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2 years ago

Find the equation of a line parallel to y=3x−6that passes through the point (1,-5).

Mathematics

1 answer:

k0ka [10]2 years ago

7 0

Answer: here answer give brainlist -2 ,5

Step-by-step explanation:

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HELP MEEEEEEE PLEASE

saw5 [17]

Answer:

3 option

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

Solve for b: 16b = 12

Zina [86]

Answer:

b = 3/4

Step-by-step explanation:

Hi!!!

You want the value of B. To get that, we have to divide both sides by 16:

$\frac{16}{16}b =\frac{12}{16}$

$\frac{12}{16} = \frac{3}{4}$

Thus the answer to your question is $$\boxed{b = \frac{3}{4} }$ .

<u>Check:</u>

3/4 * 16 = 3 * 4 = 12.

12 = 12

So we are correct!!!

5 0

3 years ago

Does it makes sense that sin(-)=-sin?

Andre45 [30]

No, you cannot used that because you won't be able to solve that.

6 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

Expand and simplify (x+5)(x-2)

lukranit [14]

Answer:

x² + 3x - 10

Step-by-step explanation:

Each factor in the second factor is multiplied by each term in the first factor, that is

x(x - 2) + 5(x - 2) ← distribute each parenthesis

= x² - 2x + 5x - 10 ← collect like terms

= x² + 3x - 10

7 0

3 years ago