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I think The QUESTION is INCOMPLETE. The INITIAL value for average is not given in the question so it is not POSSIBLE to get any numerical value of the answer.

Let the divisor be X than QUOTIENT be 6+5x
now a/q we have,
DIVIDEND = divisor × quotient + remainder

Let of the AP
first term= a
common difference=d
ATQ
5[a+(5-1)d]=8[a+(8-1)d]
5(a+4d)=8(a+7d)
5A + 20d = 8A + 56d
3a = 36d
a= -12d
Now 13th term
a+(13-1)d
=-12d+12d
=0
(proved)
Hope it helps
Mark me as BRAIN list

Hi ,

1 ) √( 5-2√6 )

= √[ (√3 )² + ( √2 )² - 2 × √3 × √2 ]

= √ ( √3 - √2 )²

= √3 - √2 ---( 1 )

2 ) √[ 10 - 2√21 ]

= √[ ( √7 )² + ( √3 )² - 2 × √7 × √3 ]

= √ ( √7 - √3 )²

= √7 - √3 ---( 2 )

Now ,

√( 5 - 2√6 ) + √( 10 - 2√21 )

= ( 1 ) + ( 2 )

= √3 - √2 + √7 - √3

= √7 - √2

I hope this helps you.

: )

10x:11X
1x= ₹600
.: 11x= 11*6000=₹66000

I THINK the LOGIC is that first and third number is DIVISIBLE of 15 and SECOND and four divisible of 12

Let the three part be a, b and c

GIVEN :

SUM = 150

Given part :

First part = 'X'

Second part = 5/6 of x

                    = 5x/6

Third part = 4/5 of 5x/6

                = 2x/3

To find :

We have to find

the VALUE of 'x'

As we know that sum of

three part is 150

So,

   According to the question,

Now we are going to

find the value of First

part, second part and

the third part. By putting

the value of 'x'.

First part = x

              = 60

Second part = 5x/6

                    = 5 × 60 / 6

                    = 50

Third part = 2x/3

                = 2 × 60 /3

               =  40

Hey

Here is your ANSWER,

In attachment

Hope it HELPS you!

Given that the POPULATION earlier was = 24500
Population now = 25000

So, INCREASE in population = 25000 - 24500 = 500

We KNOW that, 
Percentage increase in population= CHANGED population/original population *100

⇒ 500 / 24500 * 100
⇒ 500/245
⇒ 2.04%

Your ANSWER is 28.
here is explanation...

after 4 year
5 x + 7 x + 4 + 4= 56
12x+8=56
12x=56-8= 48
x= 48/12=4
then age f RAHUL= 5×4=20
age of RAVI =7×4=28

Hola!
________________________

[ refer above attachment. ]

Given : TA and TB are tangent to a CIRCLE with CENTRE O from point T

Consider,
ΔOAT and ΔOBT

Here,

OA = OB [Radii of circle]
OT = OT [Common side]
TA = TB [Tangents from an external point to a circle are equal in length]

∴ ΔOAT ≅ ΔOBT [By SSS CONGRUENCE criterion]

[ Exterior angle of a triangle is equal to sum of interior opposite angles. ]

Hence, in ΔAPT, ∠APO = ∠a + ∠b

In ΔAPO, OA = OP [Radii of circle]

∠OAP = ∠OPA = ∠a + ∠b ---(i) [Angles opposite to equal sides are equal]

From the FIGURE, ∠OAP=90°- ∠b ----(ii) [Since radius is perpendicular to tangent]

From equations (i) and (ii),

∠a + ∠b = 90° - ∠b

⇒ ∠a = 90° - 2∠b ---(iii)

In ΔOAT, ÐOAT + ÐAOT + ÐATO = 180° [Angle sum property of triangle]

90° + ∠a + ∠b+∠c+=180°

∠a+∠b+∠c=90°

90°- 2∠b + ∠b + ∠c=90°

- ∠b + ∠c = 0

∠b = ∠c

Hence,
AP bisects ∠TAB

________________________

Hope it helps! :D

Your answer is 0(zero). TAKE a look at picture with the HELP of (a+b)³= a³+b³+3ab(a+b).
formula USED for (2y+6)³

I HOPE this ANSWER MAY be CORRECT

Heya

Prime numbers are the ONES which have only two factors 1 and itself.

It cannot be expressed as a product any other numbers.

3×12×101 + 4

4 (3×3×101 + 1)

Here the number is expressed as a products of other numbers.

Hence it is not a prime number. It is a composite number

HOPE it HELPS! ^_^

Zero of ( x + 1 ) = -1.

By Remainder Theorem,

P ( x ) = x ^ 140 + 2 x ^ 151 + K

P ( 1 ) = ( -1 ) ^ 140 + 2 ( -1 ) ^ 151 + k = 0

             1 + 2 ( -1 ) + k = 0

             1 -2 + k = 0

            -1 + k = 0

              k = 1.

Answer:

FLAG Staff length = 16.785 m

Step-by-step explanation:

Given: Length of Ladder, CD = 9 m

           Lenght of staff from top to foor of ladder , AD = 9 m

To FIND: Lenght of flag staff, AB

In Δ BCD,

Using Trignometric Ratio,

   ( using VALUE of √3 = 1.73 )

⇒ AB = AD + DB

         = 9 + 7.785

         = 16.785 m

Therefore, Flag Staff length = 16.785 m

Let a is the first term and d is the COMMON DIFFERENCE of AP
∵ Sn = n/2{2a + (n - 1)d}
and
so,
And

Hence,
= (2r + 1)

First carpet,
AREA = length× breadth

a = l × b
a = 14 × 12.5
a = 175sq.m

Second carpet,
Area =lenght × breadth

a = l × b
a = 16 × 13.8
a = 220.8 sq.m

So the area of second carpet is larger than first carpet.
Hope this HELPS u........if it helps u mark me as BRAINLIEST.

Here is your ANSWER buddy:-
√2X^2 +2x + 5x + 5√2 = 0
√2x (X +√2) +5 (x+√2)= 0
(x+√2) (√2x+5) = 0

Either (x+√2) = 0 or (√2x+5) = 0
x = -√2 or x = -5/√2

Hope it helps.

H=30m
x=10 ROOT 3
tanA=30/10 root 3
tanA=root 3
A=60°

In the attachment I have answered this problem.

I have APPLIED cramers rule to solve the given system of LINEAR equations.

SEE the attachment for detailed solution.

I hope this answer HELP you

A=12
d=16-12=4
n=24
using FORMULA
an=a+(n-1)d
putting the VALUES
we GET,
an=12+(24-1)4
an=12+23×4
an=12+92
an=104
So, the 24th term of this AP is 104
I HOPE THIS WILL HELP YOU MARK ME BRAINLIEST

4M=20, =>M=4*20=80
2M=3W,
2M/3=W
3M+3W=3M+(2M*3/3)
3M+2M=5M
(M=1/80)
5M=5/80=1/16
3M and 3W can do it in 16DAYS

Heya

Initially, the TREE was 3M high

It has to grow 7/2m = 3.5m

Growth = 3.5-3=0.5m

In one MONTH, the tree grows 0.1m

No. Of MONTHS it takes to grow to 7/2m high= 0.5/0.1=5months.

Hope it HELPS! ^_^

Let EAST direction be positive and west be negative.

Delhi is mean position and so let it be 0.

Since he travels 60km east,
0+60=60.

And he travels 100km west,
60-100= -40.

Negative SIGN indicates he is west of Delhi.....

THEREFORE he is 40 km west of Delhi

So what we have to do BUDDY

-1+x =-3/7+2
x=-3/7+2+1
x=-3/7+3
x=-3+21/7
x=18/7.....

HOPE it WORKS

The ODD ONE is 697 ....
how? SEE in the attachment⬆⬆⬆

1)-

a - b - a² - b²

1(a - b) - {(a+b) (a-b)}

(a - b) { 1 - (a+b)}

(a-b)( 1 - a -b)

2)-

9 - a² +2AB -b²

-1[ -9 +a² -2ab +b²]

-1[ -9 + (a - b)²]

9 - (a -b)²

3² - (a -b)²

[3-(a-b)] [3+(a-b)]

( 3-a +b) ( 3+a -b)

3)-

1 + 2ab - (a²+b²)

-1 [ -1 -2ab +(a²+b²)]

-1 [ -1 -2ab +a²+b²]

-1 [ -1 + (a-b)²]

1 - (a-b)²

1² - (a - b)²

[ 1 + (a -b)] [ 1 - (a -b)]

[ 1 + a -b] [ 1 -a +b]





I hope this will help you


-by ABHAY

F(x)= ax^3 + 3x^2 -13
G(x)=2x^2-5x+a
Now PUT x=a in f(x)
f(2)=8a+12-13→ 8a-1
Similarly g(2)=8-10+a→a-2
Now EQUATE the two reminders i.e.
8a-1=a-2
a=-1/7

Let the digits of the no. be x and y

So according to the question :

x × y = 14

xy = 14

Let the two DIGIT BO. be 10x + y

10x + y + 45 = 10y + x

10x - x + y - 10y = - 45

9x - 9y = - 45

9 ( x - y ) = -45

x - y = - 5

x = -5 + y

x × y = 14

-5 + y ( y ) = 14

-5y + y^2 = 14

y^2 - 5y - 14 = 0

y^2 - 7y + 2y - 14 = 0

y ( y - 7 ) + 2 ( y - 7 )

( y + 2 ) ( y - 7 ) = 0

y = - 2 ( neglected)

or

y = 7

we cannot take a negative value so we must take y = 7

xy = 14

x × 7 = 14

x = 14 / 7

x = 2

So the required no. is 27

Hope it helps you .

A27=a+26d
=9+26×-5
=9-130
=-121

1 to 250 ..
250/4 = 62 + 2/4
62 no DIVISIBLE
10/4 = 2 + 2/4
2 no divisible
62-2 = 60 no between 10 to 250

Just PREPARE well and give ur 100 PERCENT in the test

I HOPE this MAY HELP you.....!!!
THANK You..............☺️☺️☺️
✌️✌️

Heya ✋

Let SEE your answer !!!

Given that

x = 2 + √5

x^2 + 1/x^2 = ?

Solution

1/x = 1/2 + √5

By rationalization

= 1 × (2 - √5)/(2 + √5) × (2 - √5)

= 2 - √5/(2)^2 - (√5)^2

= 2 - √5/4 - 5

= 2 -√5/-1

= -2 + √5

Therefore ,

x^2 + 1/x^2

= (2 + √5)^2 + (-2 + √5)^2

= (2)^2 + (√5)^2 + 2 × 2 × √5 + (-2)^2 + (√5)^2 + 2 × (-2) × √5

= 4 + 5 + 4√5 + 4 + 5 - 4√5

= 9 + 4√5 + 9 - 4√5

= 18




Thanks :)))))

Here is UR ANSWER...

The number of whole NUMBERS between 99 and 300 = (300-99) - 1
= 300 - 100
= 200


here is ur ANSWER

plz BRAINLIEST my answer

In the attachment I have ANSWERED this problem.

The problem is very SIMPLE and EASY
to understand.

See the attachment for detailed SOLUTION.