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2 years ago

Circle O and triangle OQR are shown below. What is the length of OQ?

Mathematics

1 answer:

zysi [14]2 years ago

3 0

Answer:

Step-by-step explanation:

A circle is inscribed in an equilateral triangle PQR with centre O. If angle OQR = 30°, what is the perimeter of the triangle?

This is a circle inscribed in an equilateral triangle with side s.

If you are asking for the perimeter of PQR, it is 3s.

If you are asking for the perimeter of OQR, it is (3+23–√3)s

Since OR and SR are the hypotenuses of right triangles with adjacent side equal to ½ s, their length is ½s / cos 30° = (√3) /3.

(3/3)s + ((√3) /3)s + ((√3) /3)s = ((3 + 2√3)/3)s ≈ 2.1547s

Hope it helps

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The sum of twice a number and 8 is 44. what's the number? show your work.

mezya [45]

Answer:

Step-by-step explanation:

We can work backwards with this and take 44-8=36

Then we divide it by two and thats 18

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3 years ago

Plz help asap......................................

ziro4ka [17]

Option B:

Law of cosine, two sides and the included angle are known.

Solution:

Let us first know the law of cosines and law of sines.

Law of cosine:

If we know the sides a, b and the included angle θ, then we can find the third side c. This is known as the law of cosine.

$c^2=a^2+b^2-2abcos\theta$

Law of sine:

The sides of a triangle are to one another in the same ratio as the sines of their opposite angles.

$$\frac{a}{sinA} =\frac{b}{sinB} =\frac{c}{sinC}$

Given ∠P and the sides r, q are known.

<u>To find the value of p:</u>

Option A: Law of cosines, all sides are known.

It is false by the above definition of law of cosine.

Option B: Law of cosine, two sides and the included angle are known.

It is true by the above definition of law of cosine.

Option C: Law of sines, all sides are known.

It is false, because one angle is given in question.

Option D: Law of sines, two angles and the included side are known.

It is false, because two angles are not given.

Option B is the correct answer.

Hence the answer is "Law of cosine, two sides and the included angle are known".

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3 years ago

In this equation you have to solve for x

lidiya [134]

Answer:

Step-by-step explanation:

tan 36° = x/40

x = 40 * tan 36°

x ≈ 29.1

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2 years ago

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3 years ago

A bucket that weighs 4 lb and a rope of negligible weight are used to draw water from a well that is 60 ft deep. The bucket is f

jonny [76]

Answer:

2580 ft-lb

Step-by-step explanation:

Water leaks out of the bucket at a rate of $\frac{0.15 \mathrm{lb} / \mathrm{s}}{1.5 \mathrm{ft} / \mathrm{s}}=0.1 \mathrm{lb} / \mathrm{ft}$

Work done required to pull the bucket to the top of the well is given by integral

$W=\int_{a}^{b} F(x) dx$

Here, function $F(x)$ is the total weight of the bucket and water $x$ feet above the bottom of the well. That is,

$F(x)=4+(42-0.1 x)$

$=46-0.1x$

$a$ is the initial height and $b$ is the maximum height of well. That is,

$a=0 \text { and } b=60$

Find the work done as,

$W=\int_{a}^{b} F(x) d x$

$=\int_{0}^{60}(46-0.1 x) dx$

$&\left.=46x-0.05 x^{2}\right]_{0}^{60}$

$=(2760-180)-0[$

$=2580 \mathrm{ft}-\mathrm{lb}
$

Hence, the work done required to pull the bucket to the top of the well is $2580 \mathrm{ft}- \mathrm{lb}$

4 0

3 years ago