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from right triangle BOD

BD = √(OB²-OD²) = √(13²- 8²) = √105

from right triangle POD

PD = √(OP²-OD²) = √(13²-8²) = √105

PB = BD + PD = 2√105 PB²= 420

from right triangle APB

AP = √(AB²- PB²) =√26²- 420 = √676-420 = √256 = 16

Given :

p(x) = kx² - 14x + 8 = 0

As 2 is root

p(2) = 0

2²k - 28 + 8 = 0

4k = 20

k = 5

thanks for my help

(800÷10)-{(13×7)÷(20-7)}=80-(91÷13)=80-7=73

For equal roots of quadratic equationb^2 - 4*a*c = 0

In quadratic equation kx^2 + 4x + 1=0a = k, b = 4, c = 1

4*4 - 4*k*1 = 016 = 4kK = 16/4 = 4

Value of k is 4

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= Substituting x = -4 in the equation.

= k (-4)^2 + 5 ( -4 ) - 12 = 0

= 16 k - 20 - 12 = 0

= 16 k - 32 = 0

= 16 k = 32

= k = 2.

Letp=3527^654Before answering this question let us discuss on what factors will the unit digit ofpwill depend:

For any two nature numberxandythe unit digit of of its productxywill be the product of their respective unit digit.

So forp=3527⋅3527⋅3527⋅............654timesthe unit digit will be nothing but the unit digit ofq=7⋅7⋅7⋅.....654timestimes

∴We now have to find the unit digit forq=7^654Here we need to do an important observation that the unit digit in exponent of7repeat after a particular interval (upon increasing exponent by 4).7^1→7

7^2→9

7^3→3

7^4→1

7^5→7

So now we need to find the remainder when654is divided by4654mod4=2

→q=7^654=7^4n+2 wheren=163∴∴unit digit ofq is9∴∴unit digit of3527^654is9

1 newton=

100000 dynes

Lett center of bigger semicircle(radius=3) is R1 ,the next bigger one (radius=2) is R2 and the smaller one(radius=1) is R3 and complete circle one is R (assume its radius is r )

Now consider the triangle R2 R3 R and draw a line from R to R1 it acts as a cevianon this trisngle lengths of triangle

R2 R3 = 3

R2 R =2+R

R3 R=1+R

R1 R=3-R [bigger circle is normal and complete circle are same so radius of both coincide]

R2 R1=1

R3 R1=2

now applycosine rule at point R1

(2+r)2-12-(3-r)2/2*1*(3-r) = -[(1+r)2-22-(3-r)2/2*2*(3-r)]

solving this we get r= 12/14=6/7

p+q=13

When two circles touch each other externally, 3 common tangents can be drawn to the circles.

prove that if a diameter of circle is o cord AB of bigger circle intersects the smaller circle in points p and q show. that AP =BQ

Given Diameter AB bisect chords CD and EF , So;CM = MDAndEN = NF

And we know if a line from center to chord is bisecting that chord ,then that line is also perpendicular to that chord . So;OM⊥CD and ON⊥EFSo,∠OMC =∠OMD =∠ONE =∠ONF = 90°And∠OMC +∠ ONE = 180° ------ ( 1 )

These are consecutive interior angles .And finally, equation (1) is satisfied only if;

CD | | EF , Where MN is transversal line .So,CD | | EF ( Hence proved )

Given Diameter AB bisect chords CD and EF , So

CM = MDAndEN = NF

And we know a line from center to chord is bisect that chord than that line also perpendicular on that chord . So

OM⊥CD and ON⊥EFSo,∠OMC =∠OMD =∠ONE =∠ONF = 90°

And∠OMC +∠ ONE = 180° ------ ( 1 )

These are consecutive interior angles , Only satisfied equation 1 true if

CD | | EF , Where MN is transversal line .So,CD | | EF ( Hence proved )

Given Diameter AB bisect chords CD and EF , So

CM = MDAndEN = NF

And we know a line from center to chord is bisect that chord than that line also perpendicular on that chord . So

not this give right answer

equal rootsmeans D = 0(-2k)^2-4k(-3)=04k^2+12k = 04k^2 = -12k4k = -12k = -3

Let roots beα andβA/qα = 6βnow, ifα andβ are roots then equation will be (x -α)(x -β) =0(x -α)(x -β) =0⇒ x² - (α+β)x +αβ =0now puttingα = 6β ,⇒x² - (6β +β)x + 6β×β =0⇒x² - 7βx +6β² =0now comparing with kx² -14x +8 =07β =14/k⇒β =2/k ⇒β² = 4/k²_______(1)and 6β² =8/k⇒β² =4/3k_______(2)equating (1) and (2), we get,4/k² = 4/3k⇒k =3

let the age of satish be 3x.age of satish after 8 years

Zero can be written as 0/1, here both 0 and 1 are integers it can be in p/q, both p and q should be integers.

Denominator, which is 1 here is not equal to 0 i.e., q≠0 .

So, Zero is a rational number

Yesbecause 0 can be written in form of p/q that is 0/1

Yes, 0 is arational number(provided that you understand 0 as the symbol of the natural or integernumber zero). The definition of arational numberis 'Anumberthat can be written in the form a/b, where a and b are integers with no common factors and b is NOTzero'.

Yes, because 0 can be written as 0/1 which is of the form p/q, where p and q are integers and q not=0.

origins used for moving or shifting of point it is the only point where x and y are equal to zero.

scl038 AmbitiousAnswer:

Mathematics are commonly used for calculating exact value and it is widely used in the major areas such as

•Algebra.

•Calculus and analysis.

•Combinatorics.

•Dynamical systems and differential equations.

•Geometry and topology.

•Logic.

•Mathematical physics.

•Number theory.

Hope this answer helps a lot and get collect ideas about mathematics.

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3 +12000= 77 7so 4 is 12000 71=12000÷4 =300073=1 ×3 =3000×3=90007 74 + 3 = 12000+9000 = 210007 7so there are 21000 people in the match

(a) 0.56(b) 0.27(c) 3.7(d) 40.5(e) 1.213(f) 1.109(g) dono =(h) 1.371

(a) 0.56(b) 0.27(c) 3.7(d) 40.5(e) 1.213(f) 1.109(g) 1.9(h) 1.371

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(a)0.56(b) 0.27(c) 3.7(d)40.5(e)1.213(f)1.109(g) 1.9(h) 1.371is the correct answer

Let 7+2√3 is a rational number.

7 is rational and √3 is a irrational.

(7+2√3)-7 is a rational. [ DIFFERENCE BETWEEN TWO RATIONAL NUMBER IS RATIONAL]

2√3 IS RATIONAL.

2 IS RATIONAL AND √3 IS IRRATIONAL.

2×2√3 IS RATIONAL. [ PRODUCT OF TWO RATIONAL NUMBER IS RATIONAL].

√3 IS RATIONAL.

BUT THIS CONTRADICTS THE FACT √3 IS RATIONAL..

OUR ASSUMING IS WRONG.

HENCE,

7+2√3 IS IRRATIONAL

There are 12 edges in cuboid

thank you

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Inmathematics, amatrix is arectangulararrayofnumbers,symbols, orexpressions, arranged inrowsandcolumns.

For example, the dimensions of the matrix below are 2 × 3 (read "two by three"), because there are two rows and three columns

वास्तविक अथवा सम्मिश्र संख्याओं या फलनों की आयताकार सारणी को आव्यूह अथवा मैट्रिक्स कहते हैं

Inmathematics, amatrix(plural:matrices) is a rectangular array of numbers, symbols, or expressions, arranged in rows and columns. The individual items in an m×nmatrixA, often denoted by ai,j, where i and j usually vary from 1 to m and n, respectively, are called its elements or entries.

i)

In quadrilateral ABED,

AB = DE and AB || DE

(given)

So,quadrilateral ABED is a parallelogram

[Since a pair of opposite side isequal and parallel]

(ii) In quadrilateral BEFC

Again BC = EF andBC || EF.

so,quadrilateral BEFC is a parallelogram.

[Since a pair of opposite side isequal and parallel]

(iii)Since ABED and BEFC are parallelograms.

AD = BE and BE = CF (Opposite sides of a parallelogram are equal)

Thus, AD = CF.

Also, AD || BE and BE || CF

(Oppositesides of a parallelogram are parallel)

Thus, AD ||CF

Hence , AD||CF & AD= CF

(iv)AD and CF are opposite sides of quadrilateral ACFDwhich are equal and parallel to each other.

Thus, AFCD it is a parallelogram.

(v) Since, ACFD is a parallelogram.

AC || DF and AC=DF

(vi) In ΔABC and ΔDEF,

AB = DE (Given)

BC = EF (Given)

AC = DF (Opposite sides of a parallelogram)

Thus, ΔABC ≅ ΔDEF (by SSS congruence rule)