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Step 1. Draw a line OM.Step 2.using protractor and center 'o' draw an angle MON =147°.Step 3.Now taking 'o' as center and any radius draw a arc that intersects 'OM' and 'ON' at P and Q.Step 4. now take P and Q as centers and radius more than half of PQ, draw arcs.Step 5. both the arcs intersect at 'R'Step 6. join 'OR' and produce it.Step 7. 'OR' is the required bisector of angle MON

here 3*2=6+1=75*4=20+3=237*6=42+5=479*8=72+7=7910*9=90+9=99first and second column is multiplied and added by odd numbers

Here 3*2+1=7 5*4+3=23 7*6+5=47 9*8+7=79 10*9+9=99,In the question first 1+And second 1+2=3And third 3+2=5 all word in 2 difrences

p=(-3/2,3), q=(6,-2), R=(-3,4) y 3 points. Y=1/2[x1(y2-y3)+x2(y3-y1)+x3(y1-y2)]=0; 1/2[-3/2(-2-4)+6(4-3)+(-3)[3+2]=0;. 1/2[-3/2(-6)+6(1)-3(5)=0; 1/2[9+6-15]=1/2[15-15]=0;

if the slope of PQ=QR=PR then the points lie in a same plane or are collinear

thnx

x=4/5(X+10)5x=4x+40x=40

Let the ratio be x so Arti gets 2x part andBharti gets 3x parts.According to question,

3x + 2x = 65005x = 6500x = 1300

So, Arti will get 2x = 2600Bharti will get 3x = 3900

5 is your write answers

Your correct answer is 5

Consider the numbers such that when each number is divided by 6, 8 and 9, it leaves 8 as remainder in each case.

LCM of 6, 8, 9 is 72.The required numbers are 8 + 72*n where n=1,2,3,..... and no.s are between 200 and 300.

So, possible values of n are 3 and 4.

Numbers are 224 and 296.

(b) is correct option

number s are 224&296

3²ⁿ = 9ⁿ = (8+1)ⁿ

so, the binomial expansion of this will always contain a factor of 8 , in the form of 8ⁿ .. but the last term will be 1

so, on dividing it with 8 , 1 will always be there as a reminder.

Correct option: BReason: Slope = (y2-y1)/(x2-x1)= (-1-5)/(4+3)= -6/7

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This is grade 9 question

Construction is the process of constructing a building or infrastructure. Construction differs from manufacturing in that manufacturing typically involves mass production of similar items without a designated purchaser, while construction typically takes place on location for a known client

(i) Since, BD and CE are medians.

AD = DC

AE = BE

Hence, by converse of Basic Proportionality theorem,

ED || BC

In ΔEGD and ΔCGB,

∠DEG = ∠GCB (alternate angles)

∠EGD = ∠BGC (Vertically opposite angles)

ΔEGD ~ ΔCGB (AA similarity)

28x= 20*189/2728x= 140x= 140/80x= 1.75

why p=8n-1 and q=7n-1

627÷1886526=0.000332

rational number between 3 and 4 are3.1 , 3.2, 3.3, 3.4, 3.5, 3.6

(5√2 + 3√3) - ( 6√2 - 7√3) = a√2 + b√3

(5√2 - 6√2) + (3√3 + 10√3) = a√2 + b√3

-√2 + 4√2 = a√2 + b√3

Comparing both sidesa = -1 and b = 4

4x + 5x = 6327

9x = 6327

x = 6327 / 9

x = 703

4 × 703 = 28125× 703 = 3515

Gita gets rs. 2812 and Sita gets rs. 3515

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2) is correct answer

2) is the correct answer of the following

1.slope of tangent=6also slope=6x6x=6x=1then y=72.slope of tangent=-2also slope=2x-42x-4=-22x=2x=1then y=0

People went out=8725 - 2658 = 6067

Please express your question in Math instead of English. For example,56+78, orx^2+5x+6.

56+78, orx^2+5x+6.

In 13 teams there are 390 peopleIn 1 team there are 390/13 = 30 peopleSo, Im 5 team there are 30× 5 = 150 people

4x + 5x = 6327

9x = 6327

x = 6327 / 9

x = 703

4 × 703 = 28125× 703 = 3515

Gita gets rs. 2812 and Sita gets rs. 3515

x=-5/11 and |x|=5/11hence 8 rational number =-4/11,-3/11,-2/11,-1/11,0,1/11,2/11

sinβ-cosβ=

(c) 13

Explanation:There are 6 people to either side. Therefore, total number of people = 6 + 1 + 6 = 13

Thirteen people are there in queue

10 hoga agar reminder 8 or 12 ho to

Given that the smallest number when divided by 28 and 32 leaves remainder 8 and 12 respectively.

28 - 8 = 20 and 32 - 12 = 20 are divisible by the required numbers.

Therefore the required number will be 20 less than the LCM of 28 and 32.

Prime factorization of 28 = 2 * 2 * 7

Prime factorization of 32 = 2 * 2 * 2 * 2 * 2

LCM(28,32) = 2 * 2 * 2 * 2 * 2 * 7

= 224.

Therefore the required smallest number = 224 - 20

= 204.

Verification:

204/28 = 28 * 7 = 196.

= 204 - 196 = 8

204/32 = 32 * 6 = 192

= 204 - 192

= 12.

Required number = H.C.F. of (1657 - 6) and (2037 - 5) = H.C.F. of 1651 and 2032 = 127

Required number = H.C.F. of (1657 - 6) and (2037 - 5)

= H.C.F. of 1651 and 2032

= 127

R=12cmr=5 cmradius of new circle =Rarea of new circle =sum of areas of both circles

πR^2=144π

πr^2=25π

required area is = 144π+ 25π πR°^2 = 16πR°^2=169R°=13cm

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Let O be the centre of the circle.

Given that,

OQ = 25cm and PQ = 24 cm

As the radius is perpendicular to the tangent at the point of contact,

Therefore, OP ⊥ PQ

Applying Pythagoras theorem in ΔOPQ, we obtain

OP^2 + PQ^2 = OQ^2

OP^2 + 24^2 = 25^2

OP^2 = 625 − 576

OP^2 = 49

OP = 7 cm