Question

i need correct answer with clear explanation.......fastly anyone give answer.....the answer in book angle PQO = 35 degree , angle PSQ = 55 degree , angle ORS = 35 degree..... correct answer will be marked as brainliest answer and thanks......​

Answer 1

Complementary angle- Two angles are said to be complementary if they add up/sum up to be 90°

Now, its simple, let the complementary angle of 35° be x, then,

35°+x=90

x=90°-35°

x=55°

So, the complementary angle of 35° is 55°.Firstly draw two circles with center O and O’ such that they intersect at A and B.

Draw a line PQ parallel to OO’.

In the circle with center O, we have:

OP and OB are the radii of the circle. PB is the chord with OM as its perpendicular bisector.

i.e. BM=MP....(1)

In the circle with center O’, we have:

O’B and O’Q are the radii of the circle. BQ is the chord with O’N as its perpendicular bisector.

i.e. BN=NQ....(1)BM=MP....(1)

From (1) and (2), we have:

BM+BN=MP+NQ⇒(BM+BN)+(BM+BN)=(BM+BN)+(MP+NQ)⇒2(BM+BN)=(BM+BN)+(MP+NQ)⇒2(OO’)=(BM+MP)+(BN+NQ)⇒2(OO’)=BP+BQ⇒2OO’=PQ

Hence, proved.



23. Two circles with centres O and O' intersect at two points A and B. A line

PQ is drawn parallel to 00' through A or B, intersecting the circles at P

and Q. Prove that PQ = 200'.the first order reaction is found to have a rate constant k=2 min/sec.Find the half period of the reaction?Firstly draw two circles with center O and O’ such that they intersect at A and B.

Draw a line PQ parallel to OO’.

In the circle with center O, we have:

OP and OB are the radii of the circle. PB is the chord with OM as its perpendicular bisector.

i.e. BM=MP....(1)

In the circle with center O’, we have:

O’B and O’Q are the radii of the circle. BQ is the chord with O’N as its perpendicular bisector.

i.e. BN=NQ....(1)BM=MP....(1)

From (1) and (2), we have:

BM+BN=MP+NQ⇒(BM+BN)+(BM+BN)=(BM+BN)+(MP+NQ)⇒2(BM+BN)=(BM+BN)+(MP+NQ)⇒2(OO’)=(BM+MP)+(BN+NQ)⇒2(OO’)=BP+BQ⇒2OO’=PQ

Hence, proved.



किसने किससे कहा,

1. चलो मैं भी चल कर देखता हूँ,

2आओ उठो हमें भीमा का हाथ बटाना चाहिए

3. क्या वक्ता है । class 8For the reaction 2A + B → 2C, doubling the concentration of B doubles the rate and doubling the concentration of A

also doubles the rate. The rate law is

(a) rate = k[A]”[B] (b) rate = k[A][B] (c) rate = k[A]”[B][C] (d) rate = k[A]”[B]

Rate = k[A][B] b) Rate = k[A]+[B] Rate = k[A][B] d) Rate = k[A]?[B]. 17-1. 2. ... doubled for the reaction, there is no change in the reaction rate. ... For the reaction AB → A + B, concentration of AB was measured as the ...

Answer 2

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Complementary angle- Two angles are said to be complementary if they add up/sum up to be 90°

Now, its simple, let the complementary angle of 35° be x, then,

35°+x=90

x=90°-35°

x=55°

So, the complementary angle of 35° is 55°.Firstly draw two circles with center O and O’ such that they intersect at A and B.

Draw a line PQ parallel to OO’.

In the circle with center O, we have:

OP and OB are the radii of the circle. PB is the chord with OM as its perpendicular bisector.

i.e. BM=MP....(1)

In the circle with center O’, we have:

O’B and O’Q are the radii of the circle. BQ is the chord with O’N as its perpendicular bisector.

i.e. BN=NQ....(1)BM=MP....(1)

From (1) and (2), we have:

BM+BN=MP+NQ⇒(BM+BN)+(BM+BN)=(BM+BN)+(MP+NQ)⇒2(BM+BN)=(BM+BN)+(MP+NQ)⇒2(OO’)=(BM+MP)+(BN+NQ)⇒2(OO’)=BP+BQ⇒2OO’=PQ

Hence, proved.



23. Two circles with centres O and O' intersect at two points A and B. A line

PQ is drawn parallel to 00' through A or B, intersecting the circles at P

and Q. Prove that PQ = 200'.the first order reaction is found to have a rate constant k=2 min/sec.Find the half period of the reaction?Firstly draw two circles with center O and O’ such that they intersect at A and B.

Draw a line PQ parallel to OO’.

In the circle with center O, we have:

OP and OB are the radii of the circle. PB is the chord with OM as its perpendicular bisector.

i.e. BM=MP....(1)

In the circle with center O’, we have:

O’B and O’Q are the radii of the circle. BQ is the chord with O’N as its perpendicular bisector.

i.e. BN=NQ....(1)BM=MP....(1)

From (1) and (2), we have:

BM+BN=MP+NQ⇒(BM+BN)+(BM+BN)=(BM+BN)+(MP+NQ)⇒2(BM+BN)=(BM+BN)+(MP+NQ)⇒2(OO’)=(BM+MP)+(BN+NQ)⇒2(OO’)=BP+BQ⇒2OO’=PQ

Hence, proved.



किसने किससे कहा,

1. चलो मैं भी चल कर देखता हूँ,

2आओ उठो हमें भीमा का हाथ बटाना चाहिए

3. क्या वक्ता है । class 8For the reaction 2A + B → 2C, doubling the concentration of B doubles the rate and doubling the concentration of A

also doubles the rate. The rate law is

(a) rate = k[A]”[B] (b) rate = k[A][B] (c) rate = k[A]”[B][C] (d) rate = k[A]”[B]

Rate = k[A][B] b) Rate = k[A]+[B] Rate = k[A][B] d) Rate = k[A]?[B]. 17-1. 2. ... doubled for the reaction, there is no change in the reaction rate. ... For the reaction AB → A + B, concentration of AB was measured as the ...

HOPE SO IT WILL HELP ........