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Let the rate of increase in the population is r
we know, population increases geometrical progression so, we have to USE formula

Given, P = 16,000
TIME period , n = 2
population after 2 yrs , A = 17640
use formula
so, 17640 = 16000(1 + r/100)²
=> 17640/16000 = (1 + r/100)²
=> 1764/1600 = (1 + r/100)²
=> 441/400 = (1 + r/100)²
taking square root both sides,
=> 21/20 = 1 + r/100
=> 21/20 - 1 = r/100
=> 1/20 × 100 = r
r = 5 %

Given Principal Amount= P= ?

Rates of INTEREST Per annum= i =16%

Time= n =2years

COMPOUND amount= P(1+i)^n

4036.80 = P [ (1+16/100)^2

4036.80 = P [(1 + 0.16)^2

4036.80 =P (1.16)^2

4036.80 =P[1.3456]

4036.80/1.3456 =P

=3000 is the Principal amount..

Then Compounded interest = Compounded amount - Principal Amount

=4036.80 - 3000 =1036.80 is the Compound Interest.

(a+3)(a-3)(a²+9)

= (a²-3²)(a²+9) USING (a+b)(a-b)=a²-b²

= (a²-9)(a²+9)

= ((a²)²-9²) Using (a+b)(a-b)=a²-b²

= (a²×²-81)

= (a RAISE to the POWER 4-81)

HOPE YOU ARE SATISFIED

Ratio of weights of copper and zinc in brass is 13 : 7 .
Let proportionality constant is x
then, weight of copper = 13x
weight of zinc = 7x

A/C to question,
GIVEN, weight of brass = 700 gm
weight of copper in brass + weight of zinc in brass = 700 gm
=> 13x + 7x = 700
=> 20x = 700
=> x = 35

hence, weight of copper in brass = 13 × 35 = 455 gm

Answer:

Step-by-step EXPLANATION:

Given Expression:

We have to SIMPLIFY the given expression.

first we simplify the √12 and √27

12 = 2 × 2 × 3

⇒ √12 = 2√3

27 = 3 × 3 × 3

⇒ √27 = 3√3

Therefore,

Hypo2 = (5)2 +(6)2 = 25 + 36 = 61

D=St
= 12×20
distance=240 km

s=d/t
= 240/15
speed=16km/hr

HEY HELLO !
P(X)=X^2-1
P(-1)=(-1)^2 -1
= 1-1 =0 ANSWER

I HOPE this HELPS you

It is very easy.

First of all subtract the NUMBERS horizontally and then find the product of their differences.

Don't be bothered of negative sign, you've to concerned with mod of the difference.

Let say for the first case: (12 - 4) x (7 - 3) = 8 x 4 = 32. (i.e. the number at the centre.)

Similarly for the second case: (11 - 5) x (8 - 1) = 6 x 7 = 42.

Finally for the third case: (9 - 6) x (3 - 2) = 3 x 1 = 3 (Ans)

ANSWER:

Step-by-step explanation:

MULTIPLY NUMERATOR and denominator by ,we get

/* From(1)*/

•••♪

Theta us written as A,

Given, cosecA = 5 / 3



We KNOW, cosec∅ = hypotenuse / height

HENCE, 5 / 3 = hypotenuse / height



So, let hypotenuse = ( 5x ) and height be ( 3X )


By Pythagoras theorem, base will be ( 4x )



Hence,
COSA = base / hypotenuse

cosA = 4x / 5x

cosA = 4 / 5



tanA = height / base

tanA = 3x / 4x

tanA = 3 / 4





Then, cosA + tanA


=> 4 / 5 + 3 / 4

=> ( 16 + 15 ) /20

=> 31 / 20

1 4 5 6 8 10 11 12 13 14 15 16 17 18 20 21
HOPE it will help you. ..... plz mark me as brainlist..

Answer-3/7

As KRISHNA has SEVEN LETTERS so denominator=7

And 3 straight lines

Hello DEAR,

CP = RS 400
SP = rs 460

CP < SP
so, there is profit

Now profit = SP - CP
profit = 460 - 400
profit = rs 60

now,
profit % = [ profit/CP ] × 100
profit % = [ 60/400 ]×100
profit % = 15%

A = 535$

R% = 3.5%

T = 2 Years

SI = A - P

SI = 535 - P

P = 535 - SI

By SI formula

SI = (P*T*R)/100

SI = ((535-SI)*2*3.5)/100

100SI = (535-SI)*7

100SI = 3745 - 7SI

100SI + 7SI = 3745

107SI = 3745

SI = 3745/107

SI = 35$

P = A - SI

P = 535 - SI

P = 535 - 35

P = 500$

HEY MATE ☺
Rajveer Here ✌✌
_________________________________

First Term (a) = 8
Common Difference (d) = 8
Number of TERMS (n) = 15

So, Sn = n/2 [2A + (n-1)d ]
Sn = 15/2 [2(8) + 14(8)]
Sn = 15/2 [16 + 112]
Sn = 15/2 [128].
Sn = 15 [64]
Sn = 960.....
So,Sum of first 15 multiples of 8 is 960✔✔
_______________________________
HOPE IT HELPS U✌✌
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Here,
a=9
b=3K
c=4
b^2-4ac=0 (Since the roots are SUPPOSED to be equal)
=>{3k}^2-4(9)(4)=0
=>9k^2-144=0
=>9k^2=144
=>k^2=144/9=16
Hence k=4 or -4.

SOLUTION :

Let ONE number be 1X

another number be 3x

According to the question :

if 5 is added to both numbers the RATIO becomes 1:2

so,

one number = 1x = 1 × 5 = 5

another number = 3x = 3 × 5 = 15

Hence,

TWO numbers are 5 and 15

I THINK it WOULD be 125

In triangle ABD AND FEC
AB=FE ( given )

BC = DE
in whichCD is common part coming in both triangles
BC + CD = CD + DE
= > BD = CE
THEREFORE triangle ABD is congruent to triangle FEC by SAS rule of congruence

Hope its helpfull mark it as BRAINLIEST

Two Triangles on the same base and between the same parallels are EQUAL in area.

Given:ABCDE is a Pentagon & BF||AC.

To show:(i) AR (ACB) = ar (ACF) ii) ar (AEDF) = ar (ABCDE)

PROOF:

i) △ACB and △ACF lie on the same base AC and between the same parallels AC and BF

.∴ ar(△ACB) = ar(△ ACF)

(ii)ar(△ACB) = ar(△ACF)

ar(△ACB)+ar(△ACDE) =ar(△ACF) + ar(△ACDE)

[ On adding ar(△ACDE) on both sides]

ar(ABCDE) = ar(△AEDF)

Hope this will help you...

Hiii!!!

here's ur answer...

GIVEN the SIDE of the cube = 11cm

lateral SURFACE of the cube = 4 × side²

= 4 ( 11 )²

= 4 × 121

= 484cm²

hope this HELPS..!!

Given:- ABCD is a PARALLELOGRAM
AP and CQ are perpendiculars, THEREFORE, CQD = BPA = 90°
proof:- in ∆APB and ∆CQD,
CQD = BPA. ( each 90° )
DBA = BDC. ( alternate ANGLES )
DC = AB. ( opposite sides of a parallelogram are equal )
Therefore, by AAS criteria ∆ABP is CONGURENT ∆BPA
Now by cpct AP = CQ

Given: In right angled ∆ABC,

∠C = 90°,M is the mid-point of AB i.e, AM=MB & DM = CM. 

To Prove: i) ΔAMC ≅ ΔBMD

ii) ∠DBC is a right angle. 

(iii) ΔDBC ≅ ΔACB

CM=1/2AB

Proof:(i)   In ΔAMC & ΔBMD,

AM = BM                                [M is the mid-point]

∠CMA = ∠DMB                           (Vertically opposite angles)

CM = DM                                           (Given)

Hence, ΔAMC ≅ ΔBMD[ by SAS congruence rule]

ii) since, ΔAMC ≅ ΔBMD

AC=DB. (by CPCT)

∠ACM = ∠BDM (by CPCT)

Hence, AC || BD as alternate interior angles are equal.

Then,∠ACB + ∠DBC = 180°              (co-interiors angles)

90° + ∠B = 180°

∠DBC = 90°

Hence, ∠DBC = 90°

(iii)  In ΔDBC &  ΔACB,

BC = CB (Common)

∠ACB = ∠DBC (Right angles)

DB = AC ( PROVED in part ii)

Hence, ΔDBC ≅ ΔACB (by SAS congruence rule)

(iv)  DC = AB                                              (ΔDBC ≅ ΔACB)

DM + CM =AB [CD=CM+DM]

CM + CM = AB [CM= DM (given)]

2CM = AB

Hence, CM=1/2AB

HOPE THIS WILL HELP YOU...

Interesting question here,

ALTHOUGH if you want to multiply any numbers without actually doing any multiplication you can refer to the japanese method of multiplication, the LINK of which i have mentioned in the comment section

But i would also like to RECOMMEND one of these methods where you can multiply large numbers mentally

Here is the method,

Try EXPRESSING the numbers in form of numbers which are divisible by 10

Example:

9 can be EXPRESSED as 10-1

And then you can do the multiplication,

Now, let us take the sum you have mentioned above,

(1000- 1)×(1000-1)

1000000 - 2000 + 1

998001

This is how simplified multiplication can be performed

Hope this helps!!!!!!!!!!!!!

If AMIT SOLD the bicycle to Nikhil for ₹ 1134 with no loss and no profit. Then the cost price must be ₹ 1134 including repairing cost.

Now the cost price excluding repairing cost = ₹ 1134 - ₹ 54 = ₹ 1080

Let the cost price of Raju be Rs x.

Then, SELLING price of Raju = x + 8x/100 = x + 2x/25 = 27x/25

Now the cost price of Amit = 27x/25 = 1080

or, x = (1080 x 25)/27 = Rs 1000.

Therefore,

cost price of Raju = Rs 1000 and selling price = Rs 1080 (8% profit)

cost price of Amit = Rs 1080 and selling price = Rs 1134; repairing cost = Rs 54 (no profit no loss )

cost price of Nikhil = Rs 1134.

Given

List price =₹3500

PAID amount =₹2800

Discount =list price - paid amount

=3500-2800

=₹700

Discount %= (discount / list price) x 100

=(700/3500) x 100


=(7/35)x100

=100/5

=20%

This is ur ANS hope it will help you in case of any DOUBT comment below

Deepa SOLD at

SP = 1200 + 5/100 of 1200 = 1200 + 60 = 1260

Customer purchased at 1260 and taxed at 4%

So 1260 + 4/100 of 1260 = 1260 + 50.4

= 1310.4

LET Rahim was X years OLD 8 years back

rahim's age today will be x+8

So rahim's age after 32 years will be x + 8 + 32 = x+40 -----1

And rahim's age after 32 years = 5x -----2

Equating 1 and 2,

x + 40 = 5x

4x = 40

x = 10

Rahim's age today is x+8

= 10+8

18 years

Kx+3y=1
12x+ky =2
in the CASE of no SOLUTION for equation
A1/A2=B1/B2
k/12=3/k
k^2=36
k=6

If Sameerrao has taken a loan of `12500 at a rate of 12 p.c.p.a. for 3 years and the INTEREST is compounded annually then should he pay to clear his loan can be calculated as shown below:

Amount = P{1 + (r/100)}^n

or, we know the values of P, r, n.

P = RS 12500, r = 12% and n = 3yrs.

CI = 12500{1 + (12/100)^3

or, 12500{(112 x 112 x 112)/(1000000)}

or, Rs 17561.6 (Ans)

Therefore, Sameerrao has pay to pay Rs 17561.6 to clear his loan