MATHEMATICS COLLEGE

Which of the following statements about the image below is true?

Answers

Answer 1

Answer:

Answer:

d. Line UR and Line VW are parallel

Step-by-step explanation:

If they were to continue going straight, they would not touch, making them parallel.

I hope this helps!

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Sketch the region R defined by 1 ≤ x ≤ 2 and 0 ≤ y ≤ 1/x^3 .a. Find (exactly) the number a such that the line x = a divides R into two parts of equal area.
b. Then find (to 3 decimal places) the number b such that the line y = b divides R into two parts of equal area.

Answers

For part (a), you're looking to find $a$ such that

$\displaystyle\int_1^a(\mathrm dx)/(x^3)=\int_a^2(\mathrm dx)/(x^3)$

You have

$\displaystyle\int_1^a(\mathrm dx)/(x^3)=-\frac1{2x^2}\bigg|_(x=1)^(x=a)=-\frac12\left(\frac1{a^2}-1\right)$

and

$\displaystyle\int_a^2(\mathrm dx)/(x^3)=-\frac1{2x^2}\bigg|_(x=a)^(x=2)=-\frac12\left(\frac14-\frac1{a^2}\right)$

Setting these equal, you get

$\displaystyle-\frac12\left(\frac1{a^2}-1\right)=-\frac12\left(\frac14-\frac1{a^2}\right)\implies a=2√(\frac25)$

For part (b), you have

$y=\frac1{x^3}\implies x=\frac1{\sqrt[3]y}$

and you want to find $b$ such that

$\displaystyle\int_0^(1/8)\mathrm dy+\int_(1/8)^b(\mathrm dy)/(\sqrt[3]y)=\int_b^1(\mathrm dy)/(\sqrt[3]y)$

You have

$\displaystyle\int_0^(1/8)\mathrm dy+\int_(1/8)^b(\mathrm dy)/(y^(1/3))=\frac18+\frac32y^(2/3)\bigg|_(y=1/8)^(y=b)=-frac14+\frac32b^(2/3)$

and

$\displaystyle\int_b^1(\mathrm dy)/(y^(1/3))=\frac32y^(2/3)\bigg|_(y=b)^(y=1)=\frac32-\frac32b^(2/3)$

Setting them equal gives

$-\frac14+\frac32b^(2/3)=\frac32-\frac32b^(2/3)\implies b=\frac7{24}√(\frac73)\approx0.446$

What is the value of M

Answers

Answer:....... no clue ut pls mark me brainiest

Step-by-step explanation:

A rock thrown vertically upward from the surface of the moon at a velocity of 36m/sec reaches a height of s = 36t - 0.8 t^2 meters in t sec.a. Find the rock's velocity and acceleration at time t.
b. How long does it take the rock to reach its highest point?
c. How high does the rock go?
d. How long does it take the rock to reach half its maximum height?
e. How long is the rock a loft?

Answers

Answer:

a. The rock's velocity is $v(t)=36-1.6t \:{(m/s)}$ and the acceleration is $a(t)=-1.6 \:{(m/s^2)}$

b. It takes 22.5 seconds to reach the highest point.

c. The rock goes up to 405 m.

d. It reach half its maximum height when time is 6.59 s or 38.41 s.

e. The rock is aloft for 45 seconds.

Step-by-step explanation:

Velocity is defined as the rate of change of position or the rate of displacement. $v(t)=(ds)/(dt)$
Acceleration is defined as the rate of change of velocity. $a(t)=(dv)/(dt)$

The rock's velocity is the derivative of the height function $s(t) = 36t - 0.8 t^2$

$v(t)=(d)/(dt)(36t - 0.8 t^2) \n\n\mathrm{Apply\:the\:Sum/Difference\:Rule}:\quad \left(f\pm g\right)'=f\:'\pm g'\n\nv(t)=(d)/(dt)\left(36t\right)-(d)/(dt)\left(0.8t^2\right)\n\nv(t)=36-1.6t$

The rock's acceleration is the derivative of the velocity function $v(t)=36-1.6t$

$a(t)=(d)/(dt)(36-1.6t)\n\na(t)=-1.6$

b. The rock will reach its highest point when the velocity becomes zero.

$v(t)=36-1.6t=0\n36\cdot \:10-1.6t\cdot \:10=0\cdot \:10\n360-16t=0\n360-16t-360=0-360\n-16t=-360\nt=(45)/(2)=22.5$

It takes 22.5 seconds to reach the highest point.

c. The rock reach its highest point when t = 22.5 s

Thus

$s(22.5) = 36(22.5) - 0.8 (22.5)^2\ns(22.5) =405$

So the rock goes up to 405 m.

d. The maximum height is 405 m. So the half of its maximum height = $(405)/(2) =202.5 \:m$

To find the time it reach half its maximum height, we need to solve

$36t - 0.8 t^2=202.5\n36t\cdot \:10-0.8t^2\cdot \:10=202.5\cdot \:10\n360t-8t^2=2025\n360t-8t^2-2025=2025-2025\n-8t^2+360t-2025=0$

For a quadratic equation of the form $ax^2+bx+c=0$ the solutions are

$x_(1,\:2)=(-b\pm √(b^2-4ac))/(2a)$

$\mathrm{For\:}\quad a=-8,\:b=360,\:c=-2025:\n\nt=(-360+√(360^2-4\left(-8\right)\left(-2025\right)))/(2\left(-8\right))=(45\left(2-√(2)\right))/(4)\approx 6.59\n\nt=(-360-√(360^2-4\left(-8\right)\left(-2025\right)))/(2\left(-8\right))=(45\left(2+√(2)\right))/(4)\approx 38.41$

It reach half its maximum height when time is 6.59 s or 38.41 s.

e. It is aloft until s(t) = 0 again

$36t - 0.8 t^2=0\n\n\mathrm{Factor\:}36t-0.8t^2\rightarrow -t\left(0.8t-36\right)\n\n\mathrm{The\:solutions\:to\:the\:quadratic\:equation\:are:}\n\nt=0,\:t=45$

The rock is aloft for 45 seconds.

A plane leaves the airport in galisteo and flies 170 km at 68 degrees east of north; then it changes direction to fly 230 km at 36 degrees south of east, after which it makes an immediate emergency landing in a pasture. When the airport sends out a rescue crew, in which direction and how far should this crew fly to go directly to this plane?

Answers

Answer:

4.68° south east 317.36 km

Step-by-step explanation:

We can find the angle between the two distances (vectors) because according to the diagram, we can draw two right triangles between them.

The complement of the 36 degree angle is 54 (90-36=54), and the complement of the 68 angle is 22, (90-68=22) the sum of 22 and 54 is 76. So the angle between the two distances is 76.

Then we apply the cosine law

$b^(2) =a^(2) +c^(2) -2*a*c*cosB\n \nb^(2) =230^(2) +170^(2) -2*230*170*cos(76)\n\nb=\sqrt{230^(2) +170^(2) -2*230*170*cos(76)} \n\nb=317.36 km$

then we apply the sin law

$(sin(C))/(c) =(sin(B))/(b) \n\nsin(C)=c*(sin(B))/(b)\n\nsin(C)=170*(sin(76))/(317.36)\n\nsin(C)=0.52\n\narcsin(0.52)=C=31.32\n\n$

and because in any triangle, the sum of the inside angles is equal to 180

$180=76+31.32+68+y\n\ny=180-76-31.32-68\n\ny=4.68^(o)$

180= 76+C+(68+Y)

y=180-76-C-68

So the emergency plane has to travel 317.36 km, 4.68° southeast.

Triangle RST has the vertices R(2, 3), S(-2, 1), and T(-1, 5). What are the coordinates after the two transformations: Translation (x, y) --> (x - 2, y - 1) Rotation: 90 degrees counterclockwise at the origin.Immersive Reader