2. The cost of a badger is b pence. A raccoon is 5 pence more expensive than a badger and a beaver three times as expensive as a badger.

a) What is the cost of a raccoon?
b) What is the cost of the beaver?
c) What is the total cost of a raccoon and 8 beavers?

Answer:

Step-by-step explanation:

Hope this helps and i DID NOT use a Calculator!

Answer:

$226.57

Step-by-step explanation:

The definition of vector space that fails is part A and B  because it does not satisfy the any property of vector space.

The following statement can be determined if at-least one part of vector space definition that fails as:

(a) A is not a vector space because it does not contain the zero vector.

The zero vector is the unique vector that satisfies the property that when it is added to any other vector in the space, the result is the original vector.

However, in this case, the [a1, b1] + [a2, b2] = [a1 - a2, b1 - b2] is 0x² + 1, which is not an element of A. Therefore, A fails to satisfy the requirement of having a zero vector, and it is not a vector space.

(b) B is not a vector space because it does not satisfy the distributive property of scalar multiplication over vector addition.

In general, scalar multiplication must distribute over vector addition, meaning that for any scalar a and any vectors u and v in the space, a(u+v) = au + av.

However, in B, the scalar multiplication is defined as usual for R², but the vector addition is defined differently. In particular,[a1, b1] + [a2, b2] = [a1 - a2, b1 - b2].

The vector space definition fails because the vector addition is not associative, and it is also not commutative, which are the first two conditions for vector spaces. Therefore, the first and second conditions of the definition are not met.

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The value of x is given as follows:

\(x = 2.5\sqrt{2}{/tex]

The Pythagorean Theorem states that for a right triangle, the length of the hypotenuse squared is equals to the sum of the squared lengths of the sides of the triangle.

For a right triangle with an angle of 45º, we have that:

This is because the angle of 45º has an equal value of the sine and of the cosine.

Hence the side length of the top triangle is given as follows:

s = 5.

For the bottom triangle, the side length is of x, while the hypotenuse is of 5, hence:

x² + x² = 5²

2x² = 25

\(x = \sqrt{\frac{25}{2}}\)

\(x = \frac{5}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}}\)

\(x = 2.5\sqrt{2}{/tex]

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To find the arc length of the polar curve r =\(4e^θ\), where 0 ≤ θ ≤ π, we can use the formula for arc length in polar coordinates:

\(L = ∫[θ1, θ2] √(r^2 + (dr/dθ)^2) dθ\)

First, let's find the derivative of r with respect to θ, (dr/dθ):

\(dr/dθ = d/dθ (4e^θ) = 4e^θ\)

Now, let's plug the values into the arc length formula:

\(L = ∫[0, π] √(r^2 + (dr/dθ)^2) dθ\\= ∫[0, π] √((4e^θ)^2 + (4e^θ)^2) dθ\\\\= ∫[0, π] √(16e^(2θ) + 16e^(2θ)) dθ\\\\= ∫[0, π] √(32e^(2θ)) dθ\\= 4√2 ∫[0, π] e^θ dθ\\\)

Integratin\(g ∫ e^θ dθ\) gives us \(e^θ\):

\(L = 4√2 (e^θ) |[0, π]\\= 4√2 (e^π - e^0)\\= 4√2 (e^π - 1)\)

Therefore, the exact arc length of the polar curve r = \(4e^θ\), 0 ≤ θ ≤ π, is \(4√2 (e^π - 1).\)

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The uppercase A = one-half a p, the equation for 'a' is: a = 2A/p

To solve for a in the equation "uppercase A = one-half a p," we need to isolate a on one side of the equation.
First, we can multiply both sides of the equation by 2 to eliminate the fraction:
2(uppercase A) = 2(1/2 ap)
Simplifying, we get:
2(uppercase A) = ap
Next, we can divide both sides of the equation by p:
(2(uppercase A))/p = a
Therefore, the equation solved for a is:
a = (2(uppercase A))/p
This equation tells us that if we know the value of uppercase A and p, we can solve for the value of a.
It's worth noting that the equation we derived assumes that uppercase A represents the area of a circle, and p represents the circumference of that circle.

This is because the formula for the area of a circle is A = (1/2)pr^2, where r is the radius of the circle.

By rearranging this formula, we get A = (1/2)rp, which is equivalent to the equation we started with, since r = a/2.
The equation solved for a is a = (2(uppercase A))/p.

a = 2A/p

For similar question on uppercase:

Question:

The formula for the area of a regular polygon is, A = (a × p)/2, and the equation solved for a is, a = (2 × A)/p, where a is the apothem and p is the perimeter of the regular polygon. Uppercase A = one-half a p. What is the equation solved for a?

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Answer:

Step-by-step explanation:

The volume of a cone with a base radius of 4 cm and a height of 6 cm is equal to 33.51 cm³. This is calculated by using the formula V = (1/3)πr²h, where V is the volume, π is the constant Pi, r is the radius of the base, and h is the height of the cone. Plugging in the given numbers, we have V = (1/3)(3.14)(4²)(6) = 33.51 cm³.

Answer:

13

Step-by-step explanation:

18/3 = 6

6*2 = 12

12+1 = 13

The solution to the system of linear equations is (0,2).

What is equation?

In arithmetic, an equation may be a formula that expresses the equality of 2 expressions, by connecting them with the sign =.

Main body:

2x + 3y = 6 multiplied by 3

-3x + 5y = 10 multiplied by 2

6x + 9y = 18

-6x + 10y = 20

19y = 38

y = 38 / 19

y = 2

2x + 3x2 = 6

2x + 6 = 6

2x = 0

x = 0/2

x = 0

Therefore, The correct answer is (0,2)

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The value 2,649 represents the estimated population of elephants in Northwestern Namibia and Etosha National Park in 1995 and the estimated population was 618.4.

The formula for population growth can be represented as:

P = P0ert

Here, P is the population of elephants, P0 is the initial population, e is the base of the natural logarithm, r is the rate of growth or decay, and t is the time in years.

Now, as per the given data, P = 2,649P0 = initial population r = 1.045 (as per the given expression)b = t - 1995 (as per the given formula)Now, putting these values in the population growth formula,

P = P0ert2,649 = P0e1.045(b-1995)Dividing by P0 on both sides, we get:e1.045(b-1995) = 2649/P0

Taking the natural logarithm on both sides,

ln e1.045(b-1995) = ln (2649/P0)

Applying the property of logarithm,1.045(b-1995) ln e = ln (2649/P0)1.045(b-1995) = ln (2649/P0)

vAs we need to find the value of P0, substitute the value of b = 0,1.045(0-1995) = ln (2649/P0)-1.045*1995 = ln (2649/P0)ln (2649/P0) = -2069.95

Taking anti-logarithm or exponentiation on both sides,2649/P0 = eln (2649/P0) = e^(-2069.95)2649/P0 = 4.29P0 = 2649/4.29P0 = 618.4

Therefore, the estimated population of elephants in Northwestern Namibia and Etosha National Park in 1995 was 618.4.


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Answer:

c)

Step-by-step explanation:

\( - \frac{7}{ - f} \: is \: equal \: to \: \frac{7}{f} \)

\(so \: the \: answer \: is \: none \: of \: the \: above\)

Answer:

see explanation

Step-by-step explanation:

cos C = \(\frac{adjacent}{hypotenuse}\) = \(\frac{BC}{AC}\) = \(\frac{24}{51}\) = \(\frac{8}{17}\)

-----------------------------------------------------

tan Z = \(\frac{opposite}{adjacent}\) = \(\frac{XY}{ZY}\) = \(\frac{15}{20}\) = \(\frac{3}{4}\)

---------------------------------------------------

sin C = \(\frac{opposite}{hypotenuse}\) = \(\frac{AB}{AC}\) = \(\frac{60}{65}\) = \(\frac{12}{13}\)

Answer:The probability would be 5 out of 12.

The Brewster's angle for an air-glass (n=1.58) surface is approximately 56.309 degrees.

Brewster's angle is the angle of incidence at which light with a specific polarization, known as p-polarization, is perfectly transmitted through a transparent dielectric surface, with no reflection.

The formula for Brewster's angle is:

θB = arctan(n)

Where θB is Brewster's angle and n is the refractive index of the second medium (in this case, glass).

Substituting n=1.58, we get:

θB = arctan(1.58)

Using a calculator, we find:

θB ≈ 56.309°

Therefore, the Brewster's angle for an air-glass (n=1.58) surface is approximately 56.309 degrees.

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Degree of freedom for chi-square test for 10 categories will be 9.

What is chi- square test?

A test that evaluates how well a model matches real observed data is that the chi-square (χ²) statistic. A chi-square statistic can only be calculated with data that's random, unprocessed, mutually exclusive, obtained from independent variables, and drawn from a large enough sample.

The Formula for Chi-Square Is

χ(c)=²∑(Oₐ-Eₐ)₂/Eₐ²

where:

c=Degrees of freedom

O=Observed value(s)

E=Expected value(s)

​

Main Body:

Degree of freedom = total categories -1

so no. of categories= 10

Degree of freedom = 10 - 1

Hence degree of freedom = 9

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HELLO!!!

If the diameter is 12, this means that the radius is half of that amount (6)

In order to find area of a circle, you must solve pi x r^2 pi is 3.16 and the radius is 6. 6 ^2 is 36.

36x (pi) is 113.09 which should round off to 113.1 This means that area of the circle is  113.1 cm.

I Hope This Help & good luck.

Step-by-step explanation:

\( \sqrt{a {}^{2} } + 24 - 4 = a \)

Answer:

2x=125-10. Answer is 57.5

Step-by-step explanation:

It was asking for the length of the missing sides, and both of them are the same so its 2x= something. the other 2 sides are the same length (5) so that's where the 10 comes from. The 125 is the total perimeter. So, first you subtract 10 from 125 and you get 115. And then you divide 2 on both sides and you get x=57.5

I hope this helped :)

If this answer helped the most, give it the brainliest.

The rational function f(x) = (3 - x) / (2x^2 - 5x - 3) has a domain of all real numbers except for the values that make the denominator equal to zero. The graph does not have any vertical asymptotes, but it does have a horizontal asymptote at y = 0.

(a) The domain of a rational function is all real numbers except for the values that result in a zero denominator. In this case, the denominator is 2x^2 - 5x - 3. To find the domain, we set the denominator equal to zero and solve for x. By factoring or using the quadratic formula, we find that the denominator equals zero when x = -1/2 and x = 3/2. Therefore, the domain of f(x) is all real numbers except x = -1/2 and x = 3/2.

(b) To determine if the graph of f(x) has any holes, we check if any factors in the numerator cancel out with factors in the denominator. In this case, the factor (x - 3) in the numerator cancels out with the factor (x - 3) in the denominator. Therefore, the graph of f(x) has a hole at the coordinate (3, -2/7), where x = 3 is the x-coordinate of the hole, and -2/7 is the y-coordinate.

(c) To find vertical asymptotes, we look for values of x that make the denominator zero but do not cancel out with any factors in the numerator. However, in this case, there are no such values. The denominator factors as (2x + 1)(x - 3), and the factor (x - 3) cancels out with the numerator. Therefore, the graph of f(x) does not have any vertical asymptotes.

(d) The graph of f(x) may have horizontal asymptotes if the degree of the numerator is less than or equal to the degree of the denominator. In this case, the degree of the numerator is 1 and the degree of the denominator is 2. Since the degree of the numerator is less than the degree of the denominator, we can conclude that the graph of f(x) has a horizontal asymptote. To find the equation of the asymptote, we divide the leading terms of the numerator and denominator. The leading term of the numerator is -x, and the leading term of the denominator is 2x^2. Dividing them, we get -1/2. Therefore, the graph of f(x) has a horizontal asymptote at y = -1/2, or in other words, at the line y = 0.

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Answer:

I'm not sure but I think that's the answer

Step-by-step explanation:

D. y=⅔x+2

Answer:

idrk i think it is 16x²

Step-by-step explanation:

A person weighing 72 pounds on the nearby planet would weigh 162 pounds on Earth. Therefore, a person weighing 72 pounds on the nearby planet would weigh 162 pounds on earth if the weights are proportional.

If a person who weighs 198 pounds on earth would weigh 88 pounds on a nearby planet, then the ratio of their weight on earth to their weight on the nearby planet would be:

198/88 = 2.25

So, if we want to find out what a person weighing 72 pounds on the nearby planet would weigh on earth, we can set up a proportion:

198/88 = x/72

where x is the weight of the person on earth.

To solve for x, we can cross-multiply:

198 * 72 = 88 * x

14256 = 88x

x = 162

Therefore, a person weighing 72 pounds on the nearby planet would weigh 162 pounds on earth if the weights are proportional.

To find the weight of a person on Earth if they weigh 72 pounds on the nearby planet, we'll use proportions.

Let x be the weight of the person on Earth. We can set up the proportion as follows:

198 pounds (Earth) / 88 pounds (nearby planet) = x pounds (Earth) / 72 pounds (nearby planet)

To solve for x, cross-multiply:

198 * 72 = 88 * x

14256 = 88x

Now, divide both sides by 88 to find the weight on Earth:

x = 14256 / 88

x = 162

So, a person weighing 72 pounds on the nearby planet would weigh 162 pounds on Earth.

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Answer:

162 lb

Step-by-step explanation:

The weights are proportional, so set up a proportion and solve for the only unknown.

198 is to 88 as x is to 72

198/88 = x/72

99/44 = x/72

44x = 72 × 99

x = 7128/44

x = 162

Answer: 162 lb

The optimal solution for the given linear programming problem using the Big-M method is x₁ = 4, x₂ = 2, with a maximum value of Z = 22.

To solve the given linear programming problem using the Big-M method, we first convert it into standard form by introducing slack, surplus, and artificial variables.

The objective function is to maximize Z = 4x₁ + 3x₂. The constraints are 2x₁ + x₂ ≥ 10, -3x₁ + 2x₂ ≤ 6, x₁ + x₂ ≥ 6, and x₁, x₂ ≥ 0.

We introduce slack variables s₁, s₂, and s₃ to convert the inequalities into equalities. The initial Big-M tableau is set up with the coefficients and variables, and the artificial variables are introduced to handle the inequalities. We set a large positive value (M) for the artificial variables' coefficients.

In the first iteration, we choose the most negative coefficient in the Z-row, which is -4 corresponding to x₁. We select the s₂-row as the pivot row since it has the minimum ratio of the RHS value (6) to the coefficient in the pivot column (-3). We perform row operations to make the pivot element 1 and other elements in the pivot column 0.

After multiple iterations, we find that the optimal solution is x₁ = 4, x₂ = 2, with a maximum value of Z = 22. This means that to maximize the objective function, x₁ should be set to 4 and x₂ should be set to 2, resulting in a maximum value of Z as 22." short

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Answer:

12x6+2(6-1)

72+2x5

72+10

82

82 is the answer.

Step-by-step explanation:

Just replace the X's in the problem with 6.

Answer:

The answer to your problem is 540

Step-by-step explanation:

The answer is 180 because all triangles have 3 sides and a pentagon is 3 triangles formed as one so you need to do 180 x 3 which is 540

The average rate of change of the given function is 10π.

Given function

a = πx^2

Where a is the area of a circle with radius x

The radius changes from x = 4 to x = 6.

Calculating the average rate of change of function:

The method of finding the average rate between the function value over a certain interval is called the average rate of change.

The formula is given by :

Average = \(\frac{f(b)-f(a)}{b-a}\)

Where f(a) and f(b) are the function value over the limit a and b.

Average = \(\frac{f(6)-f(4)}{6-4}\)

= \(\frac{\pi (6)^{2} -\pi (4)^{2} }{6-4}\)

= \(\frac{\pi (36-16)}{2}\)

= \(\frac{20\pi }{2}\)

= 10π

The average rate of change of the given function is 10π.

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(a) The number of possible schedules that the student can create from the offerings is 96.

(b) the number of possible schedules that the student can create from the offerings if two of the math 100 sections are closed is 48.

(c) the number of possible schedules that the student can create from the offerings if two of the math 100 sections and four of the English 105 sections are closed is 4.

(a) To find the total number of possible schedules, we need to multiply the number of choices for each class:

Number of schedules = 4 x 2 x 3 x 4 = 96

So, there are 96 possible schedules that the student can create from the offerings.

(b) If two of the math 100 sections are closed, then there are only two choices for that class. So, we have:

Number of schedules = 2 x 2 x 3 x 4 = 48

So, there are 48 possible schedules that the student can create from the offerings with two of the math 100 sections closed.

(c) If two of the math 100 sections and four of the English 105 sections are closed, then there are only two choices for math 100 and one choice for English 105. So, we have:

Number of schedules = 2 x 2 x 1 x 1 = 4

So, there are 4 possible schedules that the student can create from the offerings with two of the math 100 sections and four of the English 105 sections closed.

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Answer:

False. The union of two events A and B has outcomes from event A or event B or both A and B.