A, R, S, and C are coplanar. True or False

If C is the current value and S is the starting value, write an

LekaFEV [45]

Is there a diagram with this question?

7 0

3 years ago

The number of endangered species in a jungle decreased from 75 to 30 in one year.

lys-0071 [83]

The percent of decrease would be 60%

7 0

2 years ago

Points A and B lle in plane R. line AB does not lle in plane R. sometimes always never

Kobotan [32]

Answer: The answer is never

Step-by-step explanation: The question has been presented in a complicated manner. I am answering the question based on my understanding. Line AB will never lie in plane R. The correct option among all the options that are given in the question is the third option. I hope that this is the answer that has come to your desired help.

Pls mark me as brainiest and Yw

4 0

2 years ago

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

2 years ago

Find all point(s) of intersection of the line y = 4x and the parabola y = x^2 - 2x + 9.

Goshia [24]

Answer:

A) (3, 12)

Step-by-step explanation:

For such a problem, I like to use a graphing calculator. It shows the answer quickly without a lot of fuss.

_____

If you want to solve this analytically, set the difference in y-values equal to zero and solve the resulting quadratic in the usual way. This will give the x-value at which the y-values are equal. (After we find x, we still need to find y.)

... y - y = 0

... x² -2x +9 -4x = 0

... x² -6x +9 = 0 . . . . . . collect terms. Recognize this as a perfect square.

... (x -3)² = 0

... x = 3 is the solution to this

... y = 4x = 4·3 = 12

The point on each of the given curves is (x, y) = (3, 12). The line is tangent to the parabola there, so there is only one point of intersection.

6 0

3 years ago

A, R, S, and C are coplanar.TrueorFalse​

A, R, S, and C are coplanar.TrueorFalse