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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

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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

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I need help I have no idea what I’m doing

myrzilka [38]

I have no idea either sorry

5 0