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marta [7]
3 years ago
5

Kelly opened a savings account 14 years ago. She has not made any

Mathematics
1 answer:
tatiyna3 years ago
4 0

Answer:

2,130

Step-by-step explanation:

get 3% of 1500 then multiply by 14 and add 1500.

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Solve for m. m + 95 /<br> 4 = 5
ExtremeBDS [4]

Answer:

m = -75/4

Step-by-step explanation:

1. Subtract 95/4 from both sides

m + 95/4 = 5

x + 95/4 - 95/4 = 5 - 95/4

2. Simplify, which will get you -75/4

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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}

7 0
3 years ago
Note: Enter your answer and show all the steps that you use to solve this problem in the space provided.
agasfer [191]

Answer:

rectangle

Step-by-step explanation:

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3 years ago
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