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Here is your answer,

kx^2 - 2X - 3K = 0

Sum of zeroes = -b/a
= -(-2)/k
= 2/k

Product of zeroes = c/a
= -3k/k
= -3

Sum of zeroes = 2 x product of zeroes
2/k = 2 x (-3)
2/k = -6
k = -1/3

Hope it HELPS you!

Perimeter=sum of all sides=20cm
perimeter of RECTANGLE =2 (l+b)
20=2 (l+4)
20÷2=l+4
10=l+4
10-4=l
6=l

P = RS. 7500

R = 16%

T = 2 years.

A = P(1 + (R/100))^T

A = 7500(1+ (16/100))^2

A = 7500*(116/100)*(116/100)

A = Rs. 10092

CI = A-P

CI = 10092 - 7500

CI = Rs. 2592

636=9*(n-1)17-9
636=9*(n-1)8
636=9*8n-8
636=9-8*8n
636=1*8n
636-1=8n
635=8n
then we will DIVIDE 635 with 8.
n=79.375

I ) Area( PQRS ) = Area( ABRS )

solution :

Parallelogram PQRS and

parallelogram ABRS are on same base

SR and between the same parallels

SR//PB.

THEREFORE ,

area(PQRS) = area( ABRS )-----( 1 )

ii ) Area( ∆AXS ) = ( 1/2 ) Area ( PQRS )

Solution :

From ( 1 ) ,

area ( PQRS ) = area ( ABRS )

and parallelogram ABRS and ∆AXS

are on the same base AS and between

the same parallels AS//BR .

Therefore ,

∆AXS = ( 1/2 ) area ( ABRS )

= ( 1/2 ) Area ( PQRS ) [ from ( 1 ) ]

HENCE PROVED .

••••

We use PROBABILITY TREES in this CASE.

Probability that JOHN will be APPOINTED and ganguly appointed is :

0.5 and 0.7

This is : 0.5 × 0.7 = 0.35

The probability that Greg will be appointed and Ganguly be appointed is :

0.3 and 0.6

This is: 0.3 × 0.6 = 0.18

The probability that Gary will be appointed and Ganguly be appointed is :

0.2 and 0.5

This is : 0.2 × 0.5 = 0.1

The probability of Ganguly appointed is :

0.35 or 0.18 or 0.1

This is :

0.35 + 0.18 + 0.1 = 0.63

Let the no. of STUDENTS in room A be x.
And those Room B be y
Now,
ACCORDING to 1st CONDITION,
x-10= y+10

→ x=y+20

According to the 2nd condition,
2(y-20) = x+20
2y-40= x +20
x= 2y-60

So from both equations we get
y+20= 2y-60

y= 80

So x= 80+20=100

I.e. no. of students in room A are 100.
And no. of students in room B are 80.

HOPE THIS HELPS YOU:)
★★★★★★★

To find,

The no. of SQAURES that are MADE on a board of 5*5 i.e. a board having 5 towns and 5 columns.

Main solution :

Total no. of sqaures FOUND on this board = ( 5× 5)+(4×4)+(3×3)+(2×2)+(1×1)=25+16+9+4+1=55 SQUARES ( Answer ).

Note :Note : Here in this question NUMBER of sqaures don't only consider the small sqaures inside the board , but also several sqaures which are formed by grouping these basic ( small ) sqaures

Hey!!

PRODUCT of two NUMBERS = HCF × LCM

=> 100 × 190

=> 19000

Hope it will helps you ✌

Solution:

Given:

In quadrilateral ABCD, AB smallest & CD is longest sides.

To PROVE: ∠A>∠C

& ∠B>∠D

Construction: Join AC.

Mark the angles as shown in the FIGURE..

PROOF:

In △ABC , AB is the SHORTEST side.

BC > AB

∠2>∠4 …(i)

[Angle opposite to longer side is greater]

In △ADC , CD is the longest side

CD > AD

∠1 >∠3 …(ii)

[Angle opposite to longer side is greater]

Adding (i) and (ii), we have

∠2+∠1 >∠4+∠3

∠A >∠C

Similarly, by joining BD, we can prove that

∠B >∠D

Hope this will help you......

Cosß+sinß=1
or, (cosß+sinß)^2 = 1
or, cosß^2+sinß^2 + 2cosßsinß= 1
or, cosßsinß=0

Then, (Cosß - sinß)^2 = cosß^2+sinß^2 - 2cosßsinß
= 1 - 0
= 1
Cosß - sinß = +-1 (PROVED)

Given Equation is TANA + COTA = 4.

On SQUARING both SIDES, we get

= > (tanA + cotA)^2 = (4)^2

= > tan^2A + cot^2A + 2tanAcotA = 16

= > tan^2A + cot^2A + 2 * tanA * (1/tanA) = 16

= > tan^2A + cot^2A + 2 = 16

= > tan^2A + cot^2A = 16 - 2

= > tan^2A + cot^2A = 14.

On squaring both sides, we get

= > (tan^2A + cot^2A)^2 = (14)^2

= > tan^4A + cot^4A + 2 * tan^4A * cot^4a = 196

= > tan^4A + cot^4A + 2 * tan^4A * (1/tan^4A) = 196

= > tan^4A + cot^4A + 2 = 196

= > tan^4A + cot^4A = 196 - 2

= > tan^4A + cot^4A = 194.

Hope this helps!

Solution :-

Let the total number of books be x

50 % of the total books are of Marathi = x*50/100 = x/2

Books of English are 1/3rd of Marathi = (x/2)*(1/3)

= x/6

Books on MATHEMATICS are 25 % of the English books = (x/6)*(25/100)

= x/24

Remaining books = 560

Now, ACCORDING to the question.

⇒ x/2 + x/6 + x/24 + 560/1 = x/1

⇒ (12X + 4x + x + 13440)/24 = x/1

⇒ 17X + 13440 = 24x

⇒ 24x - 17x = 13440

⇒ 7x = 13440

⇒ x = 13440/7

⇒ x = 1920

So, there are 1920 books in the the LIBRARY.

Answer.

ANSWER:      

The sum of two odd primes is (5,31),(7, 29),(13, 23),(17,19).

Step-by-step explanation:

Given : Number 36.

To find : Express 36 as the sum of two odd primes.

Solution :

First we write all PRIME numbers between 1 to 36,

Prime numbers -  3, 5, 7, 11, 13, 17, 19 23, 29, 31

Now, TRY to make sum of two prime numbers equal to 36.

36–3=33 not a prime number.

36–5=31 , is a prime number.

So the first pair is (5,31).

36–7=29 , is a prime number.

So the 2ND pair is (7, 29).

36–11=25 not a prime number.

36–13=23 is a prime number.

So the 3rd pair is (13, 23).

36–17=19 ,is prime number.

So the 4th pair is  (17,19).

Therefore, The sum of two odd primes is (5,31),(7, 29),(13, 23),(17,19).

2x+5y=0

by COMPARING it to GIVEN EK. ax+by+c=0

we GET
a=2
b=5
c=0

The VALUE of all is 1

(a+b)=10

a²+b²= 58

We KNOW that

(a+b)²=a²+b²+2AB

(10)²=58+2ab

100=58+2ab

100-58= 2ab

42= 2ab

21= ab

Now ,

(a+b)³=a³+b³+3ab(a+b)

(10)³=a³+b³+3*21(10)

1000=a³+b³+63(10)

1000=a³+b³+630

1000-630=a³+b³

370=a³+b³

@:-)

SI = pnr/100
= 25000*4*9/100
=250*4*9
1000*9
=9000

724.68 will be the ANSWER

Here is ur answer.....hope it helps u



The total SELLING PRICE of 12 toffees (12 SP) = 1 
Let the cost price of each toffee be CP 
Loss% = {(CP -12SP)/CP}X100 
20 = {(CP -1)/CP}X100 
0.2 = (CP - 1)/CP 
0.2CP = CP - 1 
0.8CP = 1 
CP = 1/0.8 = 1.25 
So this is the cost of 12 toffees. 
Now the man wants to earn 20% profit on it 
So Profit = {(SP -CP)/CP}X100 
20 = {(SP - 1.25)/ 1.25} X 100 
SP -1.25 = 20X1.25 / 100 
SP - 1.25 = 0.25 
SP = 1.25+0.25 
SP = 1.5 

So to earn 20% profit on 12 toffees he has to sell the 12 toffees in 1.5 Rs 
Now the Selling price of one toffee is 1.5/12 = 0.125 
So for One rupee he can sell 1/0.125 toffees or 8.toffees.

The mode of propogation of radio  waves which travel in stagiht linesfrom transmitting antenna to receiving antenna is called space wave communication

Communications between a vehicle in outer space and Earth, using high-frequency electromagnetic radiation (radio waves).Provision for such communication is an essential requirement of any space mission. The total communication systemordinarily includes (1) command, the transmission of instructions to the spacecraft; (2) telemetry, the transmission of scientificand APPLICATIONS data from the spacecraft to Earth; and (3) tracking, the determination of the distance (range) from Earth tothe spacecraft and its RADIAL VELOCITY (range-rate) toward or away from Earth by the measurement of the round-trip radiotransmission time and Doppler frequency SHIFT (magnitude and direction).

 

(i) Ground wave or surface wave propagtion. 

(ii) Sky wave propagation or ionospheric 
propagation.  
(iii) Space wave propagation / LINE of sight 
propagation

When radio waves (frequency range 3 MHz to 30 MHz), emitted from the transmitting antenna, reach the receiving antenna after reflection from the ionosphere which acts as a reflector for radio waves the corresponding mode of propagation is known as sky wave propagation.

Let length, breadth and height be
L = y, B = 2Y and H = 3y respectively
Given,
2×(LB+LH+BH) = 1078
2×(y × 2y + y × 3y + 2y × 3y) = 1078
2×11y² = 1078
y = 7

Thus, L = 7 m, B = 14 m, H = 21 m
Therefore, volume = LBH = 7×14×21 = 2058 m³

3(x+2)-2(x-1)=7
Now....
3X + 6 - 2x + 2 = 7
x+1 = 0

Hope it RIGHT......

Given
length of side of carpet(s) = 5m

therefore AR (carpet) = s^2
= 5^2
=5×5
=25 m^2

Length of room(l) = 5m 50 cm
breadth (b) = 6

therefore ar (room) = l×b
=5.5 × 6 [ 5m 50 cm can written as 5.5 m]
= 33 m^2

area of remaining room after PUTTING carpet
= ar ( room) - ar ( carpet)
= 33 - 25
= 8m^2.

Therefore the area of remaining room = 8m^2


Hope this helps you
Plz mark this as BRAINLIEAST answer
Thanks

It MEANS a/B=3:2 and b:C=3:4 ,so the VALUE of a:C=3:4

Define x:

LET the length of the other SIDE be x

Solve x:

a² + b² = c²

24² + b² = 26²

b² = 26² - 24²

b² = 100

b = √100

b = 10 CM

Answer: The length of the other side is 10 cm

From FIGURE ( i ) ,

AO PERPENDICULAR to PQ ,

OB perpendicular to QR

Angles are supplementary .

From figure ( II ) ,

AO perpendicular to PQ ,

OB perpendicular to QR

Angles are equal.

I HOPE this helps you.

: )

Hello


Let BREADTH be x
Length be ( x+5)

So perimeter = 2(L +b)

2 ( x+x+5)

2(2x +5)

ANS

Hey there !

PRINCIPLE = 15,000

RATE of interest = 12

Number of compoundings = 3 ( One compounding per annum )

We know that ,



A = 15000 ( 1 + 12 / 100 ) ^ 3

A = 15000 ( 112/100)³

A = 15000 ( 28/25)³

A = 15000 ( 1.404928 )

A = 21073.92 Rupees .

Therefore , The LOAN amounts to 21073.92 Rupees after 3 YEARS .

Length of PQ = 7 cm
Length of PM = distance between two parallel sides = 4 cm
Length of SM = 3cm
see figure, length of SR = SM + MN + NR
because PQRS is isosceles trapezium, so, SR = NR = 3cm and MN = PQ = 7CM
so, Length of SR = 3cm + 7cm + 3cm = 13cm

now, area of PQRS = 1/2 { length of PQ + length of SR } × length of PM
= 1/2 { 7cm + 13cm} × 4CM
= 20cm × 2 cm
= 40 cm ²

for largest possible value take out HCF of 6075 and 8505

6075 = 3*3*3*3*3*5*5

8505 =3*3*3*3*3*5*7

now HCF = 3*3*3*3*3*5

= 1215

so largest possible AMOUNT is RS 1215

Hey mate!!

Third MULTIPLE of 4 is 12

so A/Q

➡12×9= 108


Hope it helps to you.

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