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olga_2 [115]
3 years ago
8

If perpendiculars from any point within an angle on its arms are equal, prove that it lies on the bisector of that angle

Mathematics
1 answer:
Oxana [17]3 years ago
4 0
Your question can be quite confusing, but I think the gist of the question when paraphrased is: P<span>rove that the perpendiculars drawn from any point within the angle are equal if it lies on the angle bisector?

Please refer to the picture attached as a guide you through the steps of the proofs. First. construct any angle like </span>∠ABC. Next, construct an angle bisector. This is the line segment that starts from the vertex of an angle, and extends outwards such that it divides the angle into two equal parts. That would be line segment AD. Now, construct perpendicular line from the end of the angle bisector to the two other arms of the angle. This lines should form a right angle as denoted by the squares which means 90° angles. As you can see, you formed two triangles: ΔABD and ΔADC. They have congruent angles α and β as formed by the angle bisector. Then, the two right angles are also congruent. The common side AD is also congruent with respect to each of the triangles. Therefore, by Angle-Angle-Side or AAS postulate, the two triangles are congruent. That means that perpendiculars drawn from any point within the angle are equal when it lies on the angle bisector

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Suppose the selling price of homes is skewed right with a mean of 350,000 and a standard deviation of 160000 If we record the se
lys-0071 [83]

Answer:

The distribution will be approximately normal, with mean 350,000 and standard deviation 25,298.

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

Population:

Suppose the selling price of homes is skewed right with a mean of 350,000 and a standard deviation of 160000

Sample of 40

Shape approximately normal

Mean 350000

Standard deviation s = \frac{160000}{\sqrt{40}} = 25298

The distribution will be approximately normal, with mean 350,000 and standard deviation 25,298.

5 0
3 years ago
Read 2 more answers
Question 21 is probably the hardest question on this homework
Snezhnost [94]
-15+5x= 20

5x= 35
x= 7

20= -15+5x

35= 5x

7 = x
7 0
3 years ago
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Which equation has a graph that passes through the points at (0,5) and (-3,-4)?
Roman55 [17]

Answer:

y = 3x + 5

Step-by-step explanation:

We are asked the equation of the graph which passes through the points at (0,5) and (-3,-4).  

Now, we know that the equation of a straight line passing through the points (x_{1},y_{1}) and (x_{2},y_{2}) is given by

\frac{y - y_{1}}{y_{1} - y_{2}} = \frac{x - x_{1}}{x_{1} - x_{2}}.

So, in our case (x_{1},y_{1}) ≡ (0,5) and (x_{2},y_{2}) ≡ (-3,-4)

Therefore, the equation of the straight line will be  

\frac{y - 5}{5 - (- 4)} = \frac{x - 0}{0 - (- 3)}

⇒ y - 5 = \frac{9}{3}x

⇒y = 3x + 5 (Answer)

4 0
3 years ago
A. Adjacent angles <br> B. Complementary angles <br> C. Supplementary angles <br> D. Vertical angles
DaniilM [7]

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4 0
3 years ago
A store has been selling 300 Blu-ray disc players a week at $600 each. A market survey indicates that for each $40 rebate offere
timurjin [86]

Answer:

75 $

Step-by-step explanation:

According to problem statement p(300) = 600

And we know that with a rebate of 40 $, numbers of units sold will increase by 80 then if x is number of units sold, the increase in units is

(  x  - 300 )  , and the price decrease

(1/80)*40  =  0,5

Then the demand function is:

D(x)  =  600  - 0,5* ( x - 300 )  (1)

And revenue function is:

R(x) =  x * (D(x)   ⇒   R(x) =  x* [  600  - 0,5* ( x - 300 )]

R(x) = 600*x  - 0,5*x * ( x - 300 )

R(x) = 600*x - 0,5*x² - 150*x

R(x) = 450*x  - (1/2)*x²

Now taking derivatives on both sides of the equation we get

R´(x) =  450  - x

R´(x) =  0       ⇒   450  - x = 0

x = 450 units

We can observe that for   0 < x  < 450  R(x) > 0 then R(x) has a maximum for x = 450

Plugging this value in demand equation, we get the rebate for maximize revenue

D(450)  =  600  - 0,5* ( x - 300 )

D(450)  =  600 - 225 + 150

D(450)  =

D(450)  =  600 - 0,5*( 150)

D(450)  =  600 - 75

D(450)  = 525

And the rebate must be

600 - 525  = 75 $

5 0
3 years ago
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