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choli [55]
2 years ago
5

Suppose F⃗ (x,y)=⟨ex,ey⟩F→(x,y)=⟨ex,ey⟩ and CC is the portion of the ellipse centered at the origin from the point (0,1)(0,1) to

the point (7,0)(7,0) centered at the origin oriented clockwise. (a) Find a vector parametric equation r⃗ (t)r→(t) for the portion of the ellipse described above for 0≤t≤π/20≤t≤π/2.
Mathematics
1 answer:
Rudik [331]2 years ago
7 0

Probably the intended ellipse is the one with equation

\dfrac{x^2}{49}+y^2=1

We can parameterize C as a piece of this curve by

\vec r(t)=\langle7\sin t,\cos t\rangle

with 0\le t\le\frac\pi2. Then

\displaystyle\int_C\vec F\cdot\mathrm d\vec r=\int_0^{\pi/2}\langle e^{7\sin t},e^{\cos t}\rangle\cdot\langle7\cos t,-\sin t\rangle\,\mathrm dt

etc

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Triangle ABC has vertices of A(–6, 7), B(4, –1), and C(–2, –9). Find the length of the median from angle B in triangle ABC
julsineya [31]

Answer:

  8

Step-by-step explanation:

The median is the segment from vertex B to the midpoint of AC. That midpoint (D) is ...

  D = (A + C)/2 = ((-6, 7) +(-2, -9))/2 = (-8, -2)/2

  D = (-4, -1)

The length of the midpoint is the length of the segment DB between (-4, -1) and (4, -1). These points both have the same y-coordinate, so the length is the difference of x-coordinates: 4 -(-4) = 8.

7 0
2 years ago
Simplify the equation -2(p + 4)2 - 3 + 5p. What is the simplified expression in standard form? ASAP PLEASE!! ILL GIVE EXTRA POIN
Katena32 [7]
So you distrubute and get

-2p-8+2-3+5p 
add like terms 
3p-9
ANSWER: 3p-9
4 0
2 years ago
Read 2 more answers
A boat on the ocean is 2 mi from the nearest point on a straight? shoreline; that point is 15 mi from a restaurant on the shore.
Sladkaya [172]

Answer:

a) x  =  0,70 miles

T  =  1/2* (√4 + x²)  +( 15 - x )/3   Objective Function to minimize

b) V(min)  = 2.59 m/hr

Step-by-step explanation:

Let assume the boat is in Point A,  she lands in point B, and R  is the restaurant

Let call  x distance between the point O in  which perpendicular line from the boat get to shoreline, and the point were she land.

We know  distance is

d = v*t       ⇒  t = d/v

She rows at 2 miles/hr  and walk at 3 miles /hr

According to that she takes

c (hypotenuse) of right triangle  AOB/ 2 rowing

and  15 - x /3  walking

Total time is:

T  =  c/2  + ( 15-x )/3          c =√(2)² + x²     ⇒  T = [√(2)² + x²  ] /2  +  ( 15-x )/3  

T  =  1/2* (√4 + x²)  +( 15 - x )/3   (1)

And that is  the Objective function to minimize

a) Taking derivatives on both sides of the equation we get

T´(t)  =  x /( √4 + x²)  - 1/3    ⇒   T´(t)  = 0     x /( √4 + x²)  - 1/3 = 0

3*x - √( √4 + x²) = 0     ⇒  3*x  =  √( √4 + x²)

Squaring both sides

9x²  =  4 + x²     ⇒  8x²  = 4       x² = 1/2      x  =  0,70 miles

If we plug this value in the Objective function we will get the minimum time

1/2* (√4 + x²)  +( 15 - x )/3    ⇒ [√ 4  + 0,5] /2 +  14,3/3 = T (min)

T(min)  =  1.06 + 4.77  = 5.83 hr

Distance L (hypotenuse of right triangle AOR)

L = √(2)²  + (15)²   L  = 15,13 miles

And that distance have to be traveled at least in 5.83 hr rowing

Then  as  v = d/t      V(min)  =  15.13/ 5.83   ⇒    V(min)  = 2.59 m/hr

6 0
2 years ago
in a random sample of 200 people, 154 said that they watched educational television. fine the 90% confidence interval of the tru
melisa1 [442]

Answer:

[0.7210,0.8189] = [72.10%, 81.89%]

Step-by-step explanation:

The sample size is  

n = 200

the proportion is

p = 154/200 = 0.77

<em>Since both np ≥ 10 and n(1-p) ≥ 10 </em>

<em>We can approximate this discrete binomial distribution with the continuous Normal distribution. As the sample size is large enough, not applying the continuity correction factor makes no significant  difference. </em>

The approximation would be to a Normal curve with this parameters:

<em>Mean </em>

p = 0.77

<em>Standard deviation </em>

\bf s=\sqrt{\frac{p(1-p)}{n}}=\sqrt{\frac{0.77*0.23}{200}}=0.0298

The 90% confidence interval for the proportion would be then  

\bf  [0.77-z^*0.0298, 0.77+z^*0.0298]

where \bf z^* is the 10% critical value for the Normal N(0,1) distribution , this is a value such that the area under the N(0,1) curve outside the interval \bf [-z^*,z^*] is 10%=0.1

We can either use a table, a calculator or a spreadsheet to get this value.

In Excel or OpenOffice Calc we use the function

<em>NORMSINV(0.95) and we get a value of 1.645 </em>

The 90% confidence interval for the proportion is then

\bf  [0.77-1.645^*0.0298, 0.77+1.645^*0.0298]=[0.7210,0.8189]

This means there is a 90% probability that the proportion of people who watch educational television is between 72.10% and 81.89%

If the television company wanted to publicize the proportion of viewers, do you think it should use the 90% confidence interval?

Yes, I do.

5 0
2 years ago
If a plant grows 1/4 cm every day for 21 days, what is the total<br> amount of growth?
Naddika [18.5K]
I think it’s 5.25 cm but you might want to double check.
5 0
2 years ago
Read 2 more answers
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