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3 years ago

I need help steps to find the answer

Mathematics

2 answers:

sleet_krkn [62]3 years ago

6 0

Step-by-step explanation:

101° = (6x + 5)° (by F angles)

Therefore 6x = 96 and x = 16.

stepan [7]3 years ago

4 0

Answer:

X=16

Step by Step explanation:

We want to know what x is

We would have to make an equation

REMEMBER: *Any 4 angles has to equal to 360 degrees*

Its pretty easy to remember

Triangles: All <u>interior</u> angles must add up to 180

4 angles shapes: All interior angles must add up to 360

If any shape has more than 4 angles, calculate how many angles there are and multiply them by 90

Okay so this “shape” has only 1 angle given And that would be 101

So we know right away that the angle vertical to it (angle 8) is also 101

Now we know two sides and since the other two unknown sides are vertical so they have to be equal, we really only have to focus on the 101 angle.

So we know that 2 ngles are 101

Thats all we need to know

make them equal

6x+5=101

Now sole and you would get 16

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

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and velocity

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

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

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

(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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