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2 years ago

Who wants free points lol XD

Mathematics

2 answers:

asambeis [7]2 years ago

7 0

Step-by-step explanation:

grax xlos puntos we ㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤ

Alenkasestr [34]2 years ago

3 0

Answer:

Hey thx broo

Step-by-step explanation:

:D)))))))))))))))))))))))))))))))))))))))))))))))))

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Verify the following trig identities.

bagirrra123 [75]

Answer:

see explanation

Step-by-step explanation:

Using the trigonometric identities

• 1 + cot² x = csc²x and csc x = $\frac{1}{sinx}$

• sin²x + cos²x = 1 ⇒ sin²x = 1 - cos²x

Consider the left side

sin²Θ( 1 + cot²Θ )

= sin²Θ × csc²Θ

= sin²Θ × 1 / sin²Θ = 1 = right side ⇔ verified

-----------------------------------------------------------------

Consider the left side

cos²Θ - sin²Θ

= cos²Θ - (1 - cos²Θ)

= cos²Θ - 1 + cos²Θ

= 2cos²Θ - 1 = right side ⇒ verified

3 0

3 years ago

Please help me !!!!!!!

NikAS [45]

I want to say the answer is A i caulated it and i end up getting 17

4 0

2 years ago

Please help me asap!!!

ololo11 [35]

G should be the correct answer

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4 0

3 years ago

Read 2 more answers

16y2 – 64<br> What is the answer to this problem

Elodia [21]

Answer:

448 is the answer to 16y2 – 64.

8 0

2 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