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3 years ago

Can anyone understand this? I certinaly don't!

Mathematics

1 answer:

lara [203]3 years ago

6 0

Answer:

y-3=6(x+8)

Step-by-step explanation:

Ugh here goes

the point-slope form of an equation is y=mx+c

we have m with is the slope and we have x and y ( -8,3) we neec C which is the y intercept so put all the information in the equation and solve for c

3=6(-8)+c

3=-48+c

0=-48-3+c

c=51

so the equation is y=6x+51

take 3 from 51 and put it on the other side

y-3=6x+48

y-3=6(x+8)

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3 years ago

A small rocket is fired from a launch pad 10 m above the ground with an initial velocity left angle 250 comma 450 comma 500 righ

jonny [76]

Let $\vec r(t),\vec v(t),\vec a(t)$ denote the rocket's position, velocity, and acceleration vectors at time $t$ .

We're given its initial position

$\vec r(0)=\langle0,0,10\rangle\,\mathrm m$

and velocity

$\vec v(0)=\langle250,450,500\rangle\dfrac{\rm m}{\rm s}$

Immediately after launch, the rocket is subject to gravity, so its acceleration is

$\vec a(t)=\langle0,2.5,-g\rangle\dfrac{\rm m}{\mathrm s^2}$

where $g=9.8\frac{\rm m}{\mathrm s^2}$ .

a. We can obtain the velocity and position vectors by respectively integrating the acceleration and velocity functions. By the fundamental theorem of calculus,

$\vec v(t)=\left(\vec v(0)+\displaystyle\int_0^t\vec a(u)\,\mathrm du\right)\dfrac{\rm m}{\rm s}$

$\vec v(t)=\left(\langle250,450,500\rangle+\langle0,2.5u,-gu\rangle\bigg|_0^t\right)\dfrac{\rm m}{\rm s}$

(the integral of 0 is a constant, but it ultimately doesn't matter in this case)

$\boxed{\vec v(t)=\langle250,450+2.5t,500-gt\rangle\dfrac{\rm m}{\rm s}}$

and

$\vec r(t)=\left(\vec r(0)+\displaystyle\int_0^t\vec v(u)\,\mathrm du\right)\,\rm m$

$\vec r(t)=\left(\langle0,0,10\rangle+\left\langle250u,450u+1.25u^2,500u-\dfrac g2u^2\right\rangle\bigg|_0^t\right)\,\rm m$

$\boxed{\vec r(t)=\left\langle250t,450t+1.25t^2,10+500t-\dfrac g2t^2\right\rangle\,\rm m}$

b. The rocket stays in the air for as long as it takes until $z=0$ , where $z$ is the $z$ -component of the position vector.

$10+500t-\dfrac g2t^2=0\implies t\approx102\,\rm s$

The range of the rocket is the distance between the rocket's final position and the origin (0, 0, 0):

$\boxed{\|\vec r(102\,\mathrm s)\|\approx64,233\,\rm m}$

c. The rocket reaches its maximum height when its vertical velocity (the $z$ -component) is 0, at which point we have

$-\left(500\dfrac{\rm m}{\rm s}\right)^2=-2g(z_{\rm max}-10\,\mathrm m)$

$\implies\boxed{z_{\rm max}=125,010\,\rm m}$

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