Math, asked by salunkeswati848, 2 months ago


 \:
plz answer this question

no copied answers and irrelevant answers and step by step and I want all questions answers otherwise I'd will report ​

Attachments:

Answers

Answered by itskookiesprincess
6

{ \huge\fbox \pink {Q}\fbox \blue {u} \fbox \purple {e} \fbox \green{s} \fbox \red {t} \fbox \orange {i} \fbox{o} \fbox \gray{n}}

show how √5 on number line

{ \huge\fbox \pink {a}\fbox \blue {n} \fbox \purple {s} \fbox \green{w} \fbox \red {e} \fbox \orange {r}}

Step 1: Let line AB be of 2 unit on a number line.

Step 2: At B, draw a perpendicular line BC of length 1 unit.

Step 3: Join CA

Step 4: Now, ABC is a right angled triangle. Applying Pythagoras theorem,

AB2+BC2 = CA2

22+12 = CA2 = 5

⇒ CA = √5 . Thus, CA is a line of length √5 unit.

Step 4: Taking CA as a radius and A as a center draw an arc touching

the number line. The point at which number line get intersected by

arc is at √5 distance from 0 because it is a radius of the circle

whose center was A.

Thus, √5 is represented on the number line as shown in the figure.

{ \huge\fbox \pink {Q}\fbox \blue {u} \fbox \purple {e} \fbox \green{s} \fbox \red {t} \fbox \orange {i} \fbox{o} \fbox \gray{n}}

find six rational number between 3 and 4

{ \huge\fbox \pink {a}\fbox \blue {n} \fbox \purple {s} \fbox \green{w} \fbox \red {e} \fbox \orange {r}}

There are infinite rational numbers between 3 and 4.

As we have to find 6 rational numbers between 3 and 4, we will multiply both the numbers, 3 and 4, with 6+1 = 7 (or any number greater than 6)

i.e., 3 × (7/7) = 21/7

and, 4 × (7/7) = 28/7. The numbers in between 21/7 and 28/7 will be rational and will fall between 3 and 4.

Hence, 22/7, 23/7, 24/7, 25/7, 26/7, 27/7 are the 6 rational numbers between 3 and 4.

{ \huge\fbox \pink {Q}\fbox \blue {u} \fbox \purple {e} \fbox \green{s} \fbox \red {t} \fbox \orange {i} \fbox{o} \fbox \gray{n}}

represent root 9.3 on the number line

{ \huge\fbox \pink {a}\fbox \blue {n} \fbox \purple {s} \fbox \green{w} \fbox \red {e} \fbox \orange {r}}

Step 1: Draw a 9.3 units long line segment, AB. Extend AB to C such that BC=1 unit.

Step 2: Now, AC = 10.3 units. Let the centre of AC be O.

Step 3: Draw a semi-circle of radius OC with centre O.

Step 4: Draw a BD perpendicular to AC at point B intersecting the semicircle at D. Join OD.

Step 5: OBD, obtained, is a right angled triangle.

Here, OD 10.3/2 (radius of semi-circle), OC = 10.3/2 , BC = 1

OB = OC – BC

⟹ (10.3/2)-1 = 8.3/2

Using Pythagoras theorem,

We get,

OD2=BD2+OB2

⟹ (10.3/2)2 = BD2+(8.3/2)2

⟹ BD2 = (10.3/2)2-(8.3/2)2

⟹ (BD)2 = (10.3/2)-(8.3/2)(10.3/2)+(8.3/2)

⟹ BD2 = 9.3

⟹ BD = √9.3

Thus, the length of BD is √9.3.

Step 6: Taking BD as radius and B as centre draw an arc which touches the line segment. The point where it touches the line segment is at a distance of √9.3 from O..

{ \huge\fbox \pink {Q}\fbox \blue {u} \fbox \purple {e} \fbox \green{s} \fbox \red {t} \fbox \orange {i} \fbox{o} \fbox \gray{n}}

every natural number is a whole number

{ \huge\fbox \pink {a}\fbox \blue {n} \fbox \purple {s} \fbox \green{w} \fbox \red {e} \fbox \orange {r}}

True

Natural numbers- Numbers starting from 1 to infinity (without fractions or decimals)

i.e., Natural numbers= 1,2,3,4…

Whole numbers- Numbers starting from 0 to infinity (without fractions or decimals)

i.e., Whole numbers= 0,1,2,3…

Or, we can say that whole numbers have all the elements of natural numbers and zero.

Every natural number is a whole number; however, every whole number is not a natural number...

Similar questions