Karnataka Board Class 9 Maths Chapter 4 Polynomials Exercise 4.1

Question 1.
Which of the following expressions are polynomials in one variable and which are not ? State reasons for your answer.
i) 4x² – 3x + 7
ii) y² + \(\sqrt{2}\)
iii) \(3 \sqrt{t}+t \sqrt{2}\)
iv) \(\mathrm{y}+\frac{2}{\mathrm{y}}\)
v) x¹⁰ + y³ + t⁵⁰
Answer:
i) 4x² – 3x + 7
Here polynomial has one variable, i.e. x
ii) y² + \(\sqrt{2}\)
Here polynomial has one variable, i.e. y
iii) \(3 \sqrt{t}+t \sqrt{2}\)
This is polynmomial with one variable, because T is only one variable.
iv) \(\mathrm{y}+\frac{2}{\mathrm{y}}\)
Here polynomial has one variable, ie. y.
v) x¹⁰ + y³ + t⁵⁰
This polynomial is not having one variable because here 3 variables means ‘x’, y and ‘t’ are there.

Question 2.
Write the coefficients of x in each of the followng :
i) 2 + x² + x
ii) 2 – x² + x³
iii) \(\frac{\pi}{2}\)x² + x
v) \(\sqrt{2} \mathrm{x}\) – 1
Answer:
i) 2 + x² + x
Here, coefficient of x² is 1.
ii) 2 – x² + x³
Here coefficient of x² is -1
iii) \(\frac{\pi}{2}\)x² + x
Here coefficient of x² is \(\frac{\pi}{2}\).
iv) \(\sqrt{2} \mathrm{x}\) – 1
Here coefficint of x² is -1.

Question 3.
Give one example each of a binomial of degree 35, and of a monomial of degree 100
Answer:
i) A Bionomial of degree 35
E.g. f(x) = – x³⁵ + 10
ii) A binomial of degree 100
E.g. f(y) = – y¹⁰⁰.

Question 4.
Write the degree of each of the following polynomials :
i) 5x³ + 4x² + 7x
ii) 4 – y²
iii) 5t – \(\sqrt{7}\)
iv) 3
Answer:
i) 5x³ + 4x² + 7x Highest power (degree) 3
ii) 4 – y² Highest power degree) 2
iii) 5t – \(\sqrt{7}\) Highest power (degree) 1
iv) 3 Highest power (degree) 0

Question 5.
Classify the folloiwng as linear, quadratic and cubic polynomials :
i) x² + x
ii) x – x³
iii) y + y² + 4
iv) 1 + x
iii) 3t
iv) r²
vii) 7x³
Answer:

We hope the KSEEB Solutions for Class 9 Maths Chapter 4 Polynomials Ex 4.1 help you. If you have any query regarding Karnataka Board Class 9 Maths Chapter 4 Polynomials Exercise 4.1, drop a comment below and we will get back to you at the earliest.

Karnataka Board Class 9 Maths Chapter 4 Polynomials Ex 4.2

KSEEB Solutions For Class 9 Maths Polynomials Question 1.
Find the value of the polynomial 5x – 4x² + 3 at
i) x = 0
ii) x = -1
iii) x = 2
Answer:
i) f(x) = 5x – 4x² + 3 x = 0 then,
f(0) = 5(0) – 4(0)² + 3
=0 – 0 + 3
f(0) = 3

ii) f(x) = 5x – 4x² + 3 x = -1 then,
f(-1) = 5(-1) – 4(-1)2 + 3
= -5 – 4(+1) + 3
= -5 – 4 + 3
= -9 + 3
f(-1) = -6

iii) f(x) = 5x – 4x² + 3 x = 2 then,
f(2) = 5(2) – 4(2)² + 3
= 5(2) – 4(4) + 3
= 10 – 16 + 3
= 13 – 16
f(2) = -3

9th Maths Polynomials Exercise 4.2 Question 2.
Find p(0), p(1) and p(2) for each of the following polynomials :
i) p(y) = y² – y + 1
ii) p(t) = 2 + t + 2t² – t³
iii) p(x) = x³
iv) p(x) = (x – 1) (x + 1)
Answer:
i) (a) p(y) = y² – y + 1
p(0) = (0)² – 0 + 1
= 0 – 0 + 1
∴ p(0) = =1

(b) p(y) = y² – y + 1
p(1) = (1)² – 1 + 1
= 1 – 1 + 1
∴ p(1) = 1

(c) p(y) = y² – y +
p(2) = (2)² – 2 + 1
= 4 – 2 + 1
= 5 – 2
∴ p(2) = 3

ii) (a) p(t) = 2 + t + 2t² – t³
p(0) = 2 + 0 + 2² – (0)³
= 2 + 0 + 0 – 0
∴ p(0) = 2

(b) p(t) = 2 + t + 2t² – t³
p(1) = 2 + 1 + 2(1)² – (1)³
= 2 + 1 + 2(1) – 1
= 2 + 1 + 2 – 1
∴ p(1) = 4

(c) p(t) – 2 + t + 2t² – t³
p(2) = 2 + 2 + 2(2)² – (2)³
= 2 + 2 + 2(4) – 8
= 2 + 2 + 8 – 8
∴ p(2) = 4

iii) (a) p(x) = x³
p(0) = (0)³
p(0) = 0

(b) p(x) = x³
p(1) = (1)³
∴ p(1) = 1

(c) p(x) = x³
p(2) = (2)³
∴ p(2) = 8

iv) (a) p(x) = (x – 1)(x + 1)
p(x) = x² – 1 [∵ (a + b)(a – b) = a² – b]
p(0) = (0)² – 1
= 0 – 1
∴ p(o) = -1

(b) p(x) = (x – 1)(x + 1)
p(x) = x² – a [∵ (a + b)(a – b) = a² – b²]
p(1) = (1)² – 1
= 1 – 1
∴ p(1) = 0

(c) p(x) = (x – 1)(x + 1)
p(x) = x² – 1 [∵ (a + b)(a – b) = a² – b²]
p(2) = (2)² – 1
= 4 – 1
∴ p(0) = 3

KSEEB Solutions For Class 9 Maths Polynomials Exercise 4.2 Question 3.
Verify whether the following are zeroes of the polynomial, induced against them.
i) p(x) = 3x + 1; \(x=-\frac{1}{3}\)
ii) p(x) = 5x – π; \(x=\frac{4}{5}\)
iii) p(x) = x² – 1; x = 1, -1
iv) p(x) = (x + 1) (x – 2); x = -1, 2
v) p(x) = x²; x = 0
vi) p(x) = lx + m; \(x=-\frac{m}{l}\)
vii) p(x) = 3x² – 1; \(x=-\frac{1}{\sqrt{3}}, \frac{2}{\sqrt{3}}\)
viii) p(x) = 2x + 1; \(x=\frac{1}{2}\)
Answer:
i) p(x) = 3x + 1; \(x=-\frac{1}{3}\)

Here value of polynomial is zero.
\(x=-\frac{1}{3}\) is not zero of the polynomial

ii) p(x) = 5x – π; \( x=\frac{4}{5}\)

Here value of polynomial is not zero.
\(x=\frac{4}{5}\) is not zero of the polynomial

iii) p(x) = x² – 1; x = 1, -1
p(1) = (1)² – 1
= 1 – 1
p(1) = 0
Here value of p(x) is zero.
hence its zero is 1.
p(x) = x² – 1; x = -1
p(-1) = (-1)² – 1
= 1 – 1
p(-1) = 0
Here value of p(x) is zero.
∴ -1 is zero.

iv) p(x) = (x – 1)(x – 2); x = -1, 2
p(x) = x² – 2x + x – 2
p(x) = x² – x + 2 x = -1
p(-1) = (-1)² – (-1)² + 2
= 1 + 1 + 2
p(-1) = 4
Here value of polynomila is not zero.
∴ -1 is not zero.
p(x) = x² – x + 2 x = 2
p(2) = (2)² – (2) + 2
= 4 – 2 + 2
p(-1) = 4
Here value of polynomial is not zero.
∴ 2 is not zero.

(v) p(x) = x²; x = 0
p(0) = (0)²
p(0) = 0
Here value of p(x) is zero.
∴ 0 is its zero.

vi) p(x) = lx + m; \(x=-\frac{m}{l}\)
\(\mathrm{p}\left(-\frac{\mathrm{m}}{l}\right)=l\left(-\frac{\mathrm{m}}{l}\right)+\mathrm{m}\)
= -m + m
= 0
Here p(x) is zero.
∴ \(-\frac{\mathrm{m}}{l}\) is its zero.

(vii) p(x) = 3x² – 1; \(x=-\frac{1}{\sqrt{3}}, \frac{2}{\sqrt{3}}\)

viii) p(x) = 2x + 1; \(x=\frac{1}{2}\)

Here value of p(x) is not zero.
∴ \(\frac{1}{2}\) is not its zero.

Polynomials Class 9 Exercise 4.2 Solutions Question 4.
Find the zero of the polynomial in each of the following cases :
(i) p(x) = x + 5
(ii) p(x) = x – 5
(iii) p(x) = 2x + 5
(iv) p(x) = 3x – 2
(v) p(x) = 3x
(vi) p(x) = ax, a ≠ 0
(vii) p(x) = cx + d, c ≠ 0, c, d are real numbers.
Answer:
i) p(x) = x + 5
Let p(x) =0, then,
p(x) = x + 5 = 0
x = 0 – 5
∴ x = -5
-5 is zero of p(x).

ii) p(x) = x – 5
If p(x) = 0, then
p(x) = x – 5 = 0
x = 0 + 5
∴ x = 5
5 is the zero of p(x).

iii) p(x) = 2x + 5
If p(x)= 0, then
p(x) = 2x + 5 = 0
2x = – 5
∴ \(x=\frac{-5}{2}\)
\(\frac{-5}{2}\) is the zero of p(x).

iv) p(x) = 3x – 2
If p(x)= 0, then
p(x) = 3x – 2 = 0
3x = 2
∴ \(x=\frac{2}{3}\)
\(\frac{2}{3}\) is the zero of p(x).

v) p(x) = 3x
If p(x) = 0, then
p(x) = 3x = 0
∴ \(x=\frac{0}{3}\)
\(\frac{0}{3}\) is the zero of p(x)

vi) p(x) = ax, a ≠ 0
If p(x)= 0, then
p(x) = ax = 0
∴ \(x=\frac{0}{a}\) ∴ x = ∞(infinity)
∞ is the zero of p(x).

vii) p(x) = cx + d, c ≠ 0, c, d are real numbers
If p(x)= 0, then
p(x) = cx + d = 0
cx = 0 – d
cx = -d
∴ \(x=-\frac{d}{c}\)
\(-\frac{\mathrm{d}}{\mathrm{c}}\) is the zero of p(x).

We hope the KSEEB Solutions for Class 9 Maths Chapter 4 Polynomials Ex 4.2 help you. If you have any query regarding Karnataka Board Class 9 Maths Chapter 4 Polynomials Exercise 4.2, drop a comment below and we will get back to you at the earliest.

Karnataka SSLC Class 10 Science Solutions Chapter 7 Control and Coordination

KSEEB SSLC Class 10 Science Chapter 7 Intext Questions

Text Book Part I Page No. 85

KSEEB Solutions For Class 10 Science Control And Coordination Question 1.
What is the difference between a reflex action and walking?
Answer:
A reflex action is an automatic reaction for each stimulation in our body initiated by our sense responses e.g., we move our hand immediately after a contact with hot object. It is a direct controlled action. Walking is completely controlled by our brain. On the other hand, is a voluntary action. It requires complete coordination of muscles, bones, eyes etc.

Control And Coordination Class 10 KSEEB Solutions Question 2.
What happens at the synapse between two neurons?
Answer:
Synapse allows delivery of impulses (in chemical form) from neurons to other neurons and the target point, such as muscle cells. Tiny gap between the last portion of axon of one neuron and the dendron of the other neuron is known as a synapse. At a time, it acts as a one way path and transmits impulses in one direction only.

Control And Coordination Class 10 Notes KSEEB Question 3.
Which part of the brain maintains posture and equilibrium of the body?
Answer:
A part of hindbrain-cerebellum is responsible for maintaining posture and equilibrium of the body.

KSEEB Solutions For Class 10 Science Question 4.
How do we detect the smell of an agarbatti (incense stick)?
Answer:
Forebrain is responsible for thinking work. It has separate areas that are specialized for hearing, smelling, sight, taste, touch etc. The _ forebrain also has regions that collect information or impulses from various receptors. When the smell of an incense stick reaches us, out forebrain detects it. Then, the forebrain interprets it by putting it together with the information received from other receptors and also with the information already stored in the brain.

KSEEB Solutions For Class 10 Question 5.
What is the role of the brain in reflex action?
Answer:
Reflex actions are sudden responses, which do not involve any thinking. A connection of detecting the signal from the nerves (input) and responding to it quickly (output) is called\a reflex arc. The reflex arcs can be considered as connections present between the input and output nerves which meets in a bundle in the spinal cord. The brain is only responsible of the signal and the response.

Text Book Part I Page No. 88

KSEEB Solutions For Class 10 Science Chapter 7 Question 1.
What are plant hormones?
Answer:
Plant hormones or phytohormones are chemicals which are naturally present inside the plants similar to those substances as hormones in animals. These hormones have specific functions and reach the target point to conduct the messages and initiate proper functions like growth, coordinations etc. These are synthesized in one part of the plant body (in minute quantities) and are translocated to other parts when required. The five important phytohormones are auxins, gibberellins, cytokinins, abscisic acid and ethylene.

KSEEB Solutions 10 Science Question 2.
How is the movement of leaves of the sensitive plant different from the movement of a shoot towards light?
Answer:
The movement of shoot towards light is natural and called phototropism. This type of movement is directional and is growth oriented. But, the movement of leaves of the sensitive plants, like Mimosa pudica or “touch me not”, occurs in response to touch or contact stimuli. This movement is not growth oriented.

KSEEB 10 Science Solutions Question 3.
Give an example of a plant hormone that promotes growth.
Answer:
Auxin is a growth-promoting plant hormone.

KSEEB 10th Science Solutions Question 4.
How do auxins promote the growth of a tendril around a support?
Answer:
Auxin helps the cell grow longer and is synthesized at the shoot tip. When a tendril comes in contact with a support, auxin stimulates faster growth of the cells on the opposite side, so that the tendril forms a coil around the support.

KSEEB Solutions For Class 10 Biology Question 5.
Design an experiment to demonstrate hydrotropism.
Answer:
Take two troughs A and B. Fill them with soil. In the trough B place, a small clay pot Now .plant a small seeding in both troughs. Uniformly water the soil in trough A. but pour water in the clay pot of trough B. After a few days take out the seedling from the troughs. The seedling in trough A which used get uniform water has normal straight roots while roots of seedling planted in trough B show bent growth towards the clay pot containing water.

Text Book Part I Page No. 91

KSEEB Solutions Class 10 Science Question 1.
How does chemical coordination take place in animals?
Answer:
Hormones cause chemical coordination in animals and help in various functions. Hormone is the chemical messenger that regulates the physiological processes in living organisms. It is secreted by glands. The regulation of physiological processes and control and coordination by hormones comes under the endocrine system. The nervous system along with the endocrine system in our body controls and coordinates the physiological processes.

KSEEB Solutions For 10th Science Question 2.
Why is the use of Iodised salt advisable?
Answer:
If there is deficiency of Iodine we get disease called goitre. Hence use of iodized salt is advisable Iodine is necessary for the thyroid gland to make thyroxin hormone.

KSEEB Solutions 10th Science Question 3.
How does our body respond when adrenaline is secreted into the blood?
Answer:
Adrenaline is a hormone secreted by the adrenal glands and it is responsible to control any kind of danger or emergency or any kinds of stress. It is secreted directly into the blood and is transported to different parts of the body. It fasten the heartbeat and the breathing rate, and provides more oxygen to the muscles. It also increases the blood pressure. All these responses enable the body to deal with any stress or emergency.

KSEEB Solutions For 10 Science Question 4.
Why are some patients of diabetes treated by giving injections of insulin?
Answer:
Insulin is the hormone secreted by pancreas and helps regulating blood sugar levels. If it is not secreted in proper amounts, the sugar level in the blood rises causing many harmful effects. Hence some patients of diabetes treated by giving injections of Insulin.

KSEEB SSLC Class 10 Science Chapter 7 Textbook Exercises

KSEEB Solutions Science Class 10 Question 1.
Which of the following is a plant hormone?
(a) Insulin
(b) Thyroxin
(c) Oestrogen
(d) Cytokinin.
Answer:
(d) Cytokinin.

KSEEB Solutions For Science Class 10 Question 2.
The gap between two neurons is called a
(a) dendrite
(b) synapse
(c) axon
(d) impulse
Answer:
(b) synapse.

KSEEB Solutions Of Class 10 Question 3.
The brain is responsible for
(a) thinking.
(b) regulating the heart beat.
(c) balancing the body.
(d) all of the above.
Answer:
(d) all of the above.

10th State Syllabus Science Notes Question 4.
What is the function of receptors in our body? Think of situations where receptors do not work properly. What problems are likely to arise?
Answer:
The receptors are usually located in our sense organs such as the inner ear, the nose, the tongue and so on. So gustatory receptors will detect taste while olfactory receptors will detect smell. If these receptors are not working properly, it is harmful to our body.
Eg: If we are suffering from cold, we do not feel the taste of food.

Question 5.
Draw the structure of a neuron and explain its function.
Answer:

Function of a neuron:
At the end of the axon, the electrical impulse sets off the release of some chemicals. These chemicals cross the gap or synapse, and start a similar electrical impulse in a dendrite of the next neuron. A similar synapse finally allows delivery of such impulses from neurons to other cells, such as muscle cells or gland.

Question 6.
How does phototropism occur in plants?
Answer:
The growth movement in plants in response to light stimulus is known as phototropism. The shoots show positive phototropism and the roots show negative phototropism. This means that the shoots bend towards the source of light whereas the roots bend away from the light source. For example: The flower head of sunflower is positively phototropic and hence, it moves from east to west along with the sun.

Question 7.
Which signals will get disrupted in case of a spinal cord injury?
Answer:
The reflex arc connections between the input and output nerves meet in a bundle in the spinal cord. In fact, nerves from all over the body meet in a .bundle in the spinal cord on their way to the brain. In case of any injury to the spinal cord, the signals coming from the nerves as well as the signals coming to the receptors will be disrupted.

Question 8.
How does chemical coordination occur in plants?
Answer:
In animals, control and coordination occur with the help of nervous system. However, plants do not have a nervous system. Plants respond to stimuli by showing movements. The growth, development and responses, to the environment in plants is controlled and coordinated by a special class of chemical substances known as hormones. These hormones are produced in one part of the plant body and are translocated to other needy parts. For example, a hormone produced in roots is translocated to other parts when required. The five major types of phytohormones are auxins, gibberellins, cytokinins, abscisic acid and ethylene. These phytohormones are either growth promoters (such as auxins, gibberellins cytokinins and ethylene) or growth inhibitors such as abscisic acid.

Question 9.
What is the need for a system of control and coordination in an organism?
Answer:
Controlled movement must be connected to the recognition of various events in the enviroment, followed by only the correct movement in response. In other words, living organisms must use systems providing control and coordination. In keeping with the general principles of body organisation. In multi cellular organisms, specilised tissues are used to provide these control and coordination activities. In animals such control and coordination are provided by nervous and muscular tissues.

Question 10.
How are involuntary actions and reflex actions different from each other?
Answer:

Question 11.
Compare and contrast nervous and hormonal mechanisms for control and coordination in animals.
Answer:

Question 12.
What is the difference between the manner in which movement takes place in a sensitive plant and the movement in our legs?
Answer:

Movement in sensitive plants

1. The movement that takes place in a sensitive plant such as Mimosa pudica occurs in response to touch (stimulus).
2. For this movement, the information is transmitted from cell to cell by electric chemical signals as plants do not have any specialised tissue for conduction of impulses.
3. For this movement to occur, the plant cells change shape by changing the amount of water in them.

Movement in our legs

1. Movement in our legs is an example of voluntary actions.
2. The signal or messages for these no action are passed to the brain and hence are consciously controlled.
3. In animal muscle cells, some proteins are found which allow the movement to occur.

KSEEB SSLC Class 10 Science Chapter 7 Additional Questions and Answers

Question 1.
Draw a neat diagram showing Reflex arc and label the parts.
Answer:

Question 2.
Which organ protects the spinal cord and brain?
Answer:
Vertebral column or backbone protects the spinal cord and bony box protects the brain.

Question 3.
Give one example of chemotropism.
Answer:
Growth of pollen tubes towards omles.

Question 4.
Which hormone promote cell division?
Answer:
Cytokinins.

Question 5.
Which hormone helps in regulating blood sugar levels.
Answer:
Insulin.

We hope the given KSEEB SSLC Class 10 Science Solutions Chapter 7 Control and Coordination will help you. If you have any query regarding Karnataka SSLC Class 10 Science Solutions Chapter 7 Control and Coordination, drop a comment below and we will get back to you at the earliest.

Karnataka Board Class 9 Maths Chapter 3 Lines and Angles Ex 3.1

KSEEB Solutions For Class 9 Maths Lines And Angles Question 1.
In Fig 3.13, lines AB and CD intersect at O. If ∠AOC + ∠BOE = 70° and ∠BOD = 40°, find ∠BOE and reflex ∠COE.

Answer:

∠AOC = ∠BOD = 40°
∠BOD = 40° (Date)
∠AOC = ∠BOD = 40° because vertifically opposite angles.
∴ ∠AOC + ∠BOE = 70°
40° + ∠BOE = 70°
∴ ∠BOE = 70° – 40° = 30°, and
∠AOD = 180° – ∠BOD = 180° – 40° = 140°
Reflex angle COE = ∠COA + ∠AOD + ∠BOD + ∠BOE
= 40° + 140° + 40° + 30°
∴ Reflex angle, ∠COE = 250°.

Lines And Angles Class 9 Exercise 3.1 Question 2.
In Fig. 3.14, lines XY and MN intersect at O. If ∠POY = 90° and a : b = 2 : 3, find c.

Answer:
∠XOP + ∠POY = 180°
∵ straight supplementary
∠XOP + 90° =180°
∴ ∠XOP = 180° – 90° = 90°
But, ∠XOP = a + b = 90°
a : b = 2 : 3
2 + 3 = 5 Ratio
Ratio 5 means 90°
\(=\frac{2 \times 90}{5}=2 \times 18=36^{\circ}\)
∴ If a = 36° then, ∠b = 54°.
∠XOM = ∠YON = b = 54° (∵ Vertically opposite angles)
∠XON + ∠YON = 180° (∵ Straight angle)
∴ c + 54° = 180°
c = 180 – 54
∴ c = 126°.

KSEEB Solutions For Class 9 Maths Question 3.
In Fig. 3.15, ∠PQR = ∠PRQ, then prove that ∠PQS = ∠PRT.

Answer:

Data: In this figure, ∠PQR = ∠PRQ.
To prove: ∠PQS = ∠PRT
Proof: In ∆PQR, ∠Q = ∠R.
∴ This is an isosceles triangle.
Let ∠Q = 70°, then ∠R = 70° .
∠PQS + ∠PQR = 180°
∠PQS + 70° = 180°
∴ ∠PQS =180 – 70
∠PQS = 110° … (i)
Similarly,
∠PRT + ∠PRQ = 180°
∠PRT + 70° = 180°
∴ ∠PRT = 180 – 70
∠PRT =100 … (ii)
from (i) and (ii)
∠PQS = ∠PRT =110°
∴ ∠PQS = ∠PRS proved.

KSEEB Solutions For Class 9 Maths Chapter 3 Question 4.
If Fig. 3.16, if x + y = w + z, then prove that AOB is a line.

Answer:
Data: In this fiure, ∠BOC = x°
∠AOC = y°
∠BOD = w°
∠AOD = z° and
x + y = w + z.
To Prove: AOB is a straight line.
Proof: x + y + w + z = 360° (∵ one completre angle)
But, x + y = w + z.
∴ x + y = w + z= \(\frac{360}{2}\) = 180°
∴ x + y = 180°
∴ w + z = 180° proved.
But, ∠AOC and ∠BOC are adjacent angles,
∠AOC + ∠BOC = 180°
x = y = 180°
∴ ∠AOB = 180°.
∴ AOB is a straight line.

Lines And Angles Class 9 KSEEB Question 5.
In Fig. 3.17, POQ is a line. Ray OR is perpendicular to line PQ. OS is another ray lying between rays OP and OR. Prove that
∠ROS = \(\frac{1}{2}\) (∠QOS – ∠POS).

Answer:
Data: POQ is a straight line. Ray OR is perpendicular on straight line PQ. OS Ray is in between Rays OP and OR.
To Prove: ∠ROS = \(\frac{1}{2}\) (∠QOS – ∠POS)
Proof : ∠QOR = ∠POR = 90° (Data)
∠POR = ∠POS + ∠ROS
∴ ∠ROS = ∠POR – ∠POS
∠ROS = 90°- ∠POS (i)
Now, QOS = ∠QOR + ∠ROS
∠QOS = 90° + ∠ROS
∴ ∠ROS = ∠QOS – 90° (ii)
By adding equation (i) and (ii)

∴ ∠ROS = \(\frac{1}{2}\) (∠QOS – ∠POS)

Lines And Angles Exercise 3.1 Question 6.
It is given that ∠XYZ = 64° and XY is produced to point P. Draw a figure from the given information. If ray YQ bisects ∠ZYP, find ∠XYO and reflex ∠QYP.
Answer:

Data: ∠XYZ = 64° and XY are produced up to P. ∠ZYP is bisected.
To Prove: ∠XYQ = ? Reflex ∠QYP = ?
Proof: YQ bisects ∠ZYP
∴ Let ∠PYQ = ∠ZYQ = x°.
∠XYZ + ∠ZYQ + ∠QYP = 180° (∵ ∠XYP is straight angle)
64 + x + x = 180
∴ 64 + 2x = 180
∴ 2x = 180 – 64
2x = 116
∴ \(x=\frac{116}{2}=58^{\circ}\)
∴ ∠PYQ = ∠ZYQ = 58°

(i) ∴ ∠XYQ = ∠XYZ + ∠ZYQ
= 64 + 58
∠XYQ = 122°

(ii) Reflex ∠QYP = ∠PYX + ∠XYZ + ∠ZYQ
= 180 ° + 64° + 58°
= 180 + 122
∴ Reflex ∠QYP = 302°.

We hope the KSEEB Solutions for Class 9 Maths Chapter 3 Lines and Angles Ex 3.1 help you. If you have any query regarding Karnataka Board Class 9 Maths Chapter 3 Lines and Angles Exercise 3.1, drop a comment below and we will get back to you at the earliest.

Karnataka Board Class 9 Maths Chapter 1 Number Systems Ex 1.1

KSEEB Solutions For Class 9 Maths Question 1.
Is a zero a rational number? Can you write it in the form \(\frac{p}{q}\), where p and q are integers and q ≠ 0 ? (p, q,ϵ Z, q ≠ 0)
Answer:
Zero is a rational number.
This can be written in the form of \(\frac{p}{q}\) because \(\frac{o}{q}\) is a rational number.
E.g. \(\frac{0}{2}=0, \quad \frac{0}{5}=0\). etc.
Zero belongs to set of rational number.

KSEEB Solutions For Class 9 Maths Number System Question 2.
Find six rational numbers between 3 and 4.
Answer:
We can write six rational numbers between 3 and 4 as
\(3=\frac{21}{7} \text { and } 4=\frac{28}{7}\)
∴ rational numbers between \(\frac{21}{7}\) and \(\frac{28}{7}\).

KSEEB Solutions For Class 9 Maths Chapter 1 Question 3.
Find five rational numbers between \(\frac{3}{5}\) and \(\frac{4}{5}\)
Answer:
Rational numbers between \(\frac{3}{5}\) and \(\frac{4}{5}\) are

∴ Rational numbers between \(\frac{30}{50}\) and \(\frac{40}{50}\)

KSEEB Solutions For Class 9 Maths Chapter 1 Number System Question 4.
State whether the following statements are true or false. Give reasons for your answers :
(i) Every natural number is a whole number.
Answer:
True. Because set of natural numbers belongs to a set of whole numbers.
∴ W = {0, 1, 2, 3 …………………….}

(ii) Every integer is a whole number.
Answer:
False. Because zero belongs to a set of integers. But -2, -3, -1 are not whole numbers.

(iii) Every rational number is a whole number.
Answer:
False. Because \(\frac{1}{2}\) is a rational number but not a whole number.

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