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(x+2)/3 = (y+1)/2 = (z-3)/2=k(let)

(x+2)/3=k and (y+1)/2=k and (z-3)/2=k

given x=3k-2,y=2k-1,z=2k+3 given(x,y,z) is at a distance of 5 from(1,3,3)

sqrt[(3k - 3)^2+(2k - 4)^2+(2k)^2] = 5

squaring on both sides

9k^2 + 9 - 18k + 4k^2 + 16 - 16k+ 4k^2 = 2517k^2 - 34k = 0k(17k - 34) = 0

k=0 or k=2

substitute k=0 x=1, y=1, z=5 therefore point is (1, 1, 5)

answer = 0

since cos 90°=0, when it is multiplied by any number, it gives the result = 0

no idea this question

the value of y = 8 which each of the number is exactly divisible

Let the number of passengers travelling by Class I and Class ll be x and 50x respectively.

Then amount collected from Class I and Class II will be Rs.3x and Rs.50x respectively.

Given, 3x + 50x = 1325

=> 53x = 1325

=> x = 25

Therefore, Amount collected from Class II =50×25= Rs.1250.

a=270b=90 this the answer of this question

a=270b=90 is the right answer

There are three number which is divisible by 3 is 3,6 and 9...

1)-2.5 number is indicated by point B2)point C indicates the number

1) -2.5 no. is indicated by point B2) point c indicates the number

2 first second 1 we have the number line where as many questions

there are 2 two's

There are90 twodigit numbers 10,11, 12, ...99.

there are 91two digit numbers

A.P : 14, 21, 28,……,98

a=14

d=21–14=7

aₙ=98

By using, aₙ=a+(n-1)d

98=14+(n-1)7

98–14=(n-1)7

84÷7=n-1

12=n-1

n=12+1

n=13

There are 13 numbers divisible by 7.

Loss=CP-SP150=CP-1250CP=1400Rs

7500 + (1250÷50) = 7500 + 25 = 7525

1) value of cos(x)/(π-2x) at x = π/2 is -sin(x)/(-2 ... by.. L.H rule . so, at π/2 , the value is -1/2

so, k*(-1/2) = 3 => k = -6

2) kx² = 3 => k(2)² = 3 => k = 3/4

3) kπ+1 = cos(π) => kπ = -2 => k = -2/π

y^12 ÷ (y^6 x y^3 x y)= y^12 ÷ (y^[6+3+1])= y^12 ÷ y^10= y^[12-10]= y^2 ans

tong

oo no

rong answer

this is not a three digit their are two digitplease you read the question

point Epoint P,D,C,Q,R,Epoint A ,Bon the triangle

thanks

2x^2 + 3x - 2 = 02x^2 +4x - x - 2 = 02x(x + 2) - 1(x +2)=0(2x - 1)(x+2)x = 1/2, - 2

Both equations have common roots thenPut value of x = - 2 in equation 3x^2 +4kx +2=03(-2)*(-2) + 4k*(-2) + 2 = 012 - 8k +2 = 014-8k=08k = 14k = 7/4 ans

the A.P. will be12,15, 18,21, ………….99

We need to find n Last term of this series is 99

So, an = 99 & a =12d =15–12= 3

Now, an = a + (n – 1) d

Putting values 99 =12+ (n – 1) (3) 99 =12+ 3n – 3 99 = 9 + 3n 3n = 99 – 9 3n =90n =90/3 = 30

Therefore , there are30 twodigit numbers divisible by 3.

1

2

thnx

Two digits numbers divisible by 3 form an AP

The A.P. will be12,15,.......99We need to find number of terms of an AP Last term of this series is 99 So, an = 99 & a =12d =15–12= 3 Now, an = a + (n – 1) d

Putting values,99 =12+ (n – 1)(3) 99 =12+ 3n – 3 99 = 9 + 3n 3n = 99 – 9 3n =90n =90/3 = 30

Therefore, there are30 twodigit numbers divisible by 3.

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Area of triangle= (1/2)×base×height=(1/2)×15×24 cm²=15×12 cm²=180 cm²

Two digit numbers are : 10,11,12,....,,98,99

numbers divisible by 2 are 10, 12, 14, ...96,98. series APnumbers divisible by 5 are 15, 20,... , 90, 95 there are some common numbers in the two series: 20, 30, 40, ...,90.

A.P.Series1 = 10, 12, ..,96, 98 number of terms : (98-10)/2 + 1 = 45 S1 = (first term+last term)/2*number of terms = (10+98)/2*45 = 27*45=1215

sum of numbers divisible by 5 : number of terms = (95 - 15)/5 + 1 = 17 S2 = (95+15) / 2 * 17 = 55 * 17=935

Sum of numbers divisible by 10 : number of terms= (90-20)/10+ 1 = 8 S3 = (90+20)/2 * 8= 55 * 8=440

Answer = S1 + S2 - S3 =1215+935-440 =1710

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Pi is a non-terminating decimal, non repeating . Value of pi - 3.14159

value of pie is non terminating and non repeating because the vaule of is fix

kabhi kabhi jaldi Aata Kabhi der Mein Aata likhna answer bilkul sahi Jata

tanπ/6=1/√3 tanπ/4=1 tanπ/3=√3squaring and adding1/3+1+3=1/3+4=13/3

this function is both one- one and onto