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121/16 is the right answer......

5/2 + 81/16=5(16)+81(2)/2(16)=80+162/32=242/32= 121/16=7.56

121 /16 is a correct answer

nice but it's factorisation

Also known as “special cause”, an assignable cause is an identifiable, specific cause of variation in a given process or measurement. A cause of variation that is not random and does not occur by chance is “assignable”.chance cause

A cause for variability in a measurement process that occurs randomly and unpredictably and for unknown reasons.

please give the answer of second part...explain the theory behind the control chart technique.

Assume

the number of butterflies = x

the number of flowers = y

Given that if there is one butterfly for one flower, there is one flowerleft.

=> y = x + 1

Also, if there are two butterflies on each flower, there is one flowerleft.

=> y = 2(x - 1)

Thus,

x + 1 = 2x - 2

x - 2x = -2 -1

-x = -3

=> x = 3

The number of butterflies = x = 3.

The number of flowers = y = x + 1 = 3 + 1 = 4.

The Answer is option 4.

Given that the pipe can fill a tank without any leak in 12 hrs.

Work done without leak in 1 hour = 1/12.

Given that the waste pipe can fill a tank with leak in 16 hrs.

Work done with leak in 1 hour = 1/16.

Now,

Work done by leak in 1 hour = 1/12 - 1/16

= 4 - 3/48

= 1/48.

So, the total time takento empty it = 48 hours.

Let a be the positive odd integer which when divided by 6 gives q as quotient and r as remainder.

according to Euclid's division lemmaa=bq+r

a=6q+r

where , a=0,1,2,3,4,5then,a=6qora=6q+1ora=6q+2ora=6q+3ora=6q+4ora=6q+5

but here,a=6q+1 & a=6q+3 & a=6q+5 are odd.

x-3=0 x=3now,3×3×3-3×3×3+3×4-1227-27+12-12=0it is proved

Given that the pipe can fill a tank without any leak in 12 hrs.

Work done without leak in 1 hour = 1/12.

Given that the waste pipe can fill a tank with leak in 16 hrs.

Work done with leak in 1 hour = 1/16.

Now,

Work done by leak in 1 hour = 1/12 - 1/16

= 4 - 3/48

= 1/48.

So, the total time takento empty it = 48 hours.

The resistivity remains same for a certain material at a particular temperature

you are wrong

1

2

3

4

5

shall i give you 100 thanks?

let The Total distance = S

Time to cover the first 1/3 of S = t1 = (S/3) / v1 = S/(3 v1)

Time to cover the second 1/3 of S = t2 = (S/3)/v2 = S/(3 v2)

Time to cover the last 1/3 of S = t3 = (S/3 ) / v3 = S/ (3v3)

ER. RAVI KUMAR ROY.

Total time = t = S/3 [ 1/v1 + 1/v2 + 1/v3 ]

Let v = average speed

v = S / t

So 1/v = t / S

1/ v = 1/3 [ 1/v1 + 1/v2 + 1/v3 ]

1.Let the square of any posititve integer be in the form 6q + 2, 6q +5

x = 6q +2

Squaring both sides

Xsquare = 36(q)square + 4 + 24q

x square = 36(q) square - 6+ 2 + 24q

xsquare = 6 [ 6(q) square - 1 + 4q] + 2

put [ 6(q) square - 1 + 4q] = m

Therefore x = 6m + 2

2 .let a be any +ve integerb=6

by euclid 's division lemma,

a=bq+r, 0=< r < ba= 6q+r,0=< r

when

r=0,a = 6q= (6q)3 ---> a3 = 216q3 =6(36q3 )=6q(where m is = 6q3 )

by similar manner u can prove for

r=1,2,3,4,5

and u will get the proof..(if you need it as 1,2,3,4,5ask me)...

3.If n = 5q

n is divisible by 5

Now, n = 5q

⇒ n + 4 = 5q + 4

The number (n + 4) will leave remainder 4 when divided by 5.

Again, n = 5q

⇒ n + 8 = 5q + 8 = 5(q + 1) + 3

The number (n + 8) will leave remainder 3 when divided by 5.

Again, n = 5q

⇒ n + 12 = 5q + 12 = 5(q + 2) + 2

The number (n + 12) will leave remainder 2 when divided by 5.

Again, n = 5q

⇒ n + 16 = 5q + 16 = 5(q + 3) + 1

The number (n + 16) will leave remainder 1 when divided by 5.

Case II:

When n = 5q + 1

The number n will leave remainder 1 when divided by 5.

Now, n = 5q + 1

⇒ n + 2 = 5q + 3

The number (n + 2) will leave remainder 3 when divided by 5.

Again, n = 5q + 1

⇒ n + 4 = 5q + 5 = 5(q + 1)

The number (n + 4) will be divisible by 5.

Again, n = 5q + 1

⇒ n + 8 = 5q + 9 = 5(q + 1) + 4

The number (n + 8) will leave remainder 4 when divided by 5

Again, n = 5q + 1

⇒ n + 12 = 5q + 13 = 5(q + 2) + 3

The number (n + 12) will leave remainder 3 when divided by 5.

Again, n = 5q + 1

⇒ n + 16 = 5q + 17 = 5(q + 3) + 2

The number (n + 16) will leave remainder 2 when divided by 5.

Similarly, we can check the result for 5q + 2, 5q + 3 and 5q + 4.

In each case only one out of n, n + 2, n + 4, n + 8, n + 16 will be divisible by 5.

4.Let a be any positive integer and b = 6.

Then, by Euclids algorithm,

a = 6q + rfor some integer q ≥ 0, and r = 0, 1, 2, 3, 4, 5 because 0 ≤ r

Therefore, a = 6q or 6q + 1 or 6q + 2 or 6q + 3 or 6q + 4 or 6q + 5

Also, 6q + 1 = 2 × 3q + 1 = 2k 1 + 1, where k 1 is a positive integer

6q + 3 = (6q + 2) + 1 = 2 (3q + 1) + 1 = 2k 2 + 1, where k 2 is an integer

6q + 5 = (6q + 4) + 1 = 2 (3q + 2) + 1 = 2k 3 + 1, where k 3 is an integer

Clearly, 6q + 1, 6q + 3, 6q + 5 are of the form 2k + 1, where k is an integer.

Therefore, 6q + 1, 6q + 3, 6q + 5 are not exactly divisible by 2. Hence, these expressions of numbers are odd numbers.

And therefore, any odd integer can be expressed in the form 6q + 1, or 6q + 3, or 6q + 5....

thankyou but it seems to hard plez make it simple form and easy to understand

Let a be the positive integer and b = 6.Then, by Euclid’s algorithm, a = 6q + r for some integer q ≥ 0 and r = 0, 1, 2, 3, 4, 5 because 0 ≤ r < 5.So,a = 6q or 6q + 1 or 6q + 2 or 6q + 3 or 6q + 4 or 6q + 5.(6q)^2= 36q^2= 6(6q^2)= 6m, where m is any integer.(6q + 1)^2= 36q^2+ 12q + 1= 6(6q^2+ 2q) + 1= 6m + 1, where m is any integer.(6q + 2)^2= 36q^2+ 24q + 4= 6(6q^2+ 4q) + 4= 6m + 4, where m is any integer.(6q + 3)^2= 36q^2+ 36q + 9= 6(6q^2+ 6q + 1) + 3= 6m + 3, where m is any integer.(6q + 4)^2= 36q^2+ 48q + 16= 6(6q^2+ 7q + 2) + 4= 6m + 4, where m is any integer.(6q + 5)^2= 36q^2+ 60q + 25= 6(6q^2+ 10q + 4) + 1= 6m + 1, where m is any integer.

Hence, The square of any positive integer is of the form 6m, 6m + 1, 6m + 3, 6m + 4 and cannot be of the form 6m + 2 or 6m + 5 for any integer m.

By whole numbers we mean numbers without fractions or decimals. You can also callpositive integersyour 'counting numbers' because they are the same. You don't count with fractions or decimals or negative numbers. On a number line,positive integersare all the numbers to the right of the zero

examples of even integer

2,4,6,8,etc

-2,-4,-6 also

no no only 2,4,6,8

Let a be any positive integer andb = 2.Then, by Euclid’s division lemma there exist integers q and r such that

A = 2q + r, where 0 ≤ r ≤ 2

Now, o ≤ r ≤ 2 → r = 0 or, r = 1

.˙. a = 2q or, a = 2q + 1

If a=2q, then a is an even integer.

We know that an integer can be either even or odd. Therefore, any odd integer is of the form2q+1.

Let ‘a’ be any positive even integer and ‘b= 6’.Therefore,a = 6q +r, where 0≤r < 6.Now, by placing r = 0, we get, a = 6q + 0 = 6qBy placing r = 1, we get, a = 6q +1By placing, r = 2, we get, a = 6q + 2By placing, r = 3, we get, a = 6q + 3By placing, r = 4, we get, a = 6q + 4By placing, r = 5, we get, a = 6q +5Thus, a = 6q or, 6q +1 or, 6q + 2 or, 6q + 3 or, 6q + 4 or, 6q +5.But here, 6q +1, 6q + 3, 6q +5 are the odd integers.Therefore, 6q or, 6q + 2 or, 6q + 4 are the forms of any positive even integers.

1/4 + [ 1/2*1/2 ÷{1/2*1/2÷1/2 + (1/2÷1/2)}]= 1/4 + [1/4 ÷ {(1/2*1) + (1)}]= 1/4 + [1/4 ÷ (1/2 + 1)]= 1/4 + [1/4 ÷ 3/2]= 1/4 + 1/6= 5/12

(d) is correct option

2x²-5x-3 = 2x²-6x+x-3 = 2x(x-3)+1(x-3) = (x-3)(2x+1)

Hence, the zeros of 2x²-5x-3 are -1/2 and 3.

Zeros of the polynomial x²+px+q are double in value to -1/2 and 3,

Thus zeros are -1 and 6.

Now, quadratic expression with zeros -1 and 6 is given by

(x + 1)(x-6) = x²-5x-6. By comparison, p = -5 and q = -6

Principal = rs. 800Time = 73 days = 73/365 = 1/5 yearRate = 7.5 %S.I = (P × R × T)/100= (800 × 7.5 × 1/5)/100= 12

principal =Rs. 800, time=73 /365=0.2; Rate$7.5°\°, S.I.=(PxRxT)/100=(800+7.5+0.2)/100=12

800*7.5*1/100*51.6*7.512

12 is the correct answer of the given question

p= 800t= 73day= 73/365= 1/5r= 7.5%intrest= prt/100 = 800×7.5/5×100= ₹13

p= 800, r= 7.5t= 73;day= 73/365=1/5 years I= prt/100 = 800×7.5×1/5/100= ₹12

In 🔺PQR and 🔺SQR

Then PQ=RS (given) QR =QR (comman side and Diagonal) Then RHS 🔺 PQR is congruent 🔺SQR

in ∆ PQR and ∆ SQRangle RPQ = angle QSR ( 90°)PQ = SR ( given) QR = QR ( common)By RHS rule ∆ PQR and ∆ SQR are congurent.So, by CPCT PR = SQ

x² + 7x + 10

p + q = (-7)

pq = 10

(p+q)² = p² + q² + 2pq(-7)² = p² + q² + 2×(10)49 - 20 = p² + q² p² + q² = 29

log 10 + 2log 3- log 2=1+ 2(0.47)-(0.3)=1.64

125 = angle A + Angle B (exterior angle =sum of two opposite interior angle)125=1x+4x125=5xx=125/5×=251x=25°4×=100°3rd angle= y180-125=55 (asp)the angles are 25,100,55

find the value of x

thanks