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3)angle A = 90° and angle C = 30°=> angle B = 90° - 30° = 60°=> BC is the hypotenuse = 12 cm longAB = 12/2 =6cm (sin30° = AB/hyp => 12sin30°=ABAB = 12×1/2 = 6AC = √(BC^2 - AC^2) = √(12^2 - 6^2) = √(144-36) = =√108 cm

hit like if you find it usefulAC = 4 cmAB = 6.93 cm

if x-1 is the factor , then x = 1 will make p(x) to 0

=> p(1) = (1)²+(1)+k = 0 => k+2 = 0 => k = -2

iii) p(1) = k(1)²-√2(1)+1=0 => k = √2-1

It will last for 12 days if 20 more students join to the existing 80 students. Explanation:80 students can eat food for 15 days1 student can eat food for:15 x 80 = 1200 daysIf 20 more students join, then total will be 100 students.The food will go for 100 students: 1200/100=12 days

it will be clear for u

thanks

my pleasure sorry for sending to much answer

x^3 - 9x^2 - 12x + 20 = 0x^3 + 2x^2 - 11x^2 - 22x + 10x + 20 = 0x^2(x + 2) - 11x(x + 2) + 10(x + 2) = 0(x + 2)(x^2 - 11x + 10) = 0(x + 2)(x2 - 10x - x + 10) = 0(x + 2)(x(x - 10) - 1(x - 10)) = 0(x + 2)(x - 1)(x - 10) = 0x = - 2, 1, 10

Therefore,Other two zeroes are 1 and 10

ab mat bhejo

α² + β² can be written as (α + β)² - 2αβ

p(x) = 3x² - 7x + 6a = 3 , b = - 7 , c = 6

α and β are the zeros of p(x)

we know that ,sum of zeros = α + β = -b/a = 7/3product of zeros =c/a = 6/3 = 22α + 3β and 3α + 2β are zeros of a polynomial.

sum of zeros = 2α + 3β+ 3α + 2 = 5α + 5β= 5 [ α + β]= 5 × 7/3 = 35/3product of zeros = (2α + 3β)(3α + 2β)= 2α [ 3α + 2β] + 3β [3α + 2β]= 6α² + 4αβ + 9αβ + 6β²= 6α² + 13αβ + 6β²= 6 [ α² + β² ] + 13αβ = 6 [ (α + β)² - 2αβ ] + 13αβ = 6 [ ( 7/3)² - 2 × 3 ] + 13× 2= 6 [ 49/9 - 6 ] +26 = 6 [ 49/9 – 54/9] + 26 = 204/9 = 68/3[ simplest form ]

a quadratic polynomial is given by :-k { x² - (sum of zeros)x + (product of zeros) }k {x² - 35/3x + 68/3}k = 22 {x² - 35/3x + 68/3 ]=6x² - 35x + 68 is the required polynomial

Steps to construction :Draw a line of length 3.4 name it AB.From A draw a circle of radius 4 cm and from B draw a circle of radius 2.5cm.The point where these circles meet is the third point C.

can u please tell me the line a is passing through B and the line C is passing through A , B so where is the line b

aap ne sahi draw kiya hai

lekin jaha big circle aur small circle bisector kar raha hai waha point c hoga

lim (2x + 3) = 3x-->

Let Consider ε > 0 , we will show existence of delta| 2x + 3 - 3 | < ε

| 2x - 0 | < ε

| x - 0 | < ε/2

Choose δ = ε/2implies

| x - 0 | < δ, so we have shown that for every epsilon > 0 there exist a delta .

sqauring a number, last digit will be product of unit digit's last digit number Last digiti) 81² 1ii) 272² 4v) 1234² 6vi) 26387² 9 12796² 6x) 55555 5

Let the share of A = Rs. x

According to the question,

A : B : C

Capital -> x : 2x : (4x - 50)

(x + 2x + 4x - 50) = 13,950

7x - 50 = 13,950

7x = 14000

x = 2000

Share of A = Rs. 2000

please post u r question in English to provide a solution to it

Oooo ofcurse bro

1 is the identity for multiplication of whole numbers ormultiplicativeidentity for whole numbers. A number remain unchanged when multiplied by 1. Hence the product of two whole numbers will beequalto 1 if and only if both whole numbers are 1

carpets can be put on the floor to absorb the sound.

panels made out of sound absorbing materials are put on the walls and ceilings.

Let total questions be n

(4/100)n = 6n = 6*25 = 125

Total questions = 125

Questiond answered correctly= (78/100)*125 = 39*3= 107

answer in detail please

Mean of 16 observations is 8. Now each observation are multiply by 2 So new mean 8×2 = 16

1st part

thanks a lot

If the product of 2 number is zero, then one of them is definitely zero.eg=0×2=010×0=0If the product of two whole number is zero then both of them may be 00×0=0

yes, 0 is a rational number. eg-0/5, 0/1.

ABC SCHOOL

12.may.2019NOTICE(underlined)

The Cultural programme

This is to inform all the students of classes that our school is organising a music competition on the Occasion of the annualday. Following are the details of the programme :-Day- Wednesday Date- 17/May/2019Time- 10:00 am

Students who are interested for the same are requested to give their names to the undersigned latest by 15/May/2019

(Sign)(Name)Head boy

1. Closure Property of Addition and multiplication.Property:a + bis a real number and ab is also a real number.

2. Commutative Property of Addition and multiplication.Property:a + b = b + a and ab = ba

3. Associative Property of AdditionProperty:(a + b) + c = a + (b + c)

Given,P = 40000, R = 7.3%, T = 3

Then,SI = P*R*T/100 = 40000*7.3*3/100 = 40*3*7.3 = 120*7.3 = 876

Therefore, interest will be = Rs 876

thank you

Non-store retailingis the selling of goods and services outside the confines of a retail facility. It is a generic term describingretailingtaking place outside of shops and stores (that is, off the premises of fixed retail locations and of markets stands). The non-store distribution channel can be divided intodirect selling(off-premises sales) and distance selling, the latter including all forms ofelectronic commerce.

sec4A= cosec(A-20)cosec(90-A)=cosec(A-20)90-A=A-2090+20=A+A110=2AA=110/2=55

2 cosA- secA/cosA/ sinA- sinA/ cosA=2 cosA-5=12/7; 2cosA=12/7+5=17/7

i dont no srwoory next time

4y+(-3×3)=-54y=4y=1

3x+4y=-5 (x=-3) 3(-3) +4y=-5-9+4y=-54y=-5+94y=4y=4/4y=1hence, 1 is right answer

Some of the daily life examples ofliving thingsaround us are humanbeings, animals, plants and microorganisms.

Non-livingsthingsdo not exhibit any characteristics of life.

They do not grow, respire, need energy, move, reproduce, evolve, or maintain homeostasis.

Thesethingsare made up ofnon-livingmaterials

For finding maximum weight, we have to find H.C.F. of 90 and 69.

Factors of 90= 5*3*2*3

Factors of 69 = 3 x 69

H.C.F. = 3

Therefore the required weight is 3 kg.

ranjana purchased two bags of wheat weighing 90 kg and 69 kg find maximum value

Consider the following pair of linear equations in two variables,

x + 2y = 4

2x + 4y = 12

The solution of this pair would be a pair (x, y). Let’s find the solution, geometrically. The tables for these equations are:

x04y = (4 – x)/220

x06y = (12 – 2x)/430

Now, take a graph paper and plot the following points:

A(0, 2)

B(4, 0)

P(0, 3)

Q(6, 0)

Next, draw the lines AB and PQ as shown below.

From the figure below, you can see that the two lines are parallel to each other. Therefore, these lines don’t intersect at all. Hence, this pair of equations has no solution.

(2√2+5√3)+(√2-3√3)=2√2+5√3+√2-3√3=3√2+2√3

thanks

Let the price of a saree be x and the price of a shirt be y. Then, 2x+4y = 16000....(i)x+6y = 16000....(ii)

2(ii) - (i) => 12y - 4y = 16000=> 8y = 16000=> y = Rs. 2000 = PRICE OF 1 SHIRT.

PRICE OF 12 SHIRTS = 12*2000Rs. 24000.

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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.

Let the share of P = X

Then, Q's share = x + 30

and R's share = (x + 30) + 60 = x+90

Sum of money with P, Q and R=300

x+x+30+x+90=300

=> 3x+120 =300x= 60required ratio 60:(60+30):(60+90)2:3:5

thx

CP1= 2150CP2=2150

Loss=8%

SP1=[(100-L%)/100] *CP1=[(100-8)/100]*2150=Rs. 1978

Total Profit =Rs. 1230

Hence,

(SP1+SP2)- (CP1+CP2) =1230

(1978+SP2)- (2150+2150) =1230

1978 + SP2-4300 =1230

SP2-2322=1230

SP2=Rs. 3552

To prove:

( tanA + secA -1)/( tanA- secA+1) = (1+sinA)/cosA

Proof:

(tanA - secA) - (sec2A - tan2A)/tanA + 1 - secA

(tanA - secA) (1 - (secA - tanA))/tanA + 1 - secA

(tanA - secA) (1 - secA + tanA)/tanA + 1 - secA

sec A + tan A

1/cosA + sinA/cosA

1 + sinA/cosA

Hence, Proved

thank you

Given :

sin = cos

So, tan = 1

So, = 45°

So, value of 2 tan + cos²

2 tan (45°) + cos²(45°)

2 × 1 + (1/√2)²

2 + 1/2

5/2