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Ksju [112]
3 years ago
6

A circular race track is 62.8 miles around. A driver is moving at 5 miles an hour driving from the outside to the center of the

track. How long will it take for him to arrive?
Calculate the time it takes to travel to the center.
Mathematics
1 answer:
nikdorinn [45]3 years ago
8 0

<u>G I V E N</u>

Speed of driver = 5 miles/hours

Distance to be covered = 62.8 miles

<u />

<u>F O R M U L A  ~  U S E D</u>

Speed = Distance/Time

<u>S O L U T I O N </u>

Time = Distance/Speed

= 62.8 miles / 5 miles / hour

= 12.56 hours

It will take 12.56 hours to cover 62.8 miles, if the driver is moving

with 5 miles/hours

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liberstina [14]

Answer:

p = -3

Step-by-step explanation:

Apply algebra to the equation by subtracting 4 from each side, that way it will cancel out at least the 4 next to p. It should now look like this: p = -3

Because we can't make anymore moves and the expression clearly states what p is, we're done!

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Answer:

See below

Step-by-step explanation:

3. What are two ways that a vector can be represented?

Considering a vector \vec{v} in some vector space \mathbb R^n we have

\vec{v} = \langle a,b\rangle

This is the component form. I don't like that way. It is probably used in high school, but

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4.

For this question, I think you meant

vectors

\vec{u_1} = (-8, 12)

\vec{u_2}  = (13, 15)

Once

\cos(\theta)=\dfrac{\vec{u_1} \cdot\vec{u_2}}{||\vec{u_1}||||\vec{u_2}||}

Considering that the dot product is

\vec{u_1}\cdot \vec{u_2} = (-8)\cdot 13 + 12\cdot 15 = -104+180= 76

and the norm of \vec{u_1} is ||\vec{u_1}|| = \sqrt{(-8)^2 + 12^2} = \sqrt{64 + 144}= \sqrt{208}

and the norm of \vec{u_2} is ||\vec{u_2}|| = \sqrt{13^2 + 15^2} = \sqrt{169 + 225}= \sqrt{394}

Thus,

\cos(\theta)=\dfrac{76}{\sqrt{208} \sqrt{394}} = \dfrac{19}{\sqrt{13}\sqrt{394}}=\dfrac{19}{\sqrt{5122}}

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3 years ago
What is the horizontal asymptote of mc001-1.jpg?<br> y = –2<br> y = –1<br> y = 0<br> y = 1
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3 years ago
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Activity
alina1380 [7]

Answer:

Step-by-step explanation:

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So,

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