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Sphinxa [80]
2 years ago
12

The coordinates of the vertices of the triangle are

Mathematics
2 answers:
andrew-mc [135]2 years ago
8 0

Answer:

The height of the triangle PQR is given by:

h = 4 + 8

h = 12 units

Now, the area of the triangle PQR is:

= 108 square units.

Step-by-step explanation:

Alecsey [184]2 years ago
6 0

The area of the triangle PQR is 108 square units

<h3>How to determine the area of the triangle?</h3>

The given parameters are:

  • Height, h =  12 units
  • Base, b = 18 units

The area is then calculated using:

Area = 0.5 * base * height

So, we have:

Area = 0.5 * 18 * 12

Evaluate

Area = 108

Hence, the area of the triangle PQR is 108 square units

Read more about areas at:

brainly.com/question/24487155

#SPJ2

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svetoff [14.1K]

Step-by-step explanation:

The angles on the same side of the parallel lines and the transversal are called corresponding angles.

Corresponding angles are congruent

So 11x+1 = 10x+10

x = 10-1 = 9

11*9+1 = 100

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3 years ago
A 10 ft pole has a support rope that extends from the top of the pole to the ground. The rope and the ground form a 30 degree an
mr Goodwill [35]

Answer:

The length of rope is 20.0 ft . Hence, <u>option (1) </u> is correct.

Step-by-step explanation:

In the figure below AB represents pole having height 10 ft  and AC represents the rope that is from the top of pole to the ground. BC represent the ground distance from base of tower to the rope.

The rope and the ground form a 30 degree angle that is the angle between BC and AC is 30°.

In right angled triangle ABC with right angle at B.

Since we have to find the length of rope that is the value of side AC.

Using trigonometric ratios

\sin C=\frac{\text{perpendicular}}{\text{hypotenuse}}

\sin C=\frac{AB}{AC}

Putting values,

\sin 30^\circ} =\frac{10}{AC}

We know, \sin 30^\circ}=\frac{1}{2}

\frac{1}{2} =\frac{10}{AC}

On solving we get,

AC= 20.0 ft

Thus, the length of rope is 20.0 ft

Hence, <u>option (1)</u> is correct.

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3 years ago
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Let X be a positive continuous random variable with density fXpxq. Let Y " lnpXq. (a) Find the density fY pyq of Y in terms of t
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Answer:

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Now to find Y density FYpyq interms of the density of X we compare the density of X with Y"

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(b) at p0, fYpyq =fYp0q= 0

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3 years ago
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