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I hope you know tan formula it goes like this -
tan(A+B) = (tanA+tanB)/1-tanAtanB

Now COMING to QUESTION,
Given A+B+C+D = 360°
use tan on both sides and separate using formula.

tan{(A+B)+(C+D)} = tan(360°)
=> tan(A+B) + tan(C+D) = 0.........eq (i)

=> tan(A+B) + (tanC+tanD)/1-tanCtanD = 0
=> tan(A+B) (1-tanCtanD) + tanC + tanD = 0

Using eq(i) in above result,

{tan(A+B) (1-tanCtanD) + tanC + tanD} = 0
=> -tan(C+D) (1-tanCtanD) + tanC + tanD = 0
=> - (tanC + tanD) + (tanC + tanD) = 0
=> 0 = 0
Hence Proved.

Please IGNORE if any typos. Though it is long, I hope it will help you.

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First SEVEN ODD NUMBERS are:

1,3,5,7,9,11,13

when we ADD we get:

1+3+5+7+9+11+13=49

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Gshsjkwjwhejjwhejjwshshwhjw

This is a simple PROMBLEM just GO through formula and u will FIND the answer

(1+i)^5*(1-i)^5

(1^2+I^2)^5

(1+i^2)^5

LET the Triangle be ABC

Given: ∠A= 4x  ; ∠B= 56X   ; ∠C= ∠A+∠B

                                              ∠C= 4x+5x= 9x

By Angle Sum PROPERTY

    ∠A+∠B+∠C= 180°

    4x+5x+9x= 180°

     18X= 180°

     X= 180/18= 10°

∵  ∠A= 4x= 4(10)= 40°

    ∠B= 5x= 5(10)= 50°

    ∠C= 9x= 9(10)= 90°

Given Principal AMOUNT= P= 15626

Rates of INTEREST PER annum= i =8%

Time= n =1 1/2years ×2 = 3(because compounded half yearly)

COMPOUND Interest = P[ (1+i)^n - 1]

=15625[ (1+8/100)^3 - 1]

=15625[(1 + 0.08)^3 -1]

=15625[ (1.08)^3 - 1]

=15625[1.259712- 1]

=15625[0.259712]

=4058 is the Compound Interest...

Then compounded Amount = Principal Amount + Compound Interest

=15625+ 4058 =19683 is the Compounded amount

there are MANY such NUMBER so your QUESTION is INCOMPLETE

Here is your ANSWER...
HOPE this HELPS you..

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The PATTERN is CONTINUOUSLY adding of numbers from 7
i.e, 10+7=17
17+9=26
26+11=37
so now, we should add 13 to the NUMBER on the series
so the NEXT number is 37+13=50

ANSWER
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Adding (I) and (ii)

a + b = 76 ------- (I)
a - b = 38 -------- (ii)
_______________

2a = 114

=> a = 57
Substitution of a in any EQ

=> b = 19

Value of a ÷ b = 57 ÷ 19

= 3

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HOPE IT HELPS :):):):):)

Your ANSWER is here ✌✌✌✌✌✌✌

COS^2 180/16+cos^2(90-180/16)+cos^2(900/16)+cos(90-900/16)

=cos^2(180/16)+sin^2(180/16)+cos^2(900/16)+sin^2(900/16)

=1+1

=2

There are many sources of ancient indian history. The most important one's are the following in no particular order:Literature:Both indian and foreign literature. For example, Puranas, especially Bhavisya Puran gives an idea of early indian kingdoms though it isnot that accurate. Arthshastra written by Chanakya throws light on administrative aspects of Mauryan Empire. There are several other literature that have historical value. But reliability of literature sources is questionable. As theydon't always portray the true picture.Coins:coins tells a lotabout the financial position of the king and the region. Discovery of large number of gold coins of Gupta period led tobelief that it was quite a PROSPEROUS period in indian history. Whereas, lessnumber of gold coins DENOTE that the periodwas less prosperous .Travellers ACCOUNTS- A great deal of knowledge about the time and reign of Harshvardhan has been gained from Chinese traveller Hiuen tsang’s account. The first mention of world famous Kumbha Mela is FOUND in his accounts. Similarly Fa hien visited during Gupta period and wrote about the same. This source is also affected by author’s personal biases.Excavation- Indus valley civilization wasdiscovered through massive excavation. Several artefacts, coins and other itemsof archaeological value are discovered in excavation that UNLEASH information about that era.Archaeological Evidences- such as artefacts, rock inscriptions, pillar inscriptions. A great deal of information has been obtained from Ashoka’s pillars.We come to know of Samudragupta military campaigns though his inscription“Prayag Prasashti”.

⭐hey there!!

Given that:

➡ ∫ dx/√x+a - √x}

Now rationalization of denominator,

=>> 1/(√x+a - √x) × (√x+a + √x) /(√x+a +√x)

=> (√x+a + √x) /a ------------------(1)


Now intigrating EXPRESSION (1)

= 1/a ∫(√x+a + √x ) dx

= 1/a { ∫√(x+a) dx + ∫√x dx }

= 1/a ( 2/3 (x+a) ³/2 + 2/3 x³/2 )

=> 2/3a { (x+a)³/2 + x³/2 } ans


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⭐Hope it will HELP u

A pair of negative INTEGERS whose difference is 8...

Eg ; - 1 and - 9

ie - 1 - (-9) = - 1 + 9 = 8

A positive and negative INTEGER whose SUM is 5

Eg ; - 2 and 7

ie - 2 + 7 = 5

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Given : a cone is made from a metal sheet in the form of a sector of a circle .radius of the circle is 14 cm and the sector angle is 60⁰,

To Find :   the radius of the base ,slant height and height of the cone formed​

Solution:

cone is made from a metal sheet in the form of a sector of a circle

Arc LENGTH of sector WOULD be the circumference of the base of the cone

Radius of circle would be the slant height

Radius of circle = 14 cm

Sector angle = 60°

Arc length = ( 60/360) 2π(14) = 14π/3

radius of the base of cone = R

circumference = 2πr

2πr = 14π/3

=> r =7/3  cm

Radius = 7/3 cm

Slant height = 14 cm

Height of cone = √14² - (7/3)²

=  7√2² - (1/3)²

= 7√4 -1/9

= (7/3)√35

radius of the base  = 7/3 cm  

,slant height   = 14 cm

height of the cone =   (7/3)√35  =  13.8 cm

Learn More:

A right angled triangle PQR where angle Q is 90degree.if QR =16 ...

brainly.in/question/9123359

construct a right circular cone with given slant height and ...

brainly.in/question/13042081

your ANSWER is ONE HOPE you have HELP by me

Here is your ANSWER...58/45

The USERS ASKING the questions CHOOSE their brainliest ANSWERS...

Ist ROOT ×second root= c/a
on comparing, a=5, b=13, c=K
let first root be X then its RECIPROCAL root will be 1/x
x×1/x= K/5
K=5

Here 3X+4Y=7.........(1)

4x-y=3........(2)

eq(1) * 1 --- 3x +4y =7
.
eq(2) * 4--- 16X -4y =12

adding eq(1) with eq(2)

................................................................
19x=19

=>x=19/19=1 ........

from eq(1) 3x + 4y=7

=>3*1 + 4y=7

=>4y = 7-3=4

=> y= 4/4=1 ..........

18 x 12 ÷ 2 + 3 = 111

let the second PART be x

so the FIRST part is x/3+48

now x/3+48+x=56

(x+3x)/3=56-48

x+3x=8×3

4x=24

x=24/4=6

therefore the second part =x=6

and first part =x/3+48=6/3+48=2+48=50

hope it HELPED you

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Ur ANSWER is in the PIC....

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1) find the HCF of 54 & 117 .
2) take it EQUAL to 65m-117
3) Calculate the value of 'm'

Solution :

_________

TOTAL C.P. of Horse and carriage = 3000.

SP of these two = 102% of 3000 = 3060.

Let C.P. of Horse was X and Carriage was Y.

∴ X + Y = 3000.

⇒ Y = 3000 - X -------------------- (i)

S.P. of the Horse = 120% of X = 1.2X

S.P. of the carriage = 90% of Y = 0.90Y.

Total SP = 3060

1.2X + 0.90Y = 3060

Using Equation (i)

1.2X + 0.90 * (3000 - X) = 3060

1.2X + 2700- 0.90 = 3060

0.3X = 3060 - 2700

0.3X = 360

X = 360/0.3

= Rs. 1200

So, the Cost Price of the Horse = Rs. 1200

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If any problem in understanding he solution , please feel FREE to ask in the comment box.

The VALUE of (H) is........

Just MULTIPLY 648×45=
29,160

11

Actually I have to group the now with a group 3nos like 1. 331 we have

TAKE the 1 from ones PLACE of the SECOND group and first group contains only ONE no so it's 11

Step-by-step explanation:

1/4 = 5/20
1/5 = 4/20

1/4 and 1/5 MULTIPLIED by 10

5/20x10 = 50/20
similarly 4/20 X 10 = 40 / 20

therefore ANSWER is :

41/20, 42/20, 43/20, 44/20, 45/20, 46/20, 47/20, 48/20, 49/20