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To form a TEAM of 4 from the 15 people:

Number of choices for the first PERSON = 15 (That is, any of the 15 people)

Number of choices for the second person = 14 (That is, any of the 14, excluding the one chosen earlier)

Number of choices for the third person = 13 (That is, any of the 13, excluding the ONES chosen earlier)

Number of choices for the fourth person = 12 (That is, any of the 12, excluding the ones chosen earlier)

Find the number of possible combinations:

Number of possible combinations = 15 x 14 x 13 x 12 = 32760

Answer: 32760 different combinations

Heya!!

Here is your answer.

We KNOW that the hypotenuse is the LARGEST side in a right-angled triangle.

And, the formula of Pythagoras theorem is
a²+b² = c²

Where, a = base/PERPENDICULAR, b=base/perpendicular and c=hypotenuse

a and b are the two sides of the triangle i.e the legs.(the base and perpendicular of a triangle are together CALLED the legs)

Now, let's solve the problem USING Pythagoras theorem.

a²+b² = c²

=>c² = 40²+9²

=>c² = 1600+81

=> c² = 1681

=> c = √1681

=> c = 41

⭐ Hope it helps you ⭐

A)x+4=9
b)10a=70...................

4x²+9y²+16z²+12xy-24yz-16xz

= (2X)²+(3Y)²+(-4Z)²+2[(2x)(3y)+(3y)(-4z)+(-4z)(2x)]

=(2x+3y-4z)²

If a parallelogram and a triangle are on the same base and between the same parallels then area of the triangle is half the area of the parallelogram.

solution :

Given: PQRS is a parallelogram and A is any point RS.

Now join AP and AQ. Thus, the field will be divided in 3 parts and each part is in the shape of triangle.

Since , ∆APQ & parallelogram PQRS lie on the same base PQ and between same Parallel Lines PQ & SR.

AR(∆APQ)= ½ ar(||gm PQRS)........(i)

Then, remaining

ar(∆ASP)+ ar(∆ARQ)= ½ ar(||gm PQRS)........(ii)

From eq (i) & (ii)

ar(∆APQ)= ar(∆ASP)+ar(∆ARQ)

So FARMER has two OPTIONS:

Either the farmer can sow wheat in ∆APQ and pulses in other two TRIANGLES or pulses in ∆APQ  and wheat in other two triangles.

Hope this will help you...

Here, A n for the equation is n(n+1)(n+2)

=(n^2+n)(n+2)

=n^3+3n^2+2n

Therefore, A 1 = 1^3+(3*1^2)+(2*1)

A 2 =2^3+(3*2^2)+(2*2)…

A n =n^3+(3*n^2)+(2*n)

So,

S n

=(1^3+2^3+3^3…+n^3)

+3(1^2+2^2+3^2…+n^2)+2(1+2+3+…+n)

={「n(n+1)」/2}^2

+3{「n(n+1)(2n+1)」/6}+2{「n(n+1)/2}

=[n(n+1)]/2{[n(n+1)]/2 + [3(2n+1)]/3+2}

=[n(n+1)]/2{[n^2+n+4n+2+4]/2}

=[n(n+1)]/4{n^2+5n+6}

=[n(n+1)]/4{n^2+3n+2n+6}

=[n(n+1)(n+2)(n+3)]/4

Find more:

1.Find the square root by long division method.

(i)17956 (ii) 11025 (iii) 6889 (iv) 1764

(v) 418609​

brainly.in/question/14520335

2.Square root of 1849 by using division method

brainly.in/question/1656193

Alternate angles:When two lines are crossed by another line the pair of angles on opposite SIDES of the transversal is called alternate angles. 

Theorem 1:If a transversal intersects two PARALLEL Lines then each pair of alternate INTERIOR angles is equal.

Theorem 2 :If a transversal intersects two lines such that a pair alternate interior angle is equal then the two lines are parallel. 

SOLUTION:

Given:  ∠AGE=126°

∠AGE=∠GED=126∘ [alternate interior angles]

 (ii) ∠GED=∠GEF+∠FED=126∘

∠GEF + 90° =126°

(GIVEN that EF⊥CD) 

∠GEF=126°−90°=36°

∠GEF=36°

(iii) ∠CEG+∠GED=180°

(GIVEN ∠GED=126∘)

∠CEG+126° =180°

∠CEG=180° −126°

∠CEG=54°

∠FGE=∠CEG= 54°     (alternate angles)

HOPE THIS WILL HELP YOU...

The HONORABLE Donald J. Trump. 
The USA uses a PRESIDENTIAL system of government, which MEANS Trump is both head of government and head of state.

They will intersect at one point when they are produced in the DIRECTION of GIVEN interior ANGLES.

Let the FIRST even integer be x
Therefore
x+(x+2)+(x+4)+(x+6)=156
4x+12=156
4x=144
x=36
Therefore the four even integer are 36, 38, 40AND 42


HOPE IT 'LL BE HELPFUL FOR U

PLZZ... MARK IT AS THE BRAINLIEST

Solution

Given :

The ratio's of quadrilateral

1 : 2 : 3 : 4

To FIND :

MEASURE of each ANGLE of

quadrilateral.

Let the measures of angles

be x, 2x, 3x, and 4x

Now we have the value of

'x' which is 36°

When we put the value of

'x' we get measure of all

angles of quadrilateral.

Let us do it,

Measure of angle First

= x = 36°

Measure of angle second

=2x

= 2  × 36

= 72°

Measure of angle third

= 3x

= 3 × 36

= 108°

Measure of angle FOURTH

= 4x

= 4 × 36

= 144°

Therefore measure of all

angles of quadrilateral is

36°, 72°, 108° and 144°

2 |20,30,40
2 |10,15,20
2 |5,15,10
5 |5,15,5
|1,3,1 Least Common MULTIPLE of 20,30,40 is 120. remainder is GIVEN as 7.So ANSWER is 120+7=127.

X+2y=2 x-3y=7 ATQ x+2y+x+3y=7+2=9 2x+5y=9

Alternate angles:When two lines are crossed by ANOTHER line the pair of angles on opposite sides of the transversal is called alternate angles.

 Theorem 1 : If a transversal intersects two Parallel Lines then each pair of alternate INTERIOR angles is equal.

Theorem 2 : If a transversal intersects two lines such that a pair alternate interior ANGLE is equal then the two lines are parallel.

SOLUTION:

Construction: Draw ray BE⊥PQ & ray CF⊥RS. BE⊥PQ, CF⊥RS & PQ || RS.

 ∴BE || CF

∠EBC=∠FCB     ……(i)     (Alternate interior angles) 

∠ABE=∠EBC ………(II)       (angle of incidence= angle of reflection) 

∠FCB=∠FCD  ……….(iii)       (angle of incidence = angle of reflection). 

From (i), (ii) and (iii),

∠ABE=∠FCD….(iv) 

Adding (i) and (iv), we get

∠EBC+∠ABE=∠FCB+∠FCD

∠ABC=∠BCD

But these are alternate interior angles and they are equal.Hence, AB || CD 

[fig. is in the attachment]

HOPE THIS WILL HELP YOU...

Hey USER

Here is your ANSWER :-

तर ते पाच दिवस पुढीलप्रमाणे आहेत :-

28 फेब्रुवारी
29 फेब्रुवारी
1 मार्च
2 मार्च
3 मार्च

लीप वर्षाच्या तारीखा आहेत .

28 + 29 + 1 + 2 + 3 = 63 .

It will HOLD WATER EQUAL to its volume


mark as BRAINLIEST if HELPED

Opposite sides PARALLEL means opposite ANGLES EQUAL.

so, If ONE angle =60°

opposite angle=60°

two angles left are ALSO equal. then 360-120=240°

240°÷2=120°

Other angles are of 120°

I) as you can see 2.5^2+6^2=6.5^2
L.h.s=2.5^2+6^2
= 6.25+36
=42.25
=6.5^2=r.h.s
THEREFORE right ANGLE triangle
2) 2^2+2^2=8 and 5^2=25
They are not EQUAL, therefore not right angle triangle
3) 1.5^2+2^2=2.25+4
=6.25
=2.5^2
Therefore it is a right angle triangle

Answer:

The  VALUE of x=-2.

Step-by-step EXPLANATION:

Given : Equation

To FIND : Solve the equation and verify ?

Solution :

CROSS multiply,

The  value of x=-2.

Verification,

SUBSTITUTE x is LHS,

Hence verified.

HEYA MATE,

HERE IS UR ANSWER.

Thanks for asking this QUESTION.

sin A sin 2A + sin 3A sin 6A/ sin A cos 2A + sin 3A cos 6A = tan 5A

We will use the following formulae in this solution--

(1) Sin A . Sin B = (1/2) [ Cos (A-B) - Cos (A+B) ]
(2) Sin A Cos B = (1/2) [ Sin (A+B) + Sin (A-B) ]
(3) Sin A + Sin B = 2 Sin (1/2) (A+B) . Cos (1/2)(A-B)
(4) Sin A - Cos B = 2 Sin (1/2) (A-B) . Cos (1/2)(A+B)
(5) Sin ( -x ) = - Sin x
(6) Cos ( - x ) = Cos x

Solution --

Numerator = (1/2) [ 2 SinA Sin 2A + 2 Sin 3A. Sin 6A ] --- Use formula (1)

=> (1/2) [ Cos (A-2A) - Cos (A+2A) + Cos (3A-6A) - Cos (3A+6A) ]

=> (1/2) [ Cos A - Cos 3A + Cos 3A - Cos 9A ]

=> (1/2) [ Cos A - Cos 9A ] .... Again use formula (1)

=> Sin 5A . Sin 4A

Denominator

= Sin A Cos 2A + Sin 3A.Cos 6A

=> (1/2) [ Sin (A+2A) + Sin (A-2A) + Sin ( 3A + 6A ) + Sin ( 3A - 6A ) ]

=> (1/2) [ Sin 3A - Sin A + Sin 9A - Sin 3A ]

=> (1/2) [ Sin 9A - Sin A ]

=> (1/2) [ 2 Cos 5A . Sin 4A ]

=> Cos 5A . Sin 4A

Hence the Left HAND Side of the given identity

= Numerator / Denominator

=> Sin 5A . Sin 4A / Cos 5A . Sin 4A

=> Tan 5A = RHS ………………. QED .

I HOPE IT HELPS U.

K+6)/1=2(2k-1)/2
k+6=2k-1
2k-k=6+1
k=7

Hi ,

***************************************

We KNOW that ,

1 hour = 60 MINUTES

***************************************

50mile/hour

= 50 miles / ( 60 minutes )

= ( 5/6 ) miles per minute

I hope this helps you.

: )

N/15=4
n=15×4
n=60

n/5=?

we KNOW that n=60

so,60/5=x

x=12

The REMAINDER is 12
plz MARK as BRAINLIEST

HOPE this will HELP you

We know that the sum of all angles of a quadrilateral is EQUAL to 360
Then....
=> A+B+C+D=360
=> 110+70+80+D=360
=> 260+D=360
=> D =360-260
=> D =100

HOPE THIS WILL HELP U
♥️ PLEASE MARK AS BRAINLIST ♥️

I THINK this IMAGE will HELP you

If 15 PENS costs rs.90..then what will be the cost of 22pens? mujhe lagta HAI YE sahi hai after all yhis is your CHOICE

Here is your solution

GIVEN:-

t2+t7=30
=>a+d+a+6d=30
=>2a+7d=30..................... (1)

A/q
15th term=2×8th term-1
=>a+14d=2×(a+7d)-1
=>a+14d=2a+14d-1
=>14d-14d+1=2a-a
=>1=a

Putting the value of a in EQUATION (1)

2a×7d=30
2×1+7d=30
2+7d=30
7d=30-2
7d=28
d=4

A.p are a,a+d,a+2d,a+3d,a+4d

=>A.p are 1,5,9,13.........

HOPE IT HELPS YOU

Hi ,

It is given that ,

PRINCIPAL ( P ) = Rs 8000

Rate of interest ( R ) = 10½ % p.a = 21/2 yrs

Time ( T ) = 2 years

number of times interest paid = n = 2

Amount = A

A = P( 1 + R/100 )ⁿ


A = 8000 ( 1 + 21/200 )²


A = 8000 ( 221/200 )²


A = ( 8000 × 48841 )/ 40000

A= Rs 9768.20

Compound Interest ( C . I ) = A - P

= Rs 1768.20

I hope this HELPS you.

: )

Part of book read by boy on 1st day =1/4
So REMAINING part of the book = 1 - 1/4 = 3/4
Part of book read by boy on 2ND day = 1/3 × 3/4 = 1/4
Total part of book read by boy in TWO days = 1/4 + 1/4 = 2/4 = 1/2

Left part of the book = 1/2
According to question
100 PAGES are left
So, 1/2 = 100 pages
So, total pages in book= 2 × 100 = 200 pages

Here is your ANSWER...

The REQUIRED numbers can be 8,32,40 as their hcf is 8 and sum is 80

I think it is now CLEARED to you..............

PLZ MARK my ans as brainliest

I HOPE it can HELP you and MAKE it BRAINLIST

Let the total NUMBER of students be x.

Let the sections be A, B and C.

The ratios are :

2 : 3 : 5 respectively

The sum of the ratios is 2 + 3 + 5 = 10

The number of students in section :

A = 2/10x

B = 3/10x

C = 5/10x

When 12 students are added in each section we have :

A = 2/10x + 12

B = 3/10x + 12

C = 5/10x + 12

The new ratios of A : B : C is :

4 : 5 : 7

This is equal to :

(2/10x + 12) : (3/10x + 12) : (5/10x + 12) = 4 : 5 : 7

(2/10x + 12) : (3/10x + 12) = 4 : 5

(2/10x + 12) / (3/10x + 12) = 4/5

5{2/10x + 12} = 4{ 3/10x + 12}

x + 60 = 12/10x + 48

10x + 600 = 12x + 480

12x - 10x = 600 - 480

2x = 120

x = 60

The total number of students in the three sections before more students were added is thus 60