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Ira Lisetskai [31]
3 years ago
9

HELP write in slope intercept form through (6,-1) parallel to 3x-y=2

Mathematics
1 answer:
Murrr4er [49]3 years ago
3 0

Answer:

3x-y=2

m1=-3/-1

m1=3

Since parallel, m1=m2 so,

m1=m2=3

The slope of straight line passing through the point (6,-1) is

y-(-1)=m2(x-6)

y+1=3(x-6)

Step-by-step explanation:

I don't say you have to mark my ans as brainliest but friend if it has helped you,plz don't forget to thank me....

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For what values of p is the value of the binomial 1.5p−1 smaller than the value of the binomial 1+1.1p by 4?
77julia77 [94]

Answer:

  p = -5  (only)

Step-by-step explanation:

It looks like we want to count solutions for ...

  (1.5p -1) = (1 +1.1p) -4

  0.4p = -2

  p = -2/0.4 = -5

For p having a value of -5, the binomial (1.5p -1) = -8.5 will be smaller than the value of (1 +1.1p) = -4.5 by 4.

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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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3 years ago
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bekas [8.4K]

Answer: Option 3

Step-by-step explanation:

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Answer:

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3 years ago
I need help pls. Question 11
g100num [7]

Answer:

T = .17m + 50

Step-by-step explanation:

T = Total cost

M = Mile

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3 years ago
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