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3 years ago

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When fritz drives to work his trip takes 4848 minutes, but when he takes the train it takes 4040 minutes. find the distance fri

tz travels to work if the train travels an average of 66 miles per hour faster than his driving. assume that the train travels the same distance as the car?

1 answer:

stepan [7]3 years ago

8 0

Let n be the speed of Fritzs' car, and n+6 be the speed of the train. Then:
4/5 (n)=2/3 (n+6)
4n/5=2n+12/3
10n+60=12n
2n=60
n=30
Fritz travels 4/5(30), or 24 miles to work
☺☺☺☺

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Find the angle between u =the square root of 5i-8j and v =the square root of 5i+j.

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Answer:

The angle between vector $\vec{u} = 5\, \vec{i} - 8\, \vec{j}$ and $\vec{v} = 5\, \vec{i} + \, \vec{j}$ is approximately $1.21$ radians, which is equivalent to approximately $69.3^\circ$ .

Step-by-step explanation:

The angle between two vectors can be found from the ratio between:

their dot products, and
the product of their lengths.

To be precise, if $\theta$ denotes the angle between $\vec{u}$ and $\vec{v}$ (assume that $0^\circ \le \theta < 180^\circ$ or equivalently $0 \le \theta < \pi$ ,) then:

$\displaystyle \cos(\theta) = \frac{\vec{u} \cdot \vec{v}}{\| u \| \cdot \| v \|}$ .

<h3>Dot product of the two vectors</h3>

The first component of $\vec{u}$ is $5$ and the first component of $\vec{v}$ is also

The second component of $\vec{u}$ is $(-8)$ while the second component of $\vec{v}$ is $1$ . The product of these two second components is $(-8) \times 1= (-8)$ .

The dot product of $\vec{u}$ and $\vec{v}$ will thus be:

$\begin{aligned} \vec{u} \cdot \vec{v} = 5 \times 5 + (-8) \times1 = 17 \end{aligned}$ .

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Apply the Pythagorean Theorem to both $\vec{u}$ and $\vec{v}$ :

$\| u \| = \sqrt{5^2 + (-8)^2} = \sqrt{89}$ .
$\| v \| = \sqrt{5^2 + 1^2} = \sqrt{26}$ .

<h3>Angle between the two vectors</h3>

Let $\theta$ represent the angle between $\vec{u}$ and $\vec{v}$ . Apply the formula $\displaystyle \cos(\theta) = \frac{\vec{u} \cdot \vec{v}}{\| u \| \cdot \| v \|}$ to find the cosine of this angle:

$\begin{aligned} \cos(\theta)&= \frac{\vec{u} \cdot \vec{v}}{\| u \| \cdot \| v \|} = \frac{17}{\sqrt{89}\cdot \sqrt{26}}\end{aligned}$ .

Since $\theta$ is the angle between two vectors, its value should be between $0\; \rm radians$ and $\pi \; \rm radians$ ( $0^\circ$ and $180^\circ$ .) That is: $0 \le \theta < \pi$ and $0^\circ \le \theta < 180^\circ$ . Apply the arccosine function (the inverse of the cosine function) to find the value of $\theta$ :

$\displaystyle \cos^{-1}\left(\frac{17}{\sqrt{89}\cdot \sqrt{26}}\right) \approx 1.21 \;\rm radians \approx 69.3^\circ$ .

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A truck is 300 miles east of a car and is traveling west at the constant speed of 30 miles/hr meanwhile the car is going north a

Lerok [7]

Answer:

$d(t)=\sqrt{(300+30t)^2+(60t)^2}$

Step-by-step explanation:

Let t represents the time in hours,

We know that,

$Speed =\frac{Distance}{Time}$

$\implies Distance = Speed\times time$

Since, the speed of truck = 30 miles per hour,

So, the distance covered by the truck in t hours = 30t miles,

Similarly,

Speed of car = 60 miles per hour,

So, the distance covered by car in t hours = 60t miles,

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Now, these two vehicles are going in the directions which are at right angled ( car is going north and truck is going west )

Using the Pythagoras theorem,

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Answer:

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Answer 2.8

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