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2x² + 3x - 4 = 0a=2, b=3, c=-4

alpha + bita = -b/a = -3/2

1.45, 1.5, 1.55, 1.6, 1.65, 1.7 lies between √2 and √3

1.45, 1.5, 1.55, 1.6, 1.65, 1.7 lies between √2 and √3

1.45. 1.5, .1.55, 1.6, 1.65, 1.7 lives √2 And √3

idk cash Don

In a parallelogram ABCD, angle A and angle B are adjacent angles(sum=180)

angle B - angle A=40angle B + angle A= 180

Add

2 angle B=220angle B=110°

AngleA=70°

Therefore angle A=angle C=70°angle B=angle B=110°

two crore five lakh twenty thousand two= 2,05,20,002plz like thanks

25020002 right answer

185 is the answer of maths

37/2 metre of cloth = 1961/2

1 metre cloth = 1961/2 * 2/37

1 metre = 53 rupees

26 1/2 metres = 53*53/2

= 79 1/2 rupees

Can Rama solve this problem?

1200000*2*4/100

= 12000*8

= 96000

Population = 1200000 + 96000

= 12,96,000

Present population = 12 lakh = 1200000Time = 2 yearsRate of increase = 4%Therefore, population after 2 years is= present population(1 + r/100)²= 1200000 (1 + 4/100)²= 1200000 * (104/100)²= 1200000 * 104/100 * 104/100= 120 * 104 * 104= 1297920. {ANSWER}

Present population = 12 lakh = 1200000Time = 2 yearsRate of increase = 4%Therefore, population after 2 years is= present population(1 + r/100)²= 1200000 (1 + 4/100)²= 1200000 * (104/100)²= 1200000 * 104/100 * 104/100= 120 * 104 * 104= 1297920.

we can apply the trick of finding S.I.

We use compound interest for these

Fertilisation or Fertilization, also known as generative fertilisation, insemination, pollination, fecundation, syngamy and impregnation, is the fusion of gametes to initiate the development of a new individual organism or offspring.

Let that rama,s age is x.From eqution x+12=3xso 12=3x-x so 12=2x so x =12\2=6 condition me hamko kh rha h ki 12 varsh bad uski age present age ki 3guni ho jati h.so if his present aga ko 3guna kr denge tb eqution =hoga.

Let, Rama's present age is x year.After 12 years Rama's age will be =(x+12) year.According to given statement we can write -x+12 = 3xor, 3x = x+12or, 3x-x = 12or, 2x = 12or, x = 12/2or, x =6So, Rama's present age is 6 year.

x=3 is a correct answer

3x-4=53x=9x=9÷3=3x=3

3×-4=53×=9×=9÷3=3×=3

3x-4=53x=9x=9÷3=3x=3 is the right answer

3x-4=53x=9x=3 is the right answer of this question

In triangle PQR​A and B are mid points of sides PR and QR​where CR intersects AB at Dangle ADC=90 degreeangle ACR=angleBCRtherefore triangle PQRis a equilateral triangle

Area of triangle = 1/2*base*height

Using pythagoras theoram find height p

10*10 = 6*6 + p*p100 = 36 + p*pp*p = 100 - 36 = 64p = 8

Area of triangle = 1/2*6*8 = 24 cm^2

tanA=cot(60+A)cot(90-A)=cot(60+A)90-A=60+A90-60=A+A30=2A15°=A

15° is the right answer

LichensThey have high absorption system which cause deposit of sulphur dioxide which will destroy chlorophyll and inhibits photosynthesis

LichensThey have high absorption system which cause deposit of sulphur dioxide which will destroy chlorophyll and inhibits photosynthesis

Present population = 12 lakh = 1200000Time = 2 yearsRate of increase = 4%Therefore, population after 2 years is= present population(1 + r/100)²= 1200000 (1 + 4/100)²= 1200000 * (104/100)²= 1200000 * 104/100 * 104/100= 120 * 104 * 104= 1297920. {ANSWER}

Let Alka get monthly Rs xAmount she spend every month = 90%

Then amount she save every month = 10%

Hence,10x/100 = 120x = 1200

Therefore, monthly amount Alka get = Rs 1200

We can show thata2+ b2= c2usingAlgebra

Take a look at this diagram ... it has that "abc" triangle in it (four of them actually):

Area of Whole Square

It is a big square, with each side having a length ofa+b, so thetotal areais:

A = (a+b)(a+b)

Area of The Pieces

Now let's add up the areas of all the smaller pieces:

First, the smaller (tilted) square has an area of:c2

Each of the four triangles has an area of:ab2

So all four of them together is:4ab2= 2ab

Adding up the tilted square and the 4 triangles gives:A = c2+ 2ab

Both Areas Must Be Equal

The area of thelarge squareis equal to the area of thetilted square and the 4 triangles. This can be written as:

(a+b)(a+b) = c2+ 2ab

NOW, let us rearrange this to see if we can get the pythagoras theorem:

Start with:(a+b)(a+b) = c2+ 2ab

Expand (a+b)(a+b):a2+ 2ab + b2= c2+ 2ab

Subtract "2ab" from both sides:a2+ b2= c2

(x1 + x2 + x3 + ....... +x20)/20=yx1 + x2 + x3 + ....... +x20=20y

{(x1 -101)+(x2-101)+(x3-101)+.....+(x20-101)}/20={(x1 + x2 + x3 + ....... +x20)-(101×20)}/20=(20y-20×101)/20=y-101

-384 is the answer.......

-384 is correct answer

Your correct answer is -384

The correct answer of this question = -384.

thanks

Value of Machine = 10500As machine Depreciated by 5%

Then, Machine value after one year= 10500(1 - 5/100)= 10500(1 - 1/20)= 10500(19/20)= 525*19= Rs 9975

Formula is a = p(1-r/100)^n --------------------- (i)

Given p = 10000, r = 5, n = 2

On submitting values in (i), we get

a = 10000(1-5/100)^1

= 10000(95/100)

= 9500 rs.

Original value P= 500000Depreciated value A= 364500R = 10%No. Of years = n

A = P(1 - R/100)^n364500 = 500000(1-10/100)^n3645/5000 = (9/10)^n(9/10)^3 = (9/10)^nn = 3

No. Of years = 3 yrs

Let the worth of estate be x.=>4/5 of x=16800=>x=(16800×5)/4=21,000

:.(3/7)×21000=9,000