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SOLUTION:

In ΔPQR, ∠1 = ∠2∠PQR = ∠PRQ [GIVEN]∴ PR = PQ ……………..…(1)

[Sides opposite to equal angles of a triangle are also equal]Given: QR/QS = QT/PRQR/QS = QT/PQ

QS/QR = PQ/QT…... …(ii) [Taking reciprocals]

[From eq (i)]In ΔPQS and ΔTQR,QS/QR = PQ/QT [From eq (ii)] ∠PQS = ∠TQR [ common]∴ ΔPQS ~ ΔTQR [By SAS similarity criterion]

Hence, proved

SOLUTION:

In ΔPQR, ∠1 = ∠2∠PQR = ∠PRQ [GIVEN]∴ PR = PQ ……………..…(1)

[Sides opposite to equal angles of a triangle are also equal]Given: QR/QS = QT/PRQR/QS = QT/PQ

QS/QR = PQ/QT…... …(ii) [Taking reciprocals]

[From eq (i)]In ΔPQS and ΔTQR,QS/QR = PQ/QT [From eq (ii)] ∠PQS = ∠TQR [ common]∴ ΔPQS ~ ΔTQR [By SAS similarity criterion]

Hence, proved

let the age of Malini be x..then x+30=6x then 5x=30 then x= 6

SURESH- height 176 weight 47Tharun- 156 & 56

1437.3 is the correct answer

the best answer is the 1437.3

2.28571............x7x49

1437.3 is the right answer

1437.34 is correct answer

1437.3 is the correct answer

the question is correct answer is 1437.33

1436.75 it will be right answer

1437.3 is the right answer

Let cost of first cycle be xLet cost of second cycle be 5x

Selling price of first cycle SP1= x(1 - 10/100)= 9x/10

Selling price of second cycle SP2= 5x(1 - 10/100)= 5x*9/10= 45x/10

Now,SP2 - SP1 = 3600(45x - 9x)/10 = 360036x = 36000x = 36000/36 = 1000

Therefore, cost price of costlier cycle= 5x = 5*1000= Rs 5000

Tina has to made 10 glass of 250 ml so water needed = 250×10 = 2500 ml = 2.5L

(-0.06)³

- ( 0.06)³

- ( 6/100)³

- 216/100,00,00

-0. 000216

5(7)=35

5 (7)=35 is the correct answer

odd numbers between 3 & 8 = 5, 7.product of the numbers = 5×7 = 35 is the answer of the following question.

5(7)=35 is my correct answer

35 is the correct answer of the given question.

35 is the best answer

please like my answer then I like your answer all off you

The right answer is 35...

35 is the right answer...................

5×7 is the answer of the following

5×7 it is the correct answer of the following question.

5*7is the correct answer of this question

product of all odd no.between 3 and 8 is also odd number..

5*7........................

5 (7)=35 is the correct answer

(29+2)(6-4)÷[3*11]

31*2÷3362/33

29-(-2*2)/3*1133/33=1

let no.s be x and x+2thenx+x+2=322x=30x=15

no.s are 15 and 17product =15*17=255

Given cost price of cradle = 215

Sold to Ram for price= 215(1 + 5/100)= 215(21/20) = 225.75

Servant paid for cradle= 225.75(1 - 20/100)= 225.75(4/5)= Rs 180.6

The correct answer is option A.

Let us assume, initial quantity of air = 1

Therefore, quantity removed in first step = ¼

Remaining quantity after first step

=1−1/4=3/4

Quantity removed in second step

=3/4×1/4=3/16=3/4×1/4=3/16

Remaining quantity after second step

=3/4−3/16=12−3/16=9/16

Here each subsequent term is not obtained by adding a fixed number to the previous term.

Hence, it is not an AP.

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Let us assume, initial quantity of air = 1

Therefore, quantity removed in first step = ¼

Remaining quantity after first step

=1−14=3/4

Quantity removed in second step

=3/4×1/4=3/16

Remaining quantity after second step

=3/4−3/16=12−3/16=9/16

Here each subsequent term is not obtained by adding a fixed number to the previous term.

Hence, it is not an AP.

In themodulationprocess, two signals are used namely themodulatingsignal and the carrier . Themodulatingsignal is nothing but the baseband signal or information signal while the carrier is a high frequency sinusoidal signal . ... Thismodulatedsignal is then transmitted by the transmitter.

The monkey climb 3m and slips down 2m each time

⇒ net gain = 3 - 2 = 1m

Find the number of attempts needed to reach the 7m mark:

Each attempt = 1 m gain

Length of the pole = 10 m

Number of attempts needed to reach the 7m mark = 7 ÷1 = 7

Find the total number of attempts needed:

Number of attempts needed = 7 + 1 = 8

(the last attempt of a 3m climb made at the 7m mark will reach end of the pole)

Answer; The monkey needs 8 attempts

thank you mam

thanks

Foramplitude modulation, the modulation indexis defined as the measure of extent ofamplitude variation about an un-modulated carrier. In other words it describes the amount by which themodulated carrier envelope varies about the static level.

Given , x1=x y1 =y, x2=1,y2=2, x3=2, y3=1area= 6 sq.units

Area of triangle = 1/2 {x1(y2-y3)+x2(y3-y1)+x3(y1-y2)}

6=1/2{x(2-1)+1(1-y)+2(y-2)}

12=x+1-y+2y-4

12=x-3+y

12+3=x+y

15=x+y

Hence proved

and X+y=9

Let the first term of AP = acommon difference = dWe have to show that (m+n)th term is zero ora + (m+n-1)d = 0

mth term= a + (m-1)dnth term = a + (n-1) d

Given thatm{a +(m-1)d} = n{a + (n -1)d}⇒am +m²d -md = an + n²d - nd⇒am - an +m²d - n²d -md + nd = 0⇒a(m-n) + (m²-n²)d-(m-n)d = 0⇒a(m-n) + {(m-n)(m+n)}d -(m-n)d = 0⇒ a(m-n) + {(m-n)(m+n) -(m-n)} d = 0⇒a(m-n) +(m-n)(m+n -1) d = 0⇒ (m-n){a+ (m+n-1)d} = 0⇒a + (m+n -1)d = 0/(m-n)⇒a + (m+n -1)d = 0

Substitute x=2k-1 , y= k in the equation,

Given √3 cot^2θ - 4cotθ +√3 = 0.⇒ √3 Cot^2θ - 3 Cotθ - Cot θ+√3 = 0⇒ √3 Cotθ( Cotθ - √3) - 1( Cot θ - √3) = 0⇒ ( Cotθ - √3)(√3 Cotθ - 1) = 0Cot θ = √3andCot θ = 1 / √3.letCot θ = √3 and Tan θ = 1 / √3.NowCot^2θ +Tan^2θ= 3 + 1 / 3 = 10 / 3.

9x^2+3xy+3xy+y^2=3x(3x+y)+y(3x+y)=(3x+y)(3x+y)

9x²+6xy+y²=(3x²)+2×3x×y+(y²)=(3x+y)²

(x^a/x^b)^(a+b) * (x^b/x^c)^(b+c) * (x^c/x^a)^(c+a)

=(x)^[a^2 - b^2] * (x)^[b^2 - c^2] * (x)^[c^2 - a^2]

= (x)^[a^2 - b^2 + b^2 - c^2 + c^2 - a^2]

= (x)^0

= 1

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1

2

CP=620Profit percentage=24%Let SP=xProfit percentage=(SP-CP/CP)*10024/100=x-620/620148.8=x-620x=471.2

tan30=75/AB1/√3=75/ABAB=75√3m

Let age of P = 5x

and age of Q = 6x

According to question,

(5x + 4) + (6x + 4) = 52

11x = 44

x = 4

Present age of P = 5x + 4 = 24

Present age of Q = 6x + 4 = 28

Required ratio =2428=67=6:7

thanks

angle ACE=180-(50+25)=180-(75)=105°

Given :Let the x be th number at units place and y be the number at digits placenumber will be =x+10yReverse number will be 10x+yProduct of digit is 20Therefore,xy=20 -----{1}original number + 9=reverse number9x-9y=9x-y=1-------{2}using formula (x+y)²=(x-y)²+4xy(x+y)²=(1)²+4(20) [ by using equation 1 and 2](x+y)²=1+80=81(x+y)=9-----(3)solve equation 3 and 2x+y=9x-y=1_______by adding, we get,2x=10⇒x=5y=9-5y=4so x and y are 5 and 4so number is x +10y =5 +10x4=5+40=45

∴Required number is 45