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guapka [62]
3 years ago
8

I need help with this question !!​

Mathematics
1 answer:
nevsk [136]3 years ago
5 0

Answer:

Step-by-step explanation:

Yes, A-A-A

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Please help quickly!
bekas [8.4K]

Answer:

the answer is 2.8

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
In quadratic drag problem, the deceleration is proportional to the square of velocity
Mars2501 [29]
Part A

Given that a= \frac{dv}{dt} =-kv^2

Then, 

\int dv= -kv^2\int dt \\  \\ \Rightarrow v(t)=-kv^2t+c

For v(0)=v_0, then

v(0)=-kv^2(0)+c=v_0 \\  \\ \Rightarrow c=v_0

Thus, v(t)=-kv(t)^2t+v_0

For v(t)= \frac{1}{2} v_0, we have

\frac{1}{2} v_0=-k\left( \frac{1}{2} v_0\right)^2t+v_0 \\  \\ \Rightarrow \frac{1}{4} kv_0^2t=v_0- \frac{1}{2} v_0= \frac{1}{2} v_0 \\  \\ \Rightarrow kv_0t=2 \\  \\ \Rightarrow t= \frac{2}{kv_0}


Part B

Recall that from part A, 

v(t)= \frac{dx}{dt} =-kv^2t+v_0 \\  \\ \Rightarrow dx=-kv^2tdt+v_0dt \\  \\ \Rightarrow\int dx=-kv^2\int tdt+v_0\int dt+a \\  \\ \Rightarrow x=- \frac{1}{2} kv^2t^2+v_0t+a

Now, at initial position, t = 0 and v=v_0, thus we have

x=a

and when the velocity drops to half its value, v= \frac{1}{2} v_0 and t= \frac{2}{kv_0}

Thus,

x=- \frac{1}{2} k\left( \frac{1}{2} v_0\right)^2\left( \frac{2}{kv_0} \right)^2+v_0\left( \frac{2}{kv_0} \right)+a \\  \\ =- \frac{1}{2k} + \frac{2}{k} +a

Thus, the distance the particle moved from its initial position to when its velocity drops to half its initial value is given by

- \frac{1}{2k} + \frac{2}{k} +a-a \\  \\ = \frac{2}{k} - \frac{1}{2k} = \frac{3}{2k}
7 0
3 years ago
The graph of the function g(x) = x2 + 3x - 4 is shifted 5 units to the left. Plot the zeros of the new function on the provided
Colt1911 [192]
<span>g(x) = x^2 + 3x - 4   (please be sure to use " ^ " to denote exponentiation).

This function factors into g(x) = (x+4)(x-1), and so the zeros are -4 and +1.

If we shift this graph 5 units to the left, we get h(x) = (x+5)^2 + 3(x+5) - 4.

The new zeros will be -4-5    and    1-5,    or  -9 and -4.

</span>
5 0
3 years ago
<img src="https://tex.z-dn.net/?f=11%2B5x%3D2x-4" id="TexFormula1" title="11+5x=2x-4" alt="11+5x=2x-4" align="absmiddle" class="
stira [4]
11+4 = 2x-5x
15=-3x
X=-5
4 0
3 years ago
Read 2 more answers
I need help I have no idea what I’m doing
myrzilka [38]
I have no idea either sorry
5 0
3 years ago
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