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Please find below the solution to the asked query:

We have :  x = 5 − 3√5 + 3√ , So

x = 5 − 3√5 + 3√×5 − 3√5 − 3√ = 25 − 53√ − 5 3√ + 325 − 53√ + 5 3√ − 3= 28 − 103√22= 2(14 − 53√)22 = 14 − 53√11

And

y = 5 + 3√5 − 3√ , So

y = 5 + 3√5 − 3√×5 + 3√5 + 3√ = 25 + 53√ + 5 3√ + 325 + 53√ − 5 3√ − 3= 28 + 103√22= 2(14 + 53√)22 = 14 + 53√11

Then,

To show : x2 − Y2 = −103√11 , We take L.H.S. :x2 − y2 We KNOW : a2 − b2 =(a+b)(a−b) , So⇒(x+ y)(x− y)Substitute values of 'x' and 'y' and GET :⇒(14 − 53√11+14 + 53√11)(14 − 53√11− 14 + 53√11)⇒(14 − 53√ + 14 + 53√11)(14 − 53√ − 14 − 53√11) ⇒(28/11)(− 10√3/ 11)ButL.H.S. ≠R.H.S.

So , Kindly RECHECK given query and get back to us so that we could help you precisely .

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Hello MATE!

Let total number of votes be x

Now, 1st CANDIDATE won 48% of total votes =
48x/100

But, 8% from total voters don't give VOTE ; so votes gives actually = x - 8x/100
= 92x/100

Hmn, now 2nd candidate must have got

92x/100 - 48x/100 = 44x/100, then

Now, 2nd candidate lost by 1st candidate by
1100 votes :-

48x/100 = 44x/100 + 1100
48x/100 - 44x/100 = 1100
4x/100 = 1100
x= 100/4 × 1100
x = 27,500 votes are polled

Hope it helps☺!

Let centre of circle is (h, r) And radius of circle is r
∴equation of circle is (X - h)² + (y - r)² =

A/C to question,
Circle passing through focus and touches at the point (6,9) of parabola x² = 4Y
focus of parabola is (0,1)
so, (0,1) will satisfy the equation of circle ,
∴ (0 - h)² + (1 - k)² = r² -----(1)
Again, (6, 9) will satisfy the equation of circle ,
∴(6 - h)² + (9 - k)² = r² ------(2)

From equations (1) and (2),
-2k + 1 = -12h + 36 - 18k + 81
⇒16k + 12h = 116
⇒4h + 3h = 29 ------(3)


at (6,9) slope of tangent if circle = slope of tangent of parabola [because at (6,9) circle touches the parabola ]
slope of circle :
differentiate equation of circle with respect to x
2(x - h) + 2(y - k)dy/dx = 0
⇒ dy/dx = -(x - h)/(y - k)
At (6,9) slope of tangent of circle is dy/dx = -(6 -h)/(9 - k)

Slope of parabola :
x² = 4y, differentiate with respect to x
2x = 4dy/dx
⇒dy/dx = x/2
At (6,9) slope of tangent of parabola is dy/dx = 6/2 = 3

Hence, 3 = -(6 - h)/(9 - k)
⇒3(9 - k) + (6 - h) = 0
⇒ 27 - 3k + 6 - h = 0
⇒ 3k + h = 33 -------(4)

SOLVE equation (3) and (4) ,
h = -9 and k = 14
so, r² = (6 +9)² + (9 - 14)² = 225 + 25 = 250

Hence, equation of circle is (x + 9)² + (y - 14)² = 250
⇒x² + y² + 18x - 28y + 81 + 196 = 250
⇒x² + y² + 18x - 28y + 27 = 0

Hence, answer is x² + y² + 18x - 28y + 27 = 0

••°heya USER °••

••here is your answer•••

¤¤ 17/8=2.125

hence it's DECIMAL EXPANSION is 2.125 ¤¤

hope it HELPS U.. !!

Let a's Salary be X
let's salary be y
so x+y=Rs. 2000
implies, x=2000-y   ........ equation 1
out of rs. 100, a SAVES=100-95=rs5
out of rs. x, a saves= x*(5/100) 
similarly, B saves= y*(15/100)
5x/100=15y/100
implies x=3y........ eqn 2
replacing x from eqn to eqn 2

2000-y=3y
4y=2000
y=500
x=1500
a's salary=rs 1500

I HOPE this ANSWER HELP you

On 1st Jan,
She saved 10 Rs.
    Next Day 11 Rs.
    Next Day 12 rs

So MONEY saved by her
= 10 + 11 + 12 + 13 + .........................+ (till 31ST Dec)

Now, Above Sum is the sum of a Arithmetic Progression with
a = 10
d = 1
n = 366 ( since year 2016 is a Leap Year)

=>Sum of an A.P. = (n/2){ 2A + (n-1)d}
                                  = (366/2){ 20 + 365 }
                                  = 70455

So, she saved Rs. 70455 till the END of the year

Your ANSWER is 25%. TAKE a LOOK at PICTURE.

The answer is given below :

The sequence is defined as

an = (-1)^(n - 1) × 2^n, where n is any Natural number

Now, putting n = 1, 2, 3, 4, 5, ... we will GET the RESULTING sequence.

When n = 1,
a1 = (-1)^(1-1) × 2^1 = (-1)^0 × 2 = 1 × 2
= 2

When n = 2,
a2 = (-1)^(2-1) × 2^2 = (-1)^1 × 4 = -1 × 4
= -4

When n = 3,
A3 = (-1)^(3-1) × 2^3 = (-1)^2 × 8 = 1 × 8
= 8

When n = 4,
a4 = (-1)^(4-1) × 2^4 = (-1)^3 × 16 = -1 × 16
= -16

When n = 5,
a5 = (-1)^(5-1) × 2^5 = (-1)^4 × 32 = 1 × 32
= 32

Thus, the first five of the given sequence are
2, -4, 8, -16, 32

Thank you for your question.

A TRAIN travel in 60KM in 60 minute
So
165 km in 165 minute
165km it take time 2 hour 45 minute
Train started runing at 1:45
Train reach other station on
4:30


I hope this will HLP u
By DEVANSH

2,5,8,11,14,17,20,23,26,29,32,35,37,40,43,46,49,52,55,58,61,64,67,70,73,76,79,82,85,88,91,94, 97,100,103,106,110,113,116,119,122,125,128...

We know,

AREA of an equilateral TRIANGLE = √3/4 side×side

= √3/4 ×1600

= √3 × 400

= 1.74 ×400

= 692.82 SQ. CM

According to definition of parabola , is is the locks of the points in that planes that are equidistant from both FOCUS and directrix.

Given, focus : (-3,0)
directrix : x + 5 = 0
Let (x ,y) is the point on the parabola .
∴ DISTANCE of point from focus = distance of point from directrix
⇒ √{(x + 3)² + y²} = |x + 5|/√(1² + 0²)
⇒ √{(x + 3)² + y² } = |x + 5|
squaring both sides,
(x + 3)² + y² = (x + 5)²
⇒y² = (x + 5)² - (x + 3)²
⇒y² = (x + 5 - x - 3)(x + 5 + x + 3)
⇒y² = 2(2X + 8) = 4(x + 4)

Hence, EQUATION of parabola is y² = 4(x + 4)

Answer:

The value of k is all number EXCEPT

Step-by-step explanation:

Given : Equations

To find : The value of k for which the FOLLOWING system of equation has a unique solution.

Solution :  

When the system of equation is in FORM then the CONDITION for a unique solutions is  

Comparing, a=k , b=2 , c=-5, d=3, e=1, f=-1

Substituting the values,

Cross multiply,

Therefore, The value of k is all number except

Let the digits are x and y so..
If i write a no. xy then= 10x + y
e.g. 32= 10×3+2
so.. REVERSING of xy is yx and
yx=10y+x

Now..
A/Q : x-y=3 eq1
10x+ y+ 10y+x = 99 eq 2
so from eq 2
10x+y+10y+x= 99
11x+11y=99
x+y=9 ...... eq 3

Now
from eq 1
we have
x-y=3
x=3+y
substituing VALUE of x in eq 3
x+y=9
3+y+y=9
2y=6
y=3
Substituting value of y in eq1
x-y=3
x-3=3
x=6
so the no. is 36 or 63
plz MARK me as a brainliest

Heya ✋

Let see your answer !!!

Let the required number be x. Then,

according to question ,

x - 40 % of x = 30

=> x - 40/100 × x = 30

=> x - 2x/5 = 30

=> 5X - 2x/5 = 30

=> 3x/5 = 30

=> 3x = 30 × 5

=> 3x = 150

=> x = 150/3

=> x = 50

Hence , the required number is 50.







Thanks :))))

Alfa×x +3y-alfa+3=0
12x+alfa×y-alfa=0
for UNIQUE solution a1÷a2 is not equal TI b1÷b2
a1÷a2=alfa÷12---eq 1
b1÷b2=3÷alfa---eq 2
cross MULTIPLY (of eq 1 and eq 2)
(alfa)^2=36
alfa= 6

Hey Mate!
~~~~~~~~

Here is your answer :

==============================

Let the HEIGHT of triangle be x

Base of triangle = x + 4

Area of triangle = 1/2 bh = 48


Height = 8cm as SIDES will not be NEGATIVE.

Base = x + 4 = 8 + 4 = 12cm


HOPE THIS HELPS U...

Let DENOMINATOR be X,

(x-4)/x
____

If

(x-4-2)/(x+1)
____

Then,

x+1 =8(x-4-2)

x +1=8(x-6)

x +1 =8x - 48

1+48=8x -x

49=7x

49/7 =x

7=x

______

Fraction

(7-4)/7

3/7


I HOPE this will help you


-by ABHAY

Let,

Numerator be x
Denominator be (x +3)

FRACTION = x/(x+3)

----------------------------

According to the QUESTION,







Then,






I HOPE this will help you


-by ABHAY

LET the number be x,

5 times of x =  5x

3 less than twice of square of x = 2x² - 3 

In question,

→ 5x = 2x²- 3 

→ 2x² - 5x - 3 = 0

→ 2x² - (6 - 1)x -3 = 0

→ 2x² - 6x + x - 3 = 0 

→ 2X(x -3) +1(x-3) = 0 

→ (x -3) (2x +1) = 0 

x = 3 or x = (-1/2)

x can't be negative  (Given in the QUESTION), so number is 3 

• Given:- A strip with a cross section 150 MM x 4.5 mm is being rolled with 20% reduction of area using 450 mm diameter rolls.

• To find:- The angle subtended by the deformation zone at the roll center (in RADIANS)

• Solution:- we know,

θ = tan^(-1) [√(R∆H) / { R- (∆h/2)}]

according to the question,as we have 20% reduction in cross sectional area

150×4.5× 0.8 = 150×h

or, h = 3.6 mm

∆h= 4.5-3.6 = 0.9 mm

θ = tan^(-1) [√(225×0.9) / { 225- (0.9/2)}]= 362° = 0.06 rad

so, the angle subtended by the deformation zone at the roll center is (in radians) 0.06

Given,
First of all we have to CHECK form of the LIMIT ,
PUT x = π/6 ,
(√3 sinπ/6 - cosπ/6)(π/6 - π/6) = 0/0 is the form of limit
We know, asinA -bcosA = √(a² + b²)sin{A - tan⁻¹(b/a)}, use it here,
Then, √3sinx - cosx = √{√3² + 1²}sin{x - tan⁻¹(1/√3)}
= 2sin(x - π/6)
Now, limit converts in

Use the standard form for SOLUTION of Limit ,
Lim_{f(x)→f(a) sin{f(x)-f(a)}/{f(x) - f(a)} = 1 , use this here ,
Then,
= 2 × 1 = 2


HENCE, answer is 2

ANSWER:

Step-by-step EXPLANATION:

THEREFORE.,

•••♪

3 discount of 10,20,40% discount

we will find net EFFECTIVE discount by :
net effective discount = a+b-ab/100

Case-I 20&40% discount FIRST,,,

net discount = 20+40-(20x40)/100 = 52%

Case -II

& now effective discount of 52% and 10% is
= 52+10+(52x10)/100
= 62- 5.2%
= 56.8%

195km:65km
3:1


here it is your ANS

She had 21 RS...
LETS GO in reverse
finall she had₹1
1+3=4×2
=8+2
=10×2
=20
20+1
=21

ANSWER:

Step-by-step EXPLANATION:

Answer:

SQUARE by 1444m^2

Step-by-step explanation:

SEE the attached PICTURE

Yessssssssssssssssssssssssss

Three EQUIVALENT RATIOS are -

i) 22:18
ii) 33:27
iii) 44:36

HOPE this HELPS uh MATE!!!!

☺️☺️

Hii dear!!

Question 1.
Volume of water in the canal = L*b*h
30*12*10000 = 36,00,000 cubic meter
36,00,000 cubic meter flows in 1 h

Area of plot = l*b
volume = l*b*h = 36,00,000
l*b*0.16 = 36,00,000
l*b = 36,00,000/ 0.16
l*b = 2,25,000 sq m
Therefore an area of 2,25,000 sq/m will be irrigated.
_________________________________
Question 2.

speed of meter = 3km/h = 3000m/hr
let the time be x hour.
length of water column = 3000* m
water flow in one minutes = 3000/60= 50m/min
volume of water flow into the sea = l*b*h
= 50* 2.5* 30
= 3750 m^3
___________________________________
Question 3

Let the increase in height = h
Volume of CUBE = volume of water risen
9^3 = 15* 12* hh = 9^3/(15*12)
h = 4.05cm rise in water level.