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3 years ago

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Let TU be the directed line segment beginning at point T(4,4) and ending at point U(-11,13). Find the point P on the line segmen

t that partitions the line segment into the segments TP and PU at a ratio of 5:1.

1 answer:

worty [1.4K]3 years ago

8 0

Answer:

P(-8.5, 11.5)

Step-by-step explanation:

From Exercise we have

T(4,4) ⇒ x_1=4, y_1=4

U(-11,13) ⇒ x_2=-11, y_2=13

TP:PU=5:1=a:b ⇒ a=5, b=1

P(x_p, y_p)=?

We have a formula

x_p=x_1+\frac{a}{a+b} · (x_2-x_1)

we get

x_p=4+\frac{5}{6} ·(-11-4)

x_p=4+\frac{5}{6} ·(-15)

x_p=4-75/6

x_p=-8.5

We have a formula

y_p=y_1+\frac{a}{a+b} · (y_2-y_1)

we get

y_p=4+\frac{5}{6} ·(13-4)

y_p=4+\frac{5}{6} ·(9)

y_p=4 + 45/6

y_p=11.5

Therefore, we get P(-8.5, 11.5).

We use the site geogebra.org for drawn the graph.

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Answer: Doubling the radius.

Step-by-step explanation:

The volume of a cone can be found with the following formula:

$V=\frac{1}{3}\pi r^2h$

Where "r" is the radius and "h" is the height of the cone.

Let's find the volume of the conical tent with a radius of 10.4 feet and a height of 8.4 feet.

Identifiying that:

$r=10.4\ ft\\\\h=8.4\ ft$

You get this volume:

$V_1=\frac{1}{3}\pi (10.4\ ft)^2(8.4\ ft)\\\\V_1=951.43\ ft^3$

If you double the radius, the volume of the conical tent will be:

$V_2=\frac{1}{3}\pi (2*10.4\ ft)^2(8.4\ ft)\\\\V_2=3,805.70\ ft^3$

When you divide both volumes, you get:

$\frac{3,805.70\ ft^3}{951.43\ ft^3}=4$

Therefore, doubling the radius will quadruple the volume of the tent.

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3 years ago

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Sedaia [141]

Answer:
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Explanation:
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3 years ago

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Luis's bowling scores were 195, 194, 191, 190, 238, and 192. Which measure (median, mean, mode or range) BEST describes Luis's b

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Step-by-step explanation:

The given data is :

195, 194, 191, 190, 238, and 192

Mean = (sum of observations)/(total no. of observations)

$\text{Mean}=\dfrac{195+194+191+190+238+192}{6}\\\\=200$

So, the mean is the best measure. Also, w can find median as follows.

arranging in order : 190,191,192, 194, 195,238

Here, n = 6

$\text{Median}=\dfrac{\dfrac{n}{2}^{th}+(\dfrac{n}{2}+1)^{th}}{2}\\\\\text{Median}=\dfrac{\dfrac{6}{2}^{th}+(\dfrac{6}{2}+1)^{th}}{2}\\\\=\dfrac{3^{th}+4^{th}}{2}\\\\=\dfrac{192+194}{2}\\\\=193$

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She did not qualify the exam.

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Given data

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3 years ago

What is the relation between the sine and cosine values of angles in each quadrant? How would you use the 60° angle to find sine

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The angles, 60°, $\displaystyle \frac{\pi}{4}$ and $\displaystyle \frac{\pi}{6}$ are special angles that have known trigonometric ratio values.

First part;

The sine and cosine gives the coordinates of the tip of the radius of a unit circle as it rotates P(cos(θ), sin(θ))

Second part;

With the knowledge of the sine and cosine of 60°, we have;
sin(60°) = sin(120°), sin(240°) = -sin(60°), sin(300°) = -sin(60°)
cos(120°) = -cos(60°), cos(240°) = -cos(60°), cos(300°) = cos(60°)

Third part;

$\displaystyle \frac{\pi}{4}$ can be used to find the sine and cosine of $\displaystyle \frac{3 \cdot \pi}{4}$ , $\displaystyle \frac{5 \cdot \pi}{4}$ , and $\displaystyle \frac{7 \cdot \pi}{4}$
$\displaystyle \frac{\pi}{6}$ , can be used to find the sine and cosine of $\displaystyle \frac{5 \cdot \pi}{6}$ , $\displaystyle \frac{7 \cdot \pi}{6}$ , and $\displaystyle \frac{11 \cdot \pi}{6}$

Reasons:

First Part;

Considering a unit circle with the center at the origin of the graph, we have;

The sine of the angle, θ, rotated by the radius is the vertical distance of a point <em>P</em> on the circle which is the location of the radius, from the horizontal axis.

The cosine of the angle, θ, is the horizontal distance of <em>P</em> from the vertical axis, such that we have;

The coordinates of point <em>P</em> = (cos(θ), sin(θ))

In the four quadrant, we have;

First Quadrant; All trigonometric ratios are positive

Second Quadrant; sine is positive

Third Quadrant; Tan is positive

Fourth Quadrant; Cosine is positive

Second part;

We have; At 120°, the point <em>P</em> is the same elevation from the horizontal axis, therefore;

sin(60°) = sin(120°) = 0.5·√3

However, the x-coordinate of the point <em>P</em> is in the negative direction, therefore, we get;

cos(120°) = -cos(60°) = -0.5

Similarly from the quadrant relationship, we have;

240° is in the third quadrant, and it is 60° below the negative horizontal line, therefore;

sin(240°) = -sin(60°) = -0.5·√3

cos(240°) = -cos(60°) = -0.5

300° is in the fourth quadrant, and it is 60° below the positive x-axis, therefore;

sin(300°) is negative and cos(300°) is positive

Which gives;

sin(300°) = -sin(60°) = -0.5·√3

cos(300°) = cos(60°) = 0.5

Third part;

$\displaystyle \frac{\pi}{4} =45^{\circ}$

$\displaystyle \frac{\pi}{6} =30^{\circ}$

The sine and cosine of 45° can be used to find the sine and cosine of (180° + 45°) = 225°, (360° - 45°) = 315°

Also, due to the mid location of the angle 45° on the quadrant, we have;

Another angles is the sines and cosine of (90° + 45°) = 135°

Therefore, $\displaystyle \frac{\pi}{4}$ , can be used to find the sine and cosine of 135°, 225°, and 315°

$\displaystyle 135^{\circ} = \mathbf{\frac{3 \cdot \pi}{4}}$ , $\displaystyle 225^{\circ} = \frac{5 \cdot \pi}{4}$ , $\displaystyle 315^{\circ} = \frac{7 \cdot \pi}{4}$

Therefore,

$\displaystyle \frac{\pi}{4}$ can be used to find the sine and cosine of $\displaystyle \mathbf{\frac{3 \cdot \pi}{4}}$ , $\displaystyle \mathbf{\frac{5 \cdot \pi}{4}}$ , and $\displaystyle \mathbf{ \frac{7 \cdot \pi}{4}}$

Similarly, the sine and cosine of, $\displaystyle \frac{\pi}{6}$ = 30° can be used to find the sine and cosine of 150°, 210°, and 330°.

$\displaystyle 150^{\circ} = \frac{5 \cdot \pi}{6}$ , $\displaystyle 210^{\circ} = \frac{7 \cdot \pi}{6}$ , and $\displaystyle 330^{\circ} = \frac{11 \cdot \pi}{6}$

$\displaystyle \frac{\pi}{6}$ , can be used to find the sine and cosine of $\displaystyle \mathbf{ \frac{5 \cdot \pi}{6}}$ , $\displaystyle \mathbf{ \frac{7 \cdot \pi}{6}}$ , and $\displaystyle \mathbf{\frac{11 \cdot \pi}{6}}$

Learn more about the sine and cosine of angles here:

brainly.com/question/4372174

8 0

3 years ago