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3 years ago

Joshua has 200 tiles to put on

Mathematics

1 answer:

klemol [59]3 years ago

6 0

Answer:

How big are the other rooms

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

fenix001 [56]

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:

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

<h3>Lengths of the two vectors</h3>

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 $\theta$ :

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

3 0

3 years ago

The play started at 8:15 p.m and ended at 10:42 p.m how long did the play last?

solong [7]

Answer: 2 hours and 27 minutes

4 0

4 years ago

Can someone please help

mart [117]

B is the right I think

7 0

3 years ago

A flagpole is located at the edge of a sheer y = 70-ft cliff at the bank of a river of width x = 40 ft. See the figure below. An

Gnom [1K]

I would solve this using tangents. Let h be height of flagpole.
Set up 2 right triangles, each with a base of 40.
The larger triangle has height of "h+70"
Smaller triangle has height of 70.

Now write the tangent ratios:
$tan A = \frac{h+70}{40} , tan B = \frac{70}{40}$

Note: A-B = 9
To solve for h we need to use the "Difference Angle" formula for Tangent
$tan (A-B) = \frac{tanA - tanB}{1+tan A tan B}$
Plug in what we know:
$tan(9) = \frac{ \frac{h+70}{40} - \frac{70}{40}}{1+ (\frac{h+70}{40})(\frac{7}{4})}$
$tan (9) = \frac{ \frac{h}{40}}{ \frac{7h +650}{160}} = \frac{4h}{7h+650}$
$h = \frac{650 tan(9)}{4-7 tan(9)}$
$h = 35.6$

7 0

3 years ago

Enter the equation of the line in slope-intercept form.

ad-work [718]

Answer:

$y = \frac{7}{4}x +14$