Suppose the variable x is represented by a standard normal distribution.What value of x is at the 70th percentile of the distribution? Equivalently, what is the value for which there is a probability of 0.70 that x will be less than that value?Please round your answer to the nearest hundredth.

Answer:

125/12

Step-by-step explanation:

lets take n = -10.4166666

multiply this by 100 so we get the recurring part as the decimals

100n = -1041.66666

now we multiply our original n value by 10 for simplicity while calulating

10n = -104.16666

then we subtract 10n from 100n

90n = -1041.666 - (- 104.16666)

the recurring part will cancel out infinitely

so we get

90n = 937.5

then we solve for n

n = 937.5/90

simplifying will get us n= 125/12

A graph with an appropriate pair of axes has been used to plot the points as shown in the image attached below.

In Mathematics and Geometry, a graph is a type of visual chart that is used for the graphical representation of data points or ordered pairs on both the horizontal and vertical lines of a cartesian coordinate, which are the x-coordinate (x-axis) and y-coordinate (y-axis) respectively.

In Mathematics and Geometry, an ordered pair is sometimes referred to as a coordinate and it can be defined as a pair of two elements or data points that are commonly written in a fixed order within parentheses as (x, y), which represents the x-coordinate (abscissa) and the y-coordinate (ordinate) on the coordinate plane of any graph.

In this scenario and exercise, we would use an online graphing calculator to graphically represent the given points on a graph as shown in the image attached below.

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Since W satisfies all three conditions, it is a subspace of R4. And since we have shown that W = Span{u, v}, we can choose answer (C): "Since u and v are in R4 and W = Span{u, v}, W is a subspace of R4."

What is sub space?

In mathematics, a subspace is a subset of a vector space that is itself a vector space under the same operations of vector addition and scalar multiplication as the original space.

To show that W is a subspace of R4, we need to show that it satisfies three conditions:

The zero vector is in W.

W is closed under vector addition.

W is closed under scalar multiplication.

First, let's find vectors u and v such that W = Span{u,v}. We are given that a vector B in W has the form:

B = (2s + 5s + 3t, 4s - 5t, 2t, 45-50)

We can rewrite this as:

B = (7s, 4s, 0, 45-50) + (3t, -5t, 2t, 0)

So, we can take u = (7, 4, 0, -5) and v = (3, -5, 2, 0) to span W.

Now, let's check the three conditions:

The zero vector is in W:

Setting s = t = 0 in the expression for B gives us the vector (0, 0, 0, -5). This vector is in W, so the zero vector is in W.

W is closed under vector addition:

Let B1 and B2 be two vectors in W. Then, we have:

B1 = su1 + tv1 = a1u + b1v

B2 = su2 + tv2 = a2u + b2v

where a1, b1, a2, b2 are scalars.

Then, B1 + B2 is given by:

B1 + B2 = su1 + tv1 + su2 + tv2

= (a1u + b1v) + (a2u + b2v)

= (a1 + a2)u + (b1 + b2)v

which is also in W, since it can be expressed as a linear combination of u and v.

W is closed under scalar multiplication:

Let B be a vector in W and let k be a scalar. Then, we have:

B = su + tv = au + bv

where a, b are scalars.

Then, kB is given by:

kB = k(su + tv)

= (ks)u + (kt)v

which is also in W, since it can be expressed as a linear combination of u and v.

Therefore, since W satisfies all three conditions, it is a subspace of R4. And since we have shown that W = Span{u, v}, we can choose answer (C): "Since u and v are in R4 and W = Span{u, v}, W is a subspace of R4."

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We have transformed the left side into the right side using trigonometric identities. We start with the left side of the equation:

(1 + cos 0) (1 – sin 0)

Expanding the product, we get:

1 - sin 0 + cos 0 - sin 0 cos 0

Using the identity sin² θ + cos² θ = 1, we can replace sin² θ with 1 - cos²θ:

1 - (1 - cos² θ) + cos θ - (1 - cos² θ) cos θ

Simplifying, we get:

2 cos² θ - cos θ - 1

Now we use the identity sin² θ + cos² θ = 1 again to replace cos² θ with 1 - sin²θ:

2(1 - sin² θ) - cos θ - 1

2 - 2 sin²θ - cos θ - 1

1 - 2 sin² θ - cos θ

Finally, using the identity sin 2θ = 2 sin θ cos θ, we can write:

1 - sin 2θ - cos θ

Which is the right side of the equation. Therefore, we have transformed the left side into the right side using trigonometric identities.

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The quadratic function that models this situation is given as follows:

y = -0.05(x² - 60x +  576).

The standard definition of a quadratic function is given as follows:

y = ax² + bx + c.

The ball is kicked 12 yards from the goal and lands 48 yards from the goal, hence, the roots are given as follows:

x = 12, x = 48.

Thus the function is defined as follows:

y = a(x - 12)(x - 48)

y = a(x² - 60x +  576).

The x-coordinate of the vertex is given at the mean of the roots, hence:

x = (12 + 48)/2 = 30.

The maximum height means that when x = 30, y = 17, hence the leading coefficient a is obtained as follows:

17 = a(30² - 60 x 30 + 576)

a = 17/(30² - 60 x 30 + 576)

a = -0.05

Hence the equation is:

y = -0.05(x² - 60x +  576).

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Answer:

S₁₀ = - 838860

Step-by-step explanation:

the first term a₁ = 4

r = [tex]\frac{a_{2} }{a_{1} }[/tex] = [tex]\frac{-16}{4}[/tex] = - 4

substitute these values into [tex]S_{n}[/tex] , then

S₁₀ = [tex]\frac{4-4(-4)^{10} }{1-(-4)}[/tex]

     = [tex]\frac{4-4(1048576)}{1+4}[/tex]

     = [tex]\frac{4-4194304}{5}[/tex]

     = [tex]\frac{-4194300}{5}[/tex]

     = - 838860

The curvature of y = 5x⁴ is kappa (x) = |60x²|/[1 + (20x³)²]³/².

To find the curvature (kappa) of the function y = 5x⁴, we'll use the formula kappa (x) = |f"(x)|/[1 + (f'(x))²]³/².

1. First, find the first derivative (f'(x)) by differentiating y with respect to x: f'(x) = 20x³.
2. Next, find the second derivative (f"(x)) by differentiating f'(x) with respect to x: f"(x) = 60x².
3. Substitute f'(x) and f"(x) into the curvature formula: kappa (x) = |60x²|/[1 + (20x³)²]³/².
4. Simplify the expression to get the curvature kappa(x).

To find the curvature at a specific point, substitute the x-value into kappa(x) and evaluate the expression.

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It took Trixie 200 minutes to finish her homework.

To calculate the time Trixie took to do her homework, we can subtract the starting time from the ending time.

The starting time is 5:30pm, which is equal to 5 x 60 + 30 = 330 minutes after midnight.

The ending time is 8:50pm, which is equal to 8 x 60 + 50 = 530 minutes after midnight.

To find the duration, we can subtract the starting time from the ending time:

530 minutes - 330 minutes = 200 minutes

Therefore, it took Trixie 200 minutes to finish her homework.

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a) We can say that the point P(6,5) lies on the hyperbola.

b) The quantity (F1P − F2P)2 is approximately 122.5.

(a) To verify that the point P(6,5) lies on the hyperbola, we need to substitute x=6 and y=5 into the equation of the hyperbola and see if the equation holds true.

So, substituting x=6 and y=5, we get:

5(6)^2 - 4(5)^2 = 80

180 - 100 = 80

80 = 80

Since the equation holds true, we can say that the point P(6,5) lies on the hyperbola.

(b) To compute (F1P − F2P)2, we need to first find the coordinates of the foci F1 and F2.

5x^2 - 4y^2 = 80 can be rewritten as (x^2)/(16) - (y^2)/(20) = 1, where a^2=16 and b^2=20.

The distance between the center (0,0) and the foci is c=√(a^2+b^2)=√(336)/2. So, the foci lie on the x-axis and have coordinates (±c,0).

Therefore, F1 has coordinates (√(336)/2,0) and F2 has coordinates (-√(336)/2,0).

Now, we can calculate the distance between P(6,5) and each focus using the distance formula.

F1P = √((6-√(336)/2)^2 + (5-0)^2) ≈ 3.26

F2P = √((6+√(336)/2)^2 + (5-0)^2) ≈ 13.92

So, (F1P − F2P)^2 = (3.26 - 13.92)^2 ≈ 122.5.

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Answer:

35.03 inches

Step-by-step explanation:

We Know

A tablecloth has a circumference of 220 inches.

Circumference of circle = 2 · r · π

C = 220 inches

π = 3.14

What is the radius of the tablecloth?

We Take

220 = 2 · r · 3.14

110 = r · 3.14

r ≈ 35.03 inches

So, the radius of the tablecloth is about 35.03 inches.

The result for this percentage question is deducting 52 from 10.4 is -41.6.

How much is a percentage?

A rate, number, or amount in each hundred is referred to as a percentage. Although "pct," "pct," and occasionally "pc" are also used as abbreviations, the percent symbol "%" is most usually used to denote it.

A % lacks a measurement unit and is a dimensionless (pure) number

An accepted quantity that is used to represent a physical quantity is called a measurement unit. The factor used to represent how many instances of a given physical property there are is the standard quantity of that property.

You may get 10.4 by multiplying 52 by 0.2 (20% as a decimal),

20/100=0.2

which is 52 times 20%.

The result of deducting 52 from 10.4 is -41.6.

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What is the value of 52 times 20% minus 52?

Step-by-step explanation:

5x/20 = 45/36

x/4=5/4

x=5×4/4

x=5

hope it helps

[tex]\sf x_{1} =2;\\ \\x_{2} =-6.[/tex]

Assuming that the exercise asks to find the roots or solutions to this equation, this would the process for doing so:

Standard form: [tex]\sf ax^{2} +bx+c=0[/tex]

This equation is already written in standard form so we can skip this step, but it's important to always make sure we have the equation well written for this method.

So the coefficients are just the numbers that myltiply the different values in the formula.

For example:

Coefficient "a" is the number that multiplies "x²" within the standard form of the equation. In this case, x² is being multiplied by number "2", that's the reason we have "2x²". Thus, the value for the "a" coefficient is 2.

Note: If you only have "x²" on your standard equation, the "a" coefficient is 1.

Coefficient "b"= 8, because "x" is being multiplied by 8 on the standard equation,

Coefficient "c"= -24, because -24 is the last number before the equal symbol in the standard form of the equation.

Quadratic formula: [tex]\sf \dfrac{-b+-\sqrt{b^{2}-4ac } }{2a}[/tex]

Here, we substitute the a, b and c variables within the equation by the identified coefficients in step 2.

[tex]\sf x_{1} =\sf \dfrac{-b+\sqrt{b^{2}-4ac } }{2a}=\sf \dfrac{-(8)+\sqrt{(8)^{2}-4(2)(-24) } }{2(2)}=2[/tex]

[tex]\sf x_{2} =\sf \dfrac{-b-\sqrt{b^{2}-4ac } }{2a}=\sf \dfrac{-(8)-\sqrt{(8)^{2}-4(2)(-24) } }{2(2)}=-6[/tex]

[tex]\sf x_{1} =2;\\ \\x_{2} =-6.[/tex]

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[tex] \sf{x = 2, - 6}[/tex]

Topic: Quadratic formula exercises

[tex] \: \: \: \: \: \: \: \: \: \: \: \sf2(x {}^{2} + 4x - 12) = 0[/tex]

[tex] \: \: \: \: \: \: \: \: \: \: \: \: \sf{}2(x - 2)(x + 6) = 0[/tex]

[tex] \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf{}x = 2, - 6[/tex]

[tex] \: \: \: \: \: \: \: \: \: \: \: \boxed{ \bold{\cfrac{ - b + - \sqrt{b {}^{2} - 4ac} }{2a} }}[/tex]

In this exercise, what was done was to extract common factors, then we must multiply and subtract what is inside the parentheses and, as a last step, clear as a function of "x".

But in the exercise I solved it in another way since it is easier than doing it in fraction.

But his quadratic formula of the problem is:

[tex] \: \: \: \: \: \: \: \: \: \: \: \: \: \boxed{ \cfrac{ \sf - b + - \sqrt{b {}^{2} - 4ac } }{ \sf2a} }[/tex]

Therefore, the result of the quadratic formula is: x -2, -6.

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For the first equation, we start by assuming a solution of the form y(t) = t^r. Then, we can take the derivative of y(t) twice to get:

y'(t) = rt^(r-1)
y''(t) = r(r-1)t^(r-2)

Substituting these into the original equation, we get:

3t^2(r(r-1)t^(r-2)) - 15t(rt^(r-1)) + 27t^r = 0

Simplifying, we can divide through by t^r and factor out a common factor of 3r(r-1):

3r(r-1) - 15r + 27 = 0

This simplifies to:

r^2 - 5r + 9 = 0

Using the quadratic formula, we find that r = (5 +/- sqrt(7)i)/2. Since the equation is homogeneous, we know that the general solution must be a linear combination of the two independent solutions:

y(t) = c1*t^(5/2) + c2*t^(3/2)

However, since t < 0, we need to use the absolute value of t to get the general solution:

y(t) = c1*|t|^(5/2) + c2*|t|^(3/2)

Finally, we can simplify this to:

y(t) = -t^3 [c1 + c2 ln(-t)]

For the second equation, we can use the same method of assuming a solution of the form y(x) = x^r and taking derivatives to get:

y'(x) = rx^(r-1)
y''(x) = r(r-1)x^(r-2)

Substituting these into the original equation, we get:

x^2(r(r-1)x^(r-2)) - x(rx^(r-1)) + 5x^r = 0

Simplifying, we can divide through by x^r and factor out a common factor of r(r-1):

r(r-1) - r/x + 5 = 0

This simplifies to:

r^2 - r(1/x) + 5 = 0

Using the quadratic formula, we find that r = (1/x +/- sqrt(4-20x^2))/2. Since the equation is homogeneous, we know that the general solution must be a linear combination of the two independent solutions:

y(x) = c1*x^(1/2 + sqrt(4-20x^2)/2) + c2*x^(1/2 - sqrt(4-20x^2)/2)

We can simplify this to:

y(x) = x [c1 cos (2 ln x) + c2 sin (2 ln x)]

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To determine the probability of an event where both dice are odd, let's first list all the possible odd numbers on a die: {1, 3, 5}.

Probability is a measure of the likelihood or chance that a particular event will occur. It is expressed as a number between 0 and 1, with 0 indicating that an event is impossible and 1 indicating that an event is certain to occur.

Now, let's find all the combinations of two dice showing odd numbers:

1. (1, 1) 2. (1, 3) 3. (1, 5) 4. (3, 1) 5. (3, 3) 6. (3, 5) 7. (5, 1) 8. (5, 3) 9. (5, 5)

There are a total of 9 combinations where both dice show odd numbers.

So, the size of the set that corresponds to the event that both dice are odd is 9.

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The possible zeros of the polynomial are given as follows:

± 1/3, ± 2/3, ± 1, ±4/3, ± 2, ±8/3, ±3, ± 4, ± 6, ± 8, ± 12, ± 24.

To obtain the possible rational zeros of the function, we use the Rational Zero Theroem.

The rational zero theorem states that all the possible rational zeros of a function are given by plus/minus the factors of the constant by the factors of the leading coefficient.

The parameters for this function are given as follows:

The factors are given as follows:

Hence the possible zeros are given as follows:

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The w⊥ is the trivial subspace, consisting only of the zero vector.

To find a basis for the subspace w⊥, we need to find the vectors that are orthogonal to all vectors in w, which is the subspace spanned by the given vectors.

First, we need to find a basis for w. We can do this by putting the given vectors into a matrix and reducing it to row echelon form.

[tex]\begin{pmatrix}-4 & -4 & -12 & -4 \ 2 & 2 & 6 & 2 \ 6 & -12 & 18 & 12\end{pmatrix} $\to$[/tex]

[tex]\begin{pmatrix}2 & 2 & 6 & 2 \ 0 & -8 & -24 & -8 \ 0 & 0 & 0 & 0\end{pmatrix}[/tex]

The row echelon form shows that the first two vectors are linearly independent, so we can take them as a basis for w:

w1 = [-4, -4, -12, -4] and w2 = [2, 2, 6, 2]

Next, we need to find the vectors that are orthogonal to both w1 and w2. To do this, we can set up a system of equations:

a(-4,-4,-12,-4) + b(2,2,6,2) + c(0,0,0,0) = (0,0,0,0)

Simplifying the equation, we get:

-4a + 2b = 0

-4a + 2b = 0

-12a + 6b = 0

-4a + 2b = 0

We can see that the first two rows are identical, so we only need to use the first two rows to find a basis for w⊥.

Solving the first two equations, we get:

a = b/2

Substituting this into the third equation, we get:

-12(b/2) + 6b = 0

-6b + 6b = 0

b = 0

So a = 0 as well. This means that the only vector that is orthogonal to both w1 and w2 is the zero vector, which is not a valid basis vector.

Therefore, w⊥ is the trivial subspace, consisting only of the zero vector.

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a. The degree of the polynomial is 6.

b. Factoring the equation, we have:

x6 - 49x^4 = x^4(x^2 - 49) = x^4(x - 7)(x + 7)

a.The degree of the polynomial equation x^6 - 49x^4 = 0 is 6. This is determined by the highest exponent of x in the polynomial, which is 6.

b. The two roots of multiplicity 1 can be found by factoring the equation as x^4(x^2 - 49) = 0. Setting each factor equal to zero, we have x^4 = 0 and x^2 - 49 = 0.

From x^4 = 0, we find the root x = 0 with multiplicity 4.

From x^2 - 49 = 0, we get (x - 7)(x + 7) = 0. Therefore, the roots x = 7 and x = -7 each have multiplicity 1.

In summary, the equation x^6 - 49x^4 = 0 has a degree of 6, and the roots with multiplicity 1 are x = 0, x = 7, and x = -7.

So the roots of the equation are:

x = 0 (multiplicity 4)

x = 7 (multiplicity 1)

x = -7 (multiplicity 1)

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The expressions which are equivalent to (k^1/8)^-1 as required by virtue of the laws of indices are; k^-⅛, 1 / k^⅛ and 1 / ⁸√k.

It follows from the task content that the expressions which are equivalent to the given expression are to be determined.

Given; (k^1/8)^-1

By the power of power law of indices; we have;

= k^-⅛

Also, by the negative exponent rule; we have;

= 1 / k^⅛.

Also, by the rational exponent law of indices; we have;

= 1 / ⁸√k.

Ultimately, the equivalent expressions are; k^-⅛, 1 / k^⅛ and 1 / ⁸√k.

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The Romberg array is a table of values that is used to estimate the value of a definite integral. To compute the Romberg array, we use the Richardson extrapolation method, which is a process of successive approximation.

To compute the eight rows and columns of the Romberg array, we begin by splitting the integration interval into two equal-length subintervals. The trapezoidal method is then applied to each subinterval to produce two estimates of the integral. The Richardson extrapolation method is then used to get a better estimate of the integral based on these two estimations. This operation is continued, splitting the subintervals into smaller and smaller subintervals, until the Romberg array has the necessary number of rows and columns.

The Romberg array's general formula is as follows:

R(m,n) = (4^n R(m,n-1) - R(m-1,n-1)) / (4^n - 1)

where R(m,n) is the value of the integral estimate at row m and column n in the Romberg array.

The first column of the Romberg array contains the estimates obtained by the trapezoidal rule, while the subsequent columns are obtained by applying the Richardson extrapolation method using the values in the previous column.

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it seems that it starts from 4 then every time it gets multiplied by 3 so F(n)=4*3^n-1

Answer:

To find the volume of the container, we need to use the formula for the volume of a cylinder, which is:

V = πr^2h

where V is the volume, r is the radius, and h is the height.

Since the diameter of the ball is 9.60 inches, the radius is half of that, or 4.80 inches.

Since the height of the container is the same as the diameter of the ball, the height is also 9.60 inches.

Substituting the values into the formula, we get:

V = π(4.80)^2(9.60)

V ≈ 661.95 cubic inches

Therefore, the volume of the container is approximately 661.95 cubic inches.

It will take 5 years to recoup the extra cost of buying the Prius.

The number of years it will take to recoup the extra cost of buying the Prius will depend on several factors such as the price of the car, the cost of gas, and the average number of miles driven per year. However, according to a study by Consumer Reports, the Prius has an average payback period of about 4 years compared to a similar gas-powered vehicle. This means that if the extra cost of buying the Prius is $4,000, for example, it would take about 4 years to recoup that cost through fuel savings. Keep in mind that this is just an estimate and individual results may vary.
To determine the number of years it will take to recoup the extra cost of buying the Prius, follow these steps:

1. Identify the extra cost of buying the Prius compared to a similar non-hybrid vehicle.
2. Determine the annual fuel cost savings of the Prius compared to the non-hybrid vehicle.
3. Divide the extra cost by the annual fuel cost savings.

For example, let's say the extra cost of buying the Prius is $5,000 and the annual fuel cost savings is $1,000.

Number of years to recoup extra cost = Extra cost / Annual fuel cost savings
Number of years = $5,000 / $1,000
Number of years = 5.00

So, it will take 5.00 years to recoup the extra cost of buying the Prius.

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The area under the graph of f(x) = 1/x+1 over the interval [0,4] is approximately 0.9375.

In mathematics, "area" refers to the measure of the amount of space enclosed by a two-dimensional shape or region. It is a quantitative measure of the extent or size of a shape in terms of its length squared. Area is typically expressed in square units, such as square meters (m^2), square feet (ft^2), or square centimeters (cm^2), depending on the system of measurement used.

A rectangle is a quadrilateral with four right angles, where opposite sides are parallel and equal in length.

To estimate the area under the graph of the function f(x) = 1/(x+1) over the interval [0,4], we can use numerical integration methods such as the trapezoidal rule or Simpson's rule.

Let's use the trapezoidal rule, which approximates the area under a curve by dividing the interval into smaller trapezoids and summing their areas.

Divide the interval [0,4] into n equal subintervals.

Let's choose n = 4 for this example, which means we will have 4 subintervals of equal width. The width of each subinterval is given by Δx = (4-0)/4 = 1.

Compute the sum of the areas of the trapezoids.

The area of each trapezoid is given by the formula: (h/2) * (f(x_i) + f(x_{i+1})), where h is the width of the subinterval, f(x_i) is the value of the function at the lower endpoint, and f(x_{i+1}) is the value of the function at the upper endpoint.

Using the trapezoidal rule, we can estimate the area under the curve as follows:

Area ≈ (1/2) * (f(0) + f(1)) * 1 + (1/2) * (f(1) + f(2)) * 1 + (1/2) * (f(2) + f(3)) * 1 + (1/2) * (f(3) + f(4)) * 1

Plugging in the function f(x) = 1/(x+1) and evaluating at the endpoints, we get:

Area ≈ (1/2) * (1 + 1/2) * 1 + (1/2) * (1/2 + 1/3) * 1 + (1/2) * (1/3 + 1/4) * 1 + (1/2) * (1/4 + 1/5) * 1

Simplifying further, we get:

Area ≈ 0.9375

So, the estimated area under the graph of the function f(x) = 1/(x+1) over the interval [0,4] using the trapezoidal rule is approximately 0.9375 square units.

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Polynomial equation has y=cos(x), y=sin(x), and y=eˣ as solutions.

We can consider these functions as roots of a polynomial. Let's use the terms given to construct a polynomial equation:

Let P(y) be the polynomial, and let's denote the roots as y1 = cos(x), y2 = sin(x), and y3 = eˣ.

According to Vieta's formulas, for a cubic polynomial with roots y1, y2, and y3, we have:

P(y) = (y - y1)(y - y2)(y - y3)

Now, substitute the given roots:

P(y) = (y - cos(x))(y - sin(x))(y - eˣ)

This polynomial equation has y=cos(x), y=sin(x), and y=eˣ as solutions.

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Okay, let's solve this step-by-step:

1) The equations for the two curves are:

r = sin(θ)  and  r = cos(θ)

2) We need to find the intersection points of these two curves. This is done by setting them equal and solving for θ:

sin(θ) = cos(θ)

=>  θ = π/4

3) The intersection points are (1, π/4) and (1, 3π/4). The region lies between θ = π/4 and θ = 3π/4.

4) To find the area, we use the formula:

A = ∫θ=3π/4 θ=π/4 2πr dθ

5) Substitute r = sin(θ) or r = cos(θ):

A = ∫θ=3π/4 θ=π/4 2πsin(θ) dθ

= 2π ∫θ=3π/4 θ=π/4 sin(θ) dθ

6) Integrate:

A = 2π(cos(θ) - sin(θ) )|π/4  to  3π/4

= 2π(0 - 1) = 2π

7) Therefore, the area of the region is 2π square units.

Let me know if you have any other questions!

The two points on the curve where the tangent line is horizontal are (2,26) and (-2,26).

To find where the tangent line is horizontal, we need to find where the derivative of y with respect to x (dy/dx) equals 0.

First, we need to express y in terms of x. We can do this by eliminating t from the two parametric equations.

From x=2(sin(t)+cos(t)), we get sin(t) = (x/2) - cos(t).
From y=36−10cos2(t)−20sin(t), we substitute sin(t) with the above expression and get:
y = 36 - 10cos²(t) - 20((x/2) - cos(t))

Simplifying this expression, we get:
y = -10cos²(t) - 10x + 36

Next, we need to find the derivative of y with respect to x:
dy/dx = -10sin(2t)/(dx/dt)

From x=2(sin(t)+cos(t)), we get dx/dt = 2(cos(t)-sin(t))

Substituting this into the above equation for dy/dx, we get:
dy/dx = -5sin(2t)/(cos(t)-sin(t))

Setting dy/dx equal to 0, we get:
0 = -5sin(2t)/(cos(t)-sin(t))

This means sin(2t) = 0, or t = 0 or t = π/2.

Plugging these values into the parametric equations for x and y, we get:
When t=0: x = 2, y = 26
When t=π/2: x = -2, y = 26

Thus, the two points on the curve where the tangent line is horizontal are (2,26) and (-2,26).

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Refer to the attached images. Comment any questions you may have.

Answer:

Below in bold.

Step-by-step explanation:

Let x = cosθ, then

8(cosθ)^3 − 6cosθ + 1 = 0

---> 2(4(cosθ)^3 − 3 cosθ) + 1 = 0    

---> 2(cos3θ) + 1 = 0

---> cos3θ = -1/2

---> θ = 2π/9

Therefore cos  θ  = = cos(2π/9) = x, and

cos(2π/9) is a root of the given eqation.