an engineering team has a goal of installing new, more efficient solar panels on the international space station . which equation best describes a cost-benefit analysis the team might perform

Answer:

There were several technologically advanced cars in the 1990s, including the Mitsubishi 3000GT, Porsche 959, and Williams FW15C Formula One car. It is difficult to determine a single car as the "most technologically advanced" since advancements varied across different models and manufacturers. However, the 1993 Williams FW15C is often considered one of the most technologically advanced Formula One cars of all time , featuring advanced electronics, active suspension, and advanced aerodynamics. Additionally, the 90s were a golden decade for technologically advanced Japanese cars , including iconic models like the Nissan Skyline GT-R and Toyota Supra.

Explanation:

Answer:

While there were many advanced cars in the 90's the McLaren F1 was one of the most technologically advanced cars in the '90s.

Hope this helps:))

Answer:

yes

Explanation:

Yes, I can solve this equation by power series method. Here are the steps:

Assume a power series solution of the form. [tex]y = \sum_{n=0}^{\infty} a_n x^n[/tex]Differentiate term by term to get [tex]y' = \sum_{n=1}^{\infty} n a_n x^{n-1}[/tex]Substitute into the equation and simplify to get [tex]$$\sum_{n=1}^{\infty} n a_n x^{n-1} = 2 \sum_{n=0}^{\infty} a_n x^{n+2} + \sum_{n=0}^{\infty} a_n x^n$$[/tex]

Re-index the sums to have the same power of x and combine them to get                                              [tex]$$\sum_{n=0}^{\infty} [(n+1) a_{n+1} - 2 a_n x^2 - a_n] x^n = 0$$[/tex]

Equate the coefficients of each power of x to zero and solve for the recurrence relation [tex]$$a_{n+1} = \frac{2 a_n x^2 + a_n}{n+1}$$[/tex]

Use the initial conditions [tex]$y(0) = a_0$[/tex] and [tex]$y'(0) = a_1$[/tex] to find the values of [tex]$a_0$[/tex] and [tex]$a_1$[/tex]

Substitute the values of [tex]$a_0$[/tex] and [tex]$a_1$[/tex] into the recurrence relation and find the values of [tex]a_2[/tex],[tex]a_3[/tex], etc.

Write the solution as [tex]$$y = \sum_{n=0}^{\infty} a_n x^n$$[/tex]

[tex]For example, if we have $y(0) = 1$ and $y'(0) = 2$, then we get $a_0 = 1$ and $a_1 = 2$. Then we can find $a_2$, $a_3$, etc. by using the recurrence relation:a_2 = \frac{2 a_1 x^2 + a_1}{2} = \frac{5}{2}x^2a_3 = \frac{2 a_2 x^2 + a_2}{3} = \frac{25}{12}x^4a_4 = \frac{2 a_3 x^2 + a_3}{4} = \frac{125}{96}x^6[/tex]

The solution is then [tex]y = 1 + 2x + \frac{5}{2} x^{2} + \frac{25}{12} x^{4} +\frac{125}{96} x^{6}+...[/tex]

To solve the differential equation using the power series method, we can assume a power series representation for the function \(y(x)\) as:

\[y(x) = \sum_{n=0}^{\infty} a_n x^n\]

Let's differentiate this series with respect to \(x\):

\[\frac{dy}{dx} = \sum_{n=0}^{\infty} a_n n x^{n-1} = \sum_{n=0}^{\infty} a_n (n+1) x^n\]

Substituting this into the given differential equation, we get:

\[\sum_{n=0}^{\infty} a_n (n+1) x^n = 0.2x^2 + \sum_{n=0}^{\infty} a_n x^n\]

Comparing the coefficients of like powers of \(x\) on both sides, we have:

For the left side:
\(a_0\) term: \(a_1 = 0.2a_0\)
\(a_1\) term: \(2a_2 = 0.2a_1 + a_0\)
\(a_2\) term: \(3a_3 = 0.2a_2 + a_1\)

And so on. We can use these recurrence relations to find the values of the coefficients \(a_n\) one by one.

To get started, let's determine the first few coefficients:

\(a_1 = 0.2a_0\)

\(2a_2 = 0.2a_1 + a_0\)

\(3a_3 = 0.2a_2 + a_1\)

Once we have determined the values of \(a_0\), \(a_1\), \(a_2\), and \(a_3\), we can continue the process to find more coefficients using the recurrence relations.

I recommend using a computer algebra system or a software package such as MATLAB or Mathematica to automate this process and calculate the coefficients efficiently. It might be challenging to complete this calculation manually within the given time frame of 35 minutes.

From the Bow

Head slowly into the wind or current to a position upwind it up current of where you actually want to end up

HI hope this helps you

Given Data:- Diameter of rolling circle Φ = 50 mm

Normal and tangent point = 30 mm above straight line.

Assumption – Circle is rolling towards right side (Show the direction)

Procedure:-1) Draw a circle of 50 mm diameter.

2) Draw horizontal and vertical axis and mark

centre of the circle say C3)

Divide the circle into 12 equal parts and name each division 1, 2, 3, ….12 as per shown in fig.4)

Draw a straight horizontal line of length πD from point P at the contact surface of circle and ground . 5)

Divide the line into 12 equal parts 1’, 2’, 3’…..12’ (same no. as that of circle)

.6) Draw again a circle of 50 mm diameter at πD distance with centre C’.7)

Draw horizontal and vertical axis for 2nd circle.

8) Draw horizontal lines from points (1,11) (2,10) (3,9) (4,8) (5,7) and 6 up to vertical axis of 2nd circle .9) Draw vertical lines from point 1’, 2’, 3’,…. 12’ up to horizontal axis and name it C1, C2, C3….C12 respectively.10)

Taking C1 as centre and 25 mm radius (radius of rolling circle) cut the horizontal line passing through point on the circle near point P. Mark that point P1.11) Repeat the same procedure up to C12 and accordingly marks points up to P12.12) Draw smooth curve passing through all 12 points (P1, P2, ….. P12) and name the curve.13)

Mark a point M on the curve at a distance of 30 mm from horizontal line.14) Taking M as centre and 25 mm radius (radius of rolling circle) cut the horizontal axis and mark that point Q.15)

Draw perpendicular from Q on horizontal and mark it as N.16) Draw a line passing through M and N (NMN is normal).17) Draw a line perpendicular to normal from point Mtmt is required tangentPLEASE MARK AS BRAINLIEST AND LIKEHOPE THIS WILL HELP YOU

The locus of a point on the circumference of a circle as it rolls without slipping on a straight line is called a cycloid. In this case, the circle has a diameter of 46mm and makes 1½ revolutions.

To understand how the locus is formed, imagine a point P on the circumference of the rolling circle. As the circle rolls without slipping, point P traces a path on the ground.

To determine the shape of this path, we can divide the rolling motion into two parts: the horizontal and vertical components.

1. Horizontal Component:

During each revolution, the point P moves a distance equal to the circumference of the circle, which is π times the diameter. In this case, the diameter is 46mm, so the circumference is 46π mm.

Since the circle makes 1½ revolutions, the total horizontal distance covered by point P is 1½ times the circumference of the circle, which is 1½ * 46π mm.

2. Vertical Component:

The vertical distance covered by point P is equal to the diameter of the circle, which is 46mm.

Combining the horizontal and vertical components, we can plot the locus of point P. Each point on the locus represents the position of point P as the circle rolls.

To draw the locus, we can start by marking the initial position of point P at the starting point of the rolling circle. From there, we can calculate the coordinates of point P at regular intervals, corresponding to the distances covered by the circle.

For example, if we divide the total horizontal distance covered by point P into 10 equal intervals, we can calculate the x-coordinate for each interval by dividing the total distance by 10.

Similarly, since the vertical distance remains constant, the y-coordinate of point P remains the same throughout.

By plotting these coordinates, we can trace the shape of the locus of point P as a cycloid.

I hope this helps you. :)