Please help me, I need help ASAP!

Can you help me find the slope for #2

Flauer [41]

-3
because it goes left 1 up3
3/-1 = -3 as the slope

8 0

3 years ago

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Find the equation of the line passing through the points (2, -1) and (-6, -5)

timurjin [86]

Answer: The equation that goes through the points (2, -1) and (-6, -5) is 1/2x - 2

I know this because when I graphed this equation, it passed through those points. (My graph is shown below)

5 0

2 years ago

2x-3y=42 solve for y

PtichkaEL [24]

Answer:

y=14+(2/3x)

Step-by-step explanation:

Your main goal is to get Y by itself on one side.

To start subtract 2x from both sides. This will give you -3y=42-2x.

Then you divide both sides by -3 to get Y by itself.

That will leave you with ...

y=14+(2/3x)

(remember the negatives cancel each other out)

5 0

3 years ago

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NO LINKS OR FILES!

Archy [21]

(a) If the particle's position (measured with some unit) at time t is given by s(t), where

$s(t) = \dfrac{5t}{t^2+11}\,\mathrm{units}$

then the velocity at time t, v(t), is given by the derivative of s(t),

$v(t) = \dfrac{\mathrm ds}{\mathrm dt} = \dfrac{5(t^2+11)-5t(2t)}{(t^2+11)^2} = \boxed{\dfrac{-5t^2+55}{(t^2+11)^2}\,\dfrac{\rm units}{\rm s}}$

(b) The velocity after 3 seconds is

$v(3) = \dfrac{-5\cdot3^2+55}{(3^2+11)^2} = \dfrac{1}{40}\dfrac{\rm units}{\rm s} = \boxed{0.025\dfrac{\rm units}{\rm s}}$

(c) The particle is at rest when its velocity is zero:

$\dfrac{-5t^2+55}{(t^2+11)^2} = 0 \implies -5t^2+55 = 0 \implies t^2 = 11 \implies t=\pm\sqrt{11}\,\mathrm s \imples t \approx \boxed{3.317\,\mathrm s}$

(d) The particle is moving in the positive direction when its position is increasing, or equivalently when its velocity is positive:

$\dfrac{-5t^2+55}{(t^2+11)^2} > 0 \implies -5t^2+55>0 \implies -5t^2>-55 \implies t^2 < 11 \implies |t|$

In interval notation, this happens for t in the interval (0, √11) or approximately (0, 3.317) s.

(e) The total distance traveled is given by the definite integral,

$\displaystyle \int_0^8 |v(t)|\,\mathrm dt$

By definition of absolute value, we have

$|v(t)| = \begin{cases}v(t) & \text{if }v(t)\ge0 \\ -v(t) & \text{if }v(t)$

In part (d), we've shown that v(t) > 0 when -√11 < t < √11, so we split up the integral at t = √11 as

$\displaystyle \int_0^8 |v(t)|\,\mathrm dt = \int_0^{\sqrt{11}}v(t)\,\mathrm dt - \int_{\sqrt{11}}^8 v(t)\,\mathrm dt$

and by the fundamental theorem of calculus, since we know v(t) is the derivative of s(t), this reduces to

$s(\sqrt{11})-s(0) - s(8) + s(\sqrt{11)) = 2s(\sqrt{11})-s(0)-s(8) = \dfrac5{\sqrt{11}}-0 - \dfrac8{15} \approx 0.974\,\mathrm{units}$

7 0

2 years ago

HELP?!!!!! Name the space figure you can form from the net.

daser333 [38]

Answer:

a or b but i think a

Step-by-step explanation:

4 0

3 years ago

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