Mrs. Goldstein baked 24 cookies. 2 thirds of the cookies were oatmeal. How many oatmeal cookies did Mrs. Goldstein bake?

The remainder when Fn+1 is divided by Fn is Fn-1, and the remainder when Fn is divided by Fn-1 is Fn-2.

Let's denote the greatest common divisor of Fn+1 and Fn as d. We can use the Euclidean Algorithm to find d by repeatedly taking remainders.

First, we have:

Fn+1 = 1*Fn + Fn-1

So, the remainder when Fn+1 is divided by Fn is Fn-1.

Next, we have:

Fn = 1*Fn-1 + Fn-2

So, the remainder when Fn is divided by Fn-1 is Fn-2.

We can continue this process by repeatedly dividing the larger number by the smaller number and taking remainders until we reach a remainder of 0. The last nonzero remainder we obtain is the greatest common divisor of Fn+1 and Fn.

For example, to find the greatest common divisor of F6 = 8 and F7 = 13, we have:

F7 = 1F6 + F5, so the remainder is F5 = 5

F6 = 1F5 + F4, so the remainder is F4 = 3

F5 = 1F4 + F3, so the remainder is F3 = 2

F4 = 1F3 + F2, so the remainder is F2 = 1

F3 = 2*F2 + 0, so we stop here

Therefore, the greatest common divisor of F7 and F6 is d = 1, and the remainders in the Euclidean Algorithm are 5, 3, 2, and 1.

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The work required to pump water out of the spout is calculated by multiplying the density of water (1000 kg/m3) by the height of the tank (h) and the area of the spout (a x b x c x d). The acceleration due to gravity (g) is 9.8 m/s2.

1. Calculate the volume of the tank:

V = a x b x c x d = 4 x 4 x 9 x 4 = 576 m3

2. Calculate the mass of the water in the tank:

m = V x density = 576 x 1000 = 576000 kg

3. Calculate the height of the tank:

h = m / density = 576000 / 1000 = 576 m

4. Calculate the work required to pump the water out of the spout:

W = m x g x h = 576000 x 9.8 x 576 = 3.11.107 x 576 = 1.79.107 J

The work required to pump water out of a spout can be calculated by multiplying the density of water (1000 kg/m3) by the height of the tank (h) and the area of the spout (a x b x c x d). The acceleration due to gravity (g) is 9.8 m/s2.To calculate the work, first we need to find the volume of the tank (V) by multiplying the length (a), width (b), height (c), and depth (d). Then we can calculate the mass (m) by multiplying the volume with the density of water. We can then calculate the height of the tank (h) by dividing the mass with the density. Finally, we can calculate the work required (W) by multiplying the mass, acceleration due to gravity, and the height of the tank.

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The required inequality of the solutions of the given equation is (-∞, -6).

The idea of inequality, which is the state of not being equal, especially in terms of status, rights, and opportunities, is at the core of social justice theories. However, because it frequently has diverse meanings to different people, it is prone to misunderstanding in public discourse.

We have;

2w-1 < -13

2w < -13 + 1

2w < - 12

w < -6

Thus, required inequality of the solutions of the given equation is (-∞, -6)

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The relationship between y and t can be modeled by an exponential function of the form:

y = a x e^(-rt)

An exponential function is a mathematical function which we write as a

f(x) = aˣ, where a is constant and x is variable term. The most commonly used exponential function is eˣ , where e is constant having value 2.7182

The relationship between y and t can be modeled by an exponential function of the form:

y = a x e^(-rt)

where a is the initial area of the forest (1500 km²), r is the rate of decrease (6.25%), and t is the number of years since 2000.

To find the value of r, we can convert 6.25% to a decimal:

r = 0.0625

Now we can plug in the values for a and r into our exponential function:

y = 1500 x e^(-0.0625t)

This exponential function shows the relationship between the area of the forest and the number of years since 2000.

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The payment rate that is required to pay off the loan in 3 years is $3139.88 per year. The borrower will pay approximately $1695.64 in interest over the 3-year period.

The differential equation that models this situation is:

dy/dt = 0.10y - k

where y(t) is the amount owed at time t, 0.10 is the annual interest rate, and k is the annual payment rate. The initial condition is y(0) = $8000.

The first term on the right-hand side represents the interest that accumulates on the loan, and the second term represents the payments made by the borrower.

To determine the payment rate k that is required to pay off the loan in 3 years, we need to solve the differential equation with the initial condition y(0) = $8000 and the terminal condition y(3) = 0.

The general solution to the differential equation is:

y(t) = (8000/k) e^(0.10t) - (8000/k)

Setting t = 3 and y(3) = 0, we get:

0 = (8000/k) e^(0.30) - (8000/k)

Solving for k, we get:

k = 3139.88

Therefore, the payment rate that is required to pay off the loan in 3 years is $3139.88 per year.

To determine how much interest is paid during the 3-year period, we can integrate the interest rate over the time interval [0, 3]:

∫[0,3] 0.10y(t) dt = ∫[0,3] 0.10[(8000/k) e^(0.10t) - (8000/k)] dt

= (8000/k) [e^(0.30) - 1] - 2400

Substituting k = 3139.88, we get:

∫[0,3] 0.10y(t) dt ≈ $1695.64

Therefore, the borrower will pay approximately $1695.64 in interest over the 3-year period.

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____The given question is incomplete, the complete question is given below:

A certain college graduate borrows $8000 to buy a car. The lender charges interest at an annual rate of 10%. Assume that interest is compounded continuously and that the borrower makes payments continuously at a constant annual rate k. (a) Write a differential equation that models this situation, including the initial condi- tion. (b) Determine the payment rate k that is required to pay off the loan in 3 years. (c) Determine how much interest is paid during the 3-year period.

The component form of the vector is <3, -5, 6>, the magnitude of the vector is √70, then the unit vector is <0.386, -0.643, 0.643>.

The initial and terminal points to find the component form of the vector, calculated its magnitude, and then divided the component form by the magnitude to find a unit vector in the direction of v.

To find the component form of the vector, you can subtract the coordinates of the initial point from the coordinates of the terminal point. In this case, we have:

v = (4, 1, 6) - (1, 6, 0)

v = (3, -5, 6)

So the component form of the vector is v = <3, -5, 6>.

To find the magnitude of the vector, we can use the formula:

|v| = √(3² + (-5)² + 6²)

|v| = √70

Therefore, the magnitude of the vector is √70.

Finally, to find a unit vector in the direction of v, we can divide the component form of v by its magnitude:

u = v/|v|

u = <3/√70, -5/√70, 6/√70>

So the unit vector in the direction of v is u = <0.386, -0.643, 0.643>.

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All absolute values are greater than or equal to zero so there is no such absolute value of - 4.

We know the absolute value function of the modulus function always outputs a positive value irrespective of the sign of the input.

In piecewise terms  | x | = x for x ≥ 0 and | x | = - x for x < 0.

We know, |a| = a, |- a| = a, and |0| is 0 therefore, The least possible value of a modulus is zero.

Therefore, There is no such numbers whose modulus value is - 4.

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The probability that both prizes were won by two people of the same sex is 0.48

Let's call the number of men in the small party "m" and the number of women in the small party "w". The total number of people in the party is "m + w = 10 + 12 = 22".

The number of ways to choose two people of the same sex can be calculated as follows:

The number of ways to choose two men is "C(m,2) = C(10,2) = 45".

The number of ways to choose two women is "C(w,2) = C(12,2) = 66".

The total number of ways to choose two people without replacement is "C(22,2) = 231".

The probability that both prizes were won by two people of the same sex is then given by the sum of the probabilities for two men or two women:

P(same sex) = (C(10,2) / C(22,2)) + (C(12,2) / C(22,2)) = (45/231) + (66/231) = 111/231 = approx. 0.48

So, the probability that both prizes were won by two people of the same sex is approximately 0.48

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To determine the maximum height of a shampoo bottle that can stand upright on the middle shelf, we need to consider the height of the shelf above and below it.

Since there are three shelves, the middle shelf is located at the height of 9 inches (half of the total height).

To find the maximum height of a shampoo bottle that can stand upright on the middle shelf, we need to subtract the height of the middle shelf from the total height of the system, and then divide the result by two, since there are two spaces above and below the middle shelf.

Therefore, the maximum height of a shampoo bottle that can stand upright on the middle shelf is: (18 - 9) / 2 = 4.5 inches

So, the maximum height of a shampoo bottle that can stand upright on the middle shelf is 4.5 inches.

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The following system of liner equation when converted into equivalent matrix gives us x =2, y = 3, z = -2.

Given that the augmented matrix is [AB] = [tex]\left[\begin{array}{cccc}1&2&2&4\\0&1&-3&9\\0&0&1&-2\end{array}\right][/tex]

Since p(A) = p(AB) = 3 = n = The number of variables

The system has unique solution

x + 2y + 2z = 4

y - 3z = 9

z = -2

y = 9 + 3z = 9 + 3(-2) = 3

x = 4-2y -2z

= 4 -2(3) -2(-2)

= 4 - 6 + 4

= 2

Therefore, the solution is x =2, y = 3, z = -2.

The abecedarian idea is to add multiples of one equation to the others in order to exclude a variable and to continue this process until only one variable is left. Once this final variable is determined, its value is substituted back into the other equations in order to estimate the remaining unknowns. This system, characterized by step ‐ by ‐ step elimination of the variables, is called Gaussian elimination.

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Complete question;

The augmented matrix of a system of equations has been transformed to an equivalent matrix in​ row-echelon form. Using​ x, y, and z as​ variables, write the system of equations corresponding to the following matrix. If the system is​ consistent, solve it.

left bracket Start 3 By 3 Matrix 1st Row 1st Column 1 2nd Column 2 3rd Column 2 2nd Row 1st Column 0 2nd Column 1 3rd Column negative 3 3rd Row 1st Column 0 2nd Column 0 3rd Column 1 EndMatrix Start 3 By 1 Table 1st Row 1st Column 4 2nd Row 1st Column 9 3rd Row 1st Column negative 2 EndTable right bracket

dydt = k(y-5)(y-0) where k is a positive constant. This equation has equilibrium solutions of y=0 and y=5, and y' is positive for 0<y<5 and negative for y<0 or y>5.

We can solve this problem by rearranging the equation to separate the y terms from the y' terms. We can do this by factoring the y terms on the left side, and then writing the equation as y' = k(y-5)(y-0). This equation has an equilibrium solution at y=0 and y=5, and the sign of y' depends on the sign of k. Since k is a positive constant, y' is positive when 0<y<5, and y' is negative when y<0 or y>5.  Thus, the equation dydt = k(y-5)(y-0) meets all of the given criteria. Therefore, dydt = k(y-5)(y-0) where k is a positive constant. This equation has equilibrium solutions of y=0 and y=5, and y' is positive for 0<y<5 and negative for y<0 or y>5.

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Answer: 2.25 slices which is 9/4 slices i think im right

Step-by-step explanation:

add 1 1/4+2 3/4+3/4 which makes 4.75 slices or 19/4 slices do 7-4.75=2.25/9/4 slices

Answer:

let the Becky's age be x

21=3x

x = 7

Answer:

a = 21 / 3

Step-by-step explanation:

If he's 3 times older than her, then you divide by 3.

The system of equation that does not belong to the other two is the third system of equations; y = -4·x - 3, y + 4·x = -5, This is so because, the third system has no solutions.

A linear system of equations consists of two or more linear equations that consists of common variables.

The possible equations are;

y = x + 6, y = -x + 2

3·x + y = -1, y = 4·x + 6

y = -4·x - 3, y + 4·x = 5

Evaluation of the system of equations, we get;

First system of equations;

y = x + 6, y = -x + 2

x + 6 = -x + 2

x + x = 2 - 6 = -4

2·x = -4

x = -4/2 = -2

x = -2

y = x + 6

y = -2 + 6 = 4

y = 4

The solution is; x = -2, y = 4

Second system of equation;

3·x + y = -1, y = 4·x + 6

3·x + 4·x + 6 = -1

7·x + 6 = -1

7·x  = -1 - 6 = -7

x = -7/7 = -1

x = -1

y = 4·x + 6

y = 4 × (-1) + 6 = 2

y = 2

The solution to the second system of equation is; x = -1, y = 2

Third system of equation;

y = -4·x - 3, y + 4·x = 5

y + 4·x = 5

-4·x - 3 + 4·x = 5

-4·x + 4·x - 3 = 5

0 - 3 = 5

-3 = 5

The third system of equation has no solution

The system of equations that does not belong with the other two is the third system of equation; y = -4·x - 3, y + 4·x = 5, that has no solution.

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The graph of the system of inequalities is in the image at the end.

Here we have a system of inequalities, first we want to solve them:

-x + 3y ≤ 3

-2x + y ≥ -2

Isolating y in both of these we will get:

y  ≥ -2 + 2x

y ≤ (3 + x)/3

So we just needto graph the two lines, on the first one we will shade the region above the line and on the second one we will shade the region below the line.

The graph of the system is on the image below.

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The Probability density function for the power P is f(P) = 1/2 * (1/P) for 162 < P < 242.

The power P is equal to 2I^2, so we can find the probability density function of P by finding the distribution of I first. A uniformly distributed random variable X over an interval (a, b) has a probability density function given by:

f(x) = 1/(b - a) for a < x < b

Since I is uniformly distributed over (9, 11), its probability density function is:

f(I) = 1/(11 - 9) = 1/2

Now, to find the distribution of P, we can use the transformation function P = 2I^2:

f(P) = f(I) * |dI/dP|

Using the chain rule, we have:

dI/dP = dI/d(2I^2) * d(2I^2)/dP = 1/2 * (2I) = I/P

So:

f(P) = f(I) * (1/P) = 1/2 * (1/P)

Now, we need to find the bounds for P. The power P can be calculated for any value of I between 9 and 11, so the bounds for P are:

P_min = 2 * 9^2 = 162

P_max = 2 * 11^2 = 242

Therefore, the probability density function for the power P is:

f(P) = 1/2 * (1/P) for 162 < P < 242

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____The given question is incomplete, complete question is given below:

A fluctuating electric current I may be considered a uniformly distributed random variable over the interval (9, 11). If this current flows through a 2-ohm resistor, find the probability density function of the power P = 2I^2.

The solution is given below.

A scatter plot is a type of plot or mathematical diagram using Cartesian coordinates to display values for typically two variables for a set of data. If the points are coded, one additional variable can be displayed.

here, we have,

A scatter plot is a graph in which the values of two variables are plotted along two axes. Every data of the table is the coordinate of a point in the plot. To make a scatter plot, first assign a variable to x-axis and the other variable to  y-axis, in this question speed was assigned to x-axis and distance was assigned to y-axis. And then, locate the points as coordinates, for example, the first point is (2, 5).

See picture attached.

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The solution of the expression 2 x 2[tex]\frac{1}{5}[/tex] is 4.4.

Multiplication is a type of mathematical operation. The repetition of the same expression types is another aspect of the practice.

For instance, the expression 2 x 3 indicates that 3 has been multiplied by two.

Given:

Two fractions are 2 and 2[tex]\frac{1}{5}[/tex].

To find the product of two fractions:

Applying multiplication operation,

we get,

2 x 2[tex]\frac{1}{5}[/tex]

To simplify further;

Converting mixed fractions to improper fractions,

we get,

2 x 11/5

= 22/5

= 4.4

Therefore, 2 x 2[tex]\frac{1}{5}[/tex] = 4.4.

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The second part of Theorem 6 states that if w is any solution of the linear system Ax=b, and p is a particular solution of Ax=b, then the difference vector v=h−p is a solution of the homogeneous system Ax=0.

We will now prove this statement.

Since p is a particular solution of Ax=b, we have A*p = b. Then we can write w = p + v, where v = w - p.

To show that v is a solution of Ax=0, we need to show that Av=0.

We have:

Av = A(w-p) = Aw - Ap

Since Aw = b (by the assumption that w is a solution of Ax=b) and Ap = b (by the assumption that p is a particular solution of Ax=b), we can simplify this to:

A*v = b - b = 0

Thus, v is a solution of Ax=0, as required.

Therefore, every solution of Ax=b has the form w=p+v, where p is a particular solution of Ax=b and v is a solution of Ax=0.

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Q is a non-negative linear combination of permutation matrices.

A non-negative linear combination of permutation matrices is a matrix P that can be expressed as a linear combination of permutation matrices P1, P2, ..., Pn such that each coefficient in the linear combination is non-negative. This can be expressed as:

[tex]P = c1P1 + c2P2 + ... + cnPn[/tex]

where c1, c2, ..., cn are all non-negative real numbers. For example, consider the following matrix Q:

[tex]Q = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{bmatrix}[/tex]

We can express Q as a non-negative linear combination of permutation matrices P1, P2, P3 as follows:

[tex]Q = \frac{1}{3}\begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{bmatrix} + \frac{1}{3}\begin{bmatrix} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{bmatrix} + \frac{1}{3}\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}[/tex]

Therefore, Q is a non-negative linear combination of permutation matrices.

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The graph of the given inequality of x > 23 is attached.

The relationship between two expressions or values that are not equal to each other is called inequality.

A number line can be used to represent numbers placed on regular intervals. A number line can be used to represent an inequality.

Given that the inequality of x > 23

We are asked to plot the given inequality on a number line.

x > 23

The above inequality says that, the value of x is equal to or greater than 23.

Hence, the graph of the given inequality is attached.

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Answer:

3x^2 - x - 10

Step-by-step explanation:

Foil (First, outside, inside, last)

3x^2 + 5x + -6x - 10

combine like terms

3x^2 - x - 10

(a) For no solution, we need the two equations to be inconsistent, which means that they cannot be satisfied simultaneously. We can achieve this by making the first equation a multiple of the second equation:

4(x1 + hx2) = 8

4x1 + 4hx2 = 8

4x1 + 8x2 = k

Now, we can see that the second equation is not compatible with the first equation since they imply contradictory statements:

4x1 + 8x2 = k and 4x1 + 4hx2 = 8

(b) For a unique solution, we need the two equations to be independent, which means that they are not multiples of each other. We can achieve this by choosing different coefficients for x1 and x2 in the two equations.

x1 + hx2 = 2 and 4x1 + 8x2 = k

To find the values of h and k that give a unique solution, we can solve the system by elimination or substitution. For example, we can multiply the first equation by 4 and subtract it from the second equation:

4x1 + 8x2 = k

-4x1 - 4hx2 = -8

Simplifying and dividing by -4, we get:

x2 = (2 + h)/2

x1 = (k - 4x2)/4

Since x1 and x2 are expressed in terms of h and k, we can choose any values of h and k that satisfy these equations, and the system will have a unique solution.

(c) For many solutions, we need the two equations to be dependent, which means that they are multiples of each other or one is a linear combination of the other. We can achieve this by making the second equation a multiple of the first equation:

x1 + hx2 = 2

4(x1 + hx2) = 8 + 4hkx2

4x1 + (4h - k)x2 = 8

Now, we can see that the second equation is a linear combination of the first equation, so the system has infinitely many solutions. To find the solutions, we can choose any value of x2 and solve for x1 in terms of x2:

x1 = (8 - (4h - k)x2)/4

Since x1 and x2 are expressed in terms of h and k, we can choose any values of h and k that satisfy the equation 4h - k = 0, and the system will have many solutions.

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(a) The rate of change of the brightness after t days is dB/dt = (2π/4.4) * 0.45 * cos(2πt/4.4).

(b) The rate of increase after one day is  0.22 radians/day.

For the given case the equation for the brightness of cepheid joe is  B(t)=4.2 +0.45sin(2pit/4.4). This equation tells us that the brightness of the star is determined by the sine of the time multiplied by a constant. Since the sine of a number is always changing, the brightness of the star is always changing too.

Therefore, the rate of change of the brightness is given by the derivative of the equation, which is   2π/4.4)*0.45*cos(2πt/4.4), when t is  1 day, we can plug this value into the equation to get the rate of increase after one day, which is dB/dt = (2π/4.4) * 0.45 * cos(2π/4.4) = 0.22 radians/day.

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The equation of tangent to the curve y = 8√x  at the point (4,16) is  y = 2x + 8 .

We have to find equation of tangent line to the curve y = 8√x at the point (4, 16), we first find the slope ;

So , slope of the tangent line is  derivative of function y = 8√x at point (4, 16).

which means :  y' = 4[tex]x^{-\frac{1}{2} }[/tex]  ;

At the point (4, 16), the value of x is 4.

So , y' = 4 × [tex]4^{-\frac{1}{2} }[/tex] = 2 .

By Using the point slope form, the equation of the tangent line is ;

⇒ y - 16 = (2)(x - 4)  ;

Simplifying this equation, we get:

⇒ y - 16 = 2x - 8

⇒ y = 2x + (16 - 8)

⇒ y = 2x + 8 .

Therefore, the equation of the tangent line to the curve is y = 2x + 8 .

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The given question is incomplete , the complete question is

Find an equation for the tangent to the curve y = 8√x  at the point (4,16) .

The recursive formula for p(n) is; p(n) = p(n-2) + p(n-3) for n >= 3. Case 1: The last square is painted white. If the last square is painted white, then the second to last square must be painted gold.

There are p(n-2) ways to paint the remaining n-2 squares of the checkerboard subject to the restriction.

Case 2: The last square is painted gold. If the last square is painted gold, then the second to last square can be painted either white or gold. If the second to last square is painted white, then there are p(n-3) ways to paint the remaining n-3 squares of the checkerboard subject to the restriction.

If the second to last square is painted gold, then there are p(n-2) ways to paint the remaining n-2 squares of the checkerboard subject to the restriction.

Therefore, the recursive formula for p(n) is: p(n) = p(n-2) + p(n-3) for n >= 3

with initial conditions p(1) = 2 and p(2) = 3.

The base case for the recursion is p(1) = 2 and p(2) = 3, which are the number of ways to paint a 1 x 1 checkerboard and a 1 x 2 checkerboard subject to the restriction, respectively.

The recursive formula counts the number of ways to paint a 1 x n checkerboard subject to the restriction by considering the last column of the checkerboard.

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a. Eliminating the parameter yields the equation [tex]x2+y2=100[/tex], which is a circle centered at the origin with a radius of 10.

b. The curve is a circle with a positive orientation, going counterclockwise from the origin.

a. To eliminate the parameter, we first square both sides of the equations to obtain: [tex]x2=(−10cos2t)2 and y2=(−10sin2t)2.[/tex]Then, since cos2t and sin2t are both between -1 and 1, the terms on the right hand side of each equation can be simplified to 100. Thus, the equation [tex]x2+y2=100[/tex]is obtained.

b. This equation describes a circle centered at the origin with a radius of 10. The positive orientation of the curve is counterclockwise from the origin, i.e. it starts at the origin and moves up, then to the right, then down, and then to the left.

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The simple interest rate for the item is 13 1/3%.

Simple interest is the charge on borrowing calculated as a linear function of the amount borrowed, time and the interest rate.

Interest rate = interest / (time x amount borrowed)

Interest rate = N$16 / (N$480 x 0.25)

= 0.13333 = 13 1/3 %

To learn more about simple interest, please check: https://brainly.com/question/27328409

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Answer:

Perimeter: 31 cm

Step-by-step explanation:

18.5 + 8.5 + 4 = 31 cm.

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