the tangent lines at both points are undefined (vertical), they are perpendicular to the x-axis.

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