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Licemer1 [7]
2 years ago
13

In the diagram, P1P2 and Q1Q2 are the perpendicular bisectors of line AB and BC , respectively. A1A2 and B1B2 are the angle bise

ctors of angle A and B , respectively. The center of the inscribed circle of ΔABC is point ____ , and the center of the circumscribed circle of ΔABC is point ____ .

Mathematics
2 answers:
Aliun [14]2 years ago
4 0
I hope I can show the image, but I hope this may help you, the answer is:

The center of the inscribed circle of ΔABC is <u>point S</u> , and the center of the circumscribed circle of ΔABC is <u>point P.</u>
Hatshy [7]2 years ago
3 0

Answer:  The complete statement is

The center of the inscribed circle of ΔABC is point S , and the center of the circumscribed circle of ΔABC is point P .

Step-by-step explanation:  Given that P_1P_2 and Q_1Q_2 are the perpendicular bisectors of the lines AB and BC respectively.

And, A_1A_2 and B_1B_2 are the angle bisectors of the angles A and B respectively.

We are to find the centre of the inscribed circle of ΔABC and the centre of the circumscribed circle of ΔABC.

<u>INCENTRE</u> : The point at which the angle bisectors of the three angles of a triangle meet is called the INCENTRE of the triangle. This point is the centre of the inscribed circle of the triangle.

The angle bisectors of angles A and B, that is A_1A_2 and B_1B_2  meet at the point S.

Also, since the three angle bisectors are concurrent, so the third angle bisector must also pass through S.

Thus, the point S is the incentre of ΔABC and so, the centre of the inscribed circle of ΔABC is point S.

<u>CIRCUMCENTRE</u> : The point at which the perpendicular bisectors of the three sides of a triangle meet is called the CIRCUMCENTRE of the triangle. This point is the centre of the circumscribed circle of the triangle.

The perpendicular bisectors of sides AB and BC, that is P_1P_2 and Q_1Q_2  meet at the point P.

Also, since the three perpendicular bisectors are concurrent, so the third perpendicular bisector must also pass through P.

Thus, the point P is the circumcentre of ΔABC and so, the centre of the circumscribed circle of ΔABC is point P.

Hence, the complete statement is

The center of the inscribed circle of ΔABC is point S , and the center of the circumscribed circle of ΔABC is point P .

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Daniel and his children went into a grocery store and he bought $10.15 worth of apples and bananas. Each apple costs $1.75 and e
guapka [62]

Answer: Daniel bought 3 apples and 7 bananas.

Step-by-step explanation:

Let x represent the number of apples that Daniel bought.

Let y represent the number of bananas that Daniel bought.

He bought a total of 10 apples and bananas altogether. This means that

x + y = 10

Daniel and his children went into a grocery store and he bought $10.15 worth of apples and bananas. Each apple costs $1.75 and each banana costs $0.70. This means that

1.75x + 0.7y = 10.15 - - - - - - - - - - - 1

Substituting x = 10 - y into equation 1, it becomes

1.75(10 - y) + 0.7y = 10.15

17.5 - 1.75y + 0.7y = 10.15

- 1.75y + 0.7y = 10.15 - 17.5

- 1.05y = - 7.35

y = - 7.35/- 1.05

y = 7

x = 10 - y = 10 - 7

x = 3

4 0
2 years ago
Neptune’s average distance from the sun is 4.503 x 10e9 km. Mercury’s average distance from the sun is 5.791 x10e9 km.about how
Igoryamba

Answer:

77.76 times

Step-by-step explanation:

The average distance of Neptune  from the sun

= 4.503  ×  10 ⁹  k m .

and Mercury  =  5.791  ×  10 ⁷ k m .

Hence neptune is (  4.503  ×  10 ⁹) ÷ (5.791 × 10 ⁷  ) times farther from the sun than mercury

i.e.(  \frac{4.503}{5.791} )  × 10⁹⁻⁷ times

=   0.7776  ×  10 ² times

=   77.76  times.

4 0
2 years ago
A 0.143-Henry Inductor is connected in series with a variable resistor to a 208-volt 400-cycle source. For what value of capacit
aleksandr82 [10.1K]

Answer:

A.)359.2, B.)2.5 uf

Step-by-step explanation:

E / I = R

208 / 1.04 = 200 ohms

2*pi*f*L = Xl

6.28*400*.143 = 359.2 ohm

1 / (2*pi*f*Xc) = c

1 /(6.28*400*159.2) = 2.5 uf

8 0
2 years ago
Express f(x) = |x-2| +|x+2| in the non-modulus form. Hence, sketch the graph of f.
alexgriva [62]
Recall that

|x|=\begin{cases}x&\text{if }x\ge0\\-x&\text{if }x

There are three cases to consider:

(1) When x+2, we have |x+2|=-(x+2) and |x-2|=-(x-2), so

|x-2|+|x+2|=-(x-2)-(x+2)=-2x-4

(2) When x+2\ge0 and x-2, we get |x+2|=x+2 and |x-2|=-(x-2), so

|x-2|+|x+2|=-(x-2)+(x+2)=4

(3) When x-2\ge0, we have |x+2|=x+2 and |x-2|=x-2, so

|x-2|+|x+2|=(x-2)+(x+2)=2x

So

|x-2|+|x+2|=\begin{cases}-2x-4&\text{if }x
4 0
2 years ago
In △ABC,c=71, m∠B=123°, and a=65. Find b.<br><br> A. 101.5<br> B. 117.8<br> C. 123.0<br> D. 119.6
tia_tia [17]

Answer:

Option D

Step-by-step explanation:

The questions which involve calculating the angles and the sides of a triangle either require the sine rule or the cosine rule. In this question, the two sides that are given are adjacent to each other the given angle is the included angle. This means that the angle B is formed by the intersection of the lines a and c. Therefore, cosine rule will be used to calculate the length of b. The cosine rule is:

b^2 = a^2 + c^2 - 2*a*c*cos(B).

The question specifies that c=71, B=123°, and a=65. Plugging in the values:

b^2 = 65^2 + 71^2 - 2(65)(71)*cos(123°).

Simplifying gives:

b^2 = 14293.0182932.

Taking square root on the both sides gives b = 119.6 (rounded to the one decimal place).

This means that the Option D is the correct choice!!!

8 0
2 years ago
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