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zubka84 [21]
3 years ago
9

Simplify polynomial. 3f-5g-f-2g

Mathematics
1 answer:
ankoles [38]3 years ago
7 0
Combine like terms
3f - 5g - f - 2g...lets do some rearranging to make it easier

3f - f - 5g - 2g =
2f - 7g <===
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Velocity of the ring wraiths relative to the air:

\vec v_{W/A}=(v_{W/A}\cos45^\circ,v_{W/A}\sin45^\circ)=\left(\dfrac{v_{W/A}}{\sqrt2},\dfrac{v_{W/A}}{\sqrt2}\right)

The speed v_{W/A} is what we want to find - this is the speed at which the wraiths are flying in the air.

(Note: \vec v denotes velocity, while v denotes speed)

Velocity of air relative to the ground:

\vec v_{A/G}=((69\,\mathrm{mph})\cos180^\circ,(69\,\mathrm{mph})\sin180^\circ)=(-69,0)\,\mathrm{mph}

The ring wraiths want their trajectory to be due north, which means their velocity relative to the ground should be

\vec v_{W/G}=(v_{W/G}\cos90^\circ,v_{W/G}\sin90^\circ)=(0,v_{W/G})

To an observer on the ground, v_{W/G} is the speed at which the wraiths would appear to be moving in the air.

The relative velocities satisify

\vec v_{W/A}+\vec v_{A/G}=\vec v_{W/G}

\implies\begin{cases}\dfrac{v_{W/A}}{\sqrt2}-(69\,\mathrm{mph})=0\\\\\dfrac{v_{W/A}}{\sqrt2}=v_{W/G}\end{cases}

\implies v_{W/A}=69\sqrt2\,\mathrm{mph}\approx98\,\mathrm{mph}

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3 years ago
Please need help with this two problems
Julli [10]
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8 0
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TEA [102]

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Step-by-step explanation:

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Given : f(x) = 3x + 1

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3 years ago
Given the Trapezoid ABCD<br><br> Find the length of AD <br><br> Pls hellpppp
matrenka [14]

Given:

In trapezoid ABCD, AD\parallel BC, MN is the mid-segment of ABCD, AD=30x-10,MN=31x+1,BC=30x+28.

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The length of AD.

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In trapezoid ABCD,

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Therefore, the length of AD is 230.

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