Ask question

Ask question

All categories

Oduvanchick [21]

3 years ago

15

SOMEONE MAKE A ENDING PARAGRAPH about someone rescuing me QUICK 40 POINTS!!!!

1 answer:

scZoUnD [109]3 years ago

4 0

Answer:

Although, my life can't end here, life has so much more to offer me. I collected the last of my strength to swim up to to surface where I can be released from the horrors of the deep waters. I swallowed gulps of the chlorine water and my eyes ached from being open for to long. I wrestled the clenching fists of the water that urged me to stay with them. I swam, and swam, swam, until I reached my hands out for help, and that's when I knew I made it. I could feel hands from above grasping me and helping me climb out of the bitter cold pool. I ran into my mother's arms, as a gush of relief spread through my body. The relief of staying in my mothers arm, that helped me forget the isolated pits of the dark waters.

Sorry if it's not that good! I'm not the best writer, and you don't have to use this ending paragraph if you don't want to!!

You might be interested in

Consider the curve of the form y(t) = ksin(bt2) . (a) Given that the first critical point of y(t) for positive t occurs at t = 1

mafiozo [28]

Answer:

(a). y'(1)=0 and y'(2) = 3

(b). $y'(t)=kb2t\cos(bt^2)$

(c). $b = \frac{\pi}{2} \text{ and}\ k = \frac{3}{2\pi}$

Step-by-step explanation:

(a). Let the curve is,

$y(t)=k \sin (bt^2)$

So the stationary point or the critical point of the differential function of a single real variable , f(x) is the value $x_{0}$ which lies in the domain of f where the derivative is 0.

Therefore, y'(1)=0

Also given that the derivative of the function y(t) is 3 at t = 2.

Therefore, y'(2) = 3.

(b).

Given function, $y(t)=k \sin (bt^2)$

Differentiating the above equation with respect to x, we get

$y'(t)=\frac{d}{dt}[k \sin (bt^2)]\\ y'(t)=k\frac{d}{dt}[\sin (bt^2)]$

Applying chain rule,

$y'(t)=k \cos (bt^2)(\frac{d}{dt}[bt^2])\\ y'(t)=k\cos(bt^2)(b2t)\\ y'(t)= kb2t\cos(bt^2)$

(c).

Finding the exact values of k and b.

As per the above parts in (a) and (b), the initial conditions are

y'(1) = 0 and y'(2) = 3

And the equations were

$y(t)=k \sin (bt^2)$

$y'(t)=kb2t\cos (bt^2)$

Now putting the initial conditions in the equation y'(1)=0

$kb2(1)\cos(b(1)^2)=0$

2kbcos(b) = 0

cos b = 0 (Since, k and b cannot be zero)

$b=\frac{\pi}{2}$

And

y'(2) = 3

$\therefore kb2(2)\cos [b(2)^2]=3$

$4kb\cos (4b)=3$

$4k(\frac{\pi}{2})\cos(\frac{4 \pi}{2})=3$

$2k\pi\cos 2 \pi=3$

$2k\pi(1) = 3$$

$k=\frac{3}{2\pi}$

$\therefore b = \frac{\pi}{2} \text{ and}\ k = \frac{3}{2\pi}$

7 0

4 years ago

What is 9 - 3 divided by 1/3 + 1 <br><br> (Marking brainlest also doing random problems)

Lisa [10]

Answer:

the answer is 1 if i read this right

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

A forest ranger can see for a distance of 12 miles from a firetower.How many square miles can he observe?

mr_godi [17]

I think it is 12x12 = 144

7 0

4 years ago

PLEASE ANSWER WILL MARK BRAINIEST IF CORRECT :)

katrin [286]

1) The set of elements is,

10n where n € {.....,-3,-2,-1,1,2,3....}

Not sure if the 2nd question is complete.

3 0

3 years ago

Mr. Khans is buying staplers for his office. Each stapler costs $16.99. Part A: What does his final total cost depend upon? Part

melamori03 [73]

Answer:

his cost depends on how many staplers he gets and the input is how much each stapler cost and output is how much he has to spend

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers