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Alik [6]
3 years ago
15

Need help solving this​

Mathematics
1 answer:
Alex_Xolod [135]3 years ago
3 0

Answer:

15,625 feet and either 56.469 seconds or 87.719 seconds

Step-by-step explanation:

strap in because we need calculus buddy. we're given the displacement equation but we want velocity because at the bullets max height it will pause briefly before coming down and this implies zero velocity. derivative of that equation is -32t+1000. set this equal to zero and find t to see how long this takes. 31.25 seconds to reach zero velocity (the tip of the parabolic motion). now plug that t back into the displacement equation to find height after 31.25 seconds and it's like 15,625 which is ridiculously high. now it has to come down so gravity will be taking over here so it has it's own special equations and we know the equation for that is like x=1/2at^2 or something. we know x is 15,625. let's find time it takes to get down by solving that for t. 15,625÷4.9=t^2 so t=56.469 or something. now its unclear if they want the entire time elapsed from firing the gun or just the falling to ground time so if it's the entire time let's go ahead and add that initial 31.25 seconds to the 56.498 seconds so like 87.719 total seconds.

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Step-by-step explanation:

<em><u>Lets</u></em><em><u> </u></em><em><u>x</u></em><em><u> </u></em><em><u>=</u></em><em><u> </u></em><em><u>the</u></em><em><u> </u></em><em><u>width</u></em><em><u> </u></em><em><u>of</u></em><em><u> </u></em><em><u>the</u></em><em><u> </u></em><em><u>path</u></em><em><u> </u></em><em><u> </u></em><em><u>the</u></em><em><u> </u></em><em><u>surrounding </u></em><em><u>path</u></em><em><u> </u></em><em><u>wil</u></em><em><u>l</u></em><em><u> </u></em><em><u>add</u></em><em><u> </u></em><em><u>2x</u></em><em><u> </u></em><em><u>to</u></em><em><u> </u></em><em><u>the</u></em><em><u> </u></em><em><u>pool</u></em><em><u> </u></em><em><u>dimension</u></em><em><u>,therefore</u></em><em><u> </u></em><em><u>over</u></em><em><u> </u></em><em><u>all</u></em><em><u> </u></em><em><u>dimesion</u></em><em><u>:</u></em><em><u> </u></em><em><u>(</u></em><em><u>2x</u></em><em><u>+</u></em><em><u>4</u></em><em><u>0</u></em><em><u>)</u></em><em><u> </u></em><em><u>b</u></em><em><u>y</u></em><em><u> </u></em><em><u>(</u></em><em><u>2x</u></em><em><u>+</u></em><em><u>6</u></em><em><u>0</u></em><em><u>)</u></em><em><u> </u></em><em><u>the</u></em><em><u> </u></em><em><u>over</u></em><em><u>all</u></em><em><u> </u></em><em><u>perimeter</u></em><em><u> </u></em><em><u>(</u></em><em><u>2x</u></em><em><u> </u></em><em><u>+</u></em><em><u> </u></em><em><u>4</u></em><em><u>0</u></em><em><u>)</u></em><em><u> </u></em><em><u>+</u></em><em><u> </u></em><em><u>2</u></em><em><u>(</u></em><em><u>2x</u></em><em><u>+</u></em><em><u>6</u></em><em><u>0</u></em><em><u>)</u></em><em><u> </u></em><em><u>=</u></em><em><u> </u></em><em><u>2</u></em><em><u>4</u></em><em><u>8</u></em><em><u> </u></em><em><u>Simplify</u></em><em><u> </u></em><em><u>divide</u></em><em><u> </u></em><em><u>b</u></em><em><u>y</u></em><em><u> </u></em><em><u>2,</u></em><em><u> </u></em><em><u>result</u></em><em><u> </u></em><em><u>(</u></em><em><u>2</u></em><em><u>x</u></em><em><u>+</u></em><em><u>4</u></em><em><u>0</u></em><em><u>)</u></em><em><u> </u></em><em><u>+</u></em><em><u>(</u></em><em><u>2</u></em><em><u>x</u></em><em><u>+</u></em><em><u>6</u></em><em><u>0</u></em><em><u>)</u></em><em><u> </u></em><em><u>=</u></em><em><u> </u></em><em><u>1</u></em><em><u>2</u></em><em><u>4</u></em>

<em><u> </u></em><em><u>Combine</u></em><em><u> </u></em><em><u>like</u></em><em><u> </u></em><em><u>term</u></em><em><u>s</u></em><em><u> </u></em><em><u>2x</u></em><em><u> </u></em><em><u>+</u></em><em><u> </u></em><em><u>2</u></em><em><u>x</u></em><em><u> </u></em><em><u>+</u></em><em><u>4</u></em><em><u>0</u></em><em><u> </u></em><em><u>+</u></em><em><u>6</u></em><em><u>0</u></em><em><u> </u></em><em><u>=</u></em><em><u>1</u></em><em><u>2</u></em><em><u>4</u></em><em><u> </u></em>

<em><u>4x</u></em><em><u> </u></em><em><u>+</u></em><em><u> </u></em><em><u>1</u></em><em><u>0</u></em><em><u>0</u></em><em><u> </u></em><em><u>=</u></em><em><u> </u></em><em><u>1</u></em><em><u>2</u></em><em><u>4</u></em><em><u> </u></em>

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<em><u>4</u></em><em><u>x</u></em><em><u>=</u></em><em><u>2</u></em><em><u>4</u></em>

<em><u>x</u></em><em><u>=</u></em><em><u>2</u></em><em><u>4</u></em><em><u>/</u></em><em><u>4</u></em>

<em><u>x</u></em><em><u>=</u></em><em><u> </u></em><em><u>6</u></em><em><u>ft</u></em><em><u> </u></em><em><u>is</u></em><em><u> </u></em><em><u>th</u></em><em><u>e</u></em><em><u> </u></em><em><u>width</u></em><em><u> </u></em><em><u>of</u></em><em><u> </u></em><em><u>the</u></em><em><u> </u></em><em><u>path</u></em>

<em><u>check</u></em><em><u> </u></em><em><u>this</u></em><em><u> </u></em><em><u>by</u></em><em><u> </u></em><em><u>finding</u></em><em><u> </u></em><em><u>the</u></em><em><u> </u></em><em><u>perimeter</u></em><em><u> </u></em><em><u>with</u></em><em><u> </u></em><em><u>these</u></em><em><u> </u></em><em><u>values</u></em><em><u>;</u></em><em><u> </u></em><em><u>2</u></em><em><u>x</u></em><em><u> </u></em><em><u>=</u></em><em><u> </u></em><em><u>12</u></em><em><u> </u></em><em><u>ft</u></em><em><u> </u></em>

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5 0
3 years ago
use the guide to construct a two column proof proving △aec≅△deb, given that ca is parallel to db and e is the midpoint of ad. gi
MrMuchimi

By using the AAA congruence property of Triangles, △aec≅△deb is proved

It is given two triangles, Δaec and Δdeb, where e is the common point known as the midpoint of ad

Also, it is given that,

ca ║db

We need to prove that, △aec≅△deb

Then we'll use AAA congruence property of Triangles to prove the situation

As ca ║db

then, ∠cae = ∠ebd  (Alternate angles)

∠ace = ∠edb (Alternate angles)

and ∠aec = ∠deb (common angles)

Thus, by AAA congruence property, △aec≅△deb

Hence, proved

To learn more about, congruence property, here

brainly.com/question/2039214

#SPJ4

5 0
1 year ago
Find the length of AX. Please!!
valentinak56 [21]

Answer:

hope this helps.. look at it this way it should help!!

Step-by-step explanation:

Considered as an ideal camping axe, the Boy’s Axe (also called the ¾ Axe) usually has a length of 28 inches, ¾ of 36 inches being 27 inches or roughly 700 millimeters. Finally, the Scout Axe has an average handle length of 18-20 inches or 450-500 millimeters.

6 0
3 years ago
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