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As x = 2

kx²-14x+8 = 0

k(2)²-14(2)+8 = 0

4k = 28-8

4k = 20

k = 5

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bro I am not sure hi 😄 for

2*2*2+2k-6=2+2k=02k=-2k=-1

2x^2+kx-6=0; 2(2)^2+k(2)+6=0 2(4)+2k-6=8+2k-6=0; 2k=6-8=-2; k=-2/2=-1

Given that : AB and CD are two equal chords. And, M, N are mid point of chord AB and CD respectively.

To prove : ∠AMN=∠CNM and ∠BMN=∠DNM

Construction : Join OM and ON

Proof :Since the line segment joining the centre of a circle to the midpoint of a chord is perpendicular to the chord.

Since AB and CD are equal chords, they are equidistant from the other. i.e., OM =ON

In ΔOMN,OM=ON (Proved)∠OMN=∠ONM (Angles opposite to equal sides) .....1∠OMA=∠ONC (each 90°) .....2 ∠OMB=∠OND (each 90°) .....3Subtracting 2 from 1, we have,∠OMA-∠OMN=∠ONC-∠ONM⇒∠AMN=∠CNMAdding 1 and 3, we have,∠OMB+∠OMN=∠OND+∠ONM⇒∠BMN=∠DNMHence Proved.

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Let the statement P (n) be1 + 2 + 3 + ... + n = n (n + 1) / 2STEP 1: We first show that p (1) is true.Left Side = 1Right Side = 1 (1 + 1) / 2 = 1Both sides of the statement are equal hence p (1) is true.STEP 2: We now assume that p (k) is true1 + 2 + 3 + ... + k = k (k + 1) / 2and show that p (k + 1) is true by adding k + 1 to both sides of the above statement1 + 2 + 3 + ... + k + (k + 1) = k (k + 1) / 2 + (k + 1)= (k + 1)(k / 2 + 1)= (k + 1)(k + 2) / 2The last statement may be written as1 + 2 + 3 + ... + k + (k + 1) = (k + 1)(k + 2) / 2Which is the statement p(k + 1)

thanks sir

Theformulationof alubricantis made up of base oils and additives that combine to determine behaviour when in use, both in terms of performance and in terms of duration.

Theformulationof alubricantis made up of base oils and additives that combine to determine behaviour when in use, both in terms of performance and in terms of duration.

Thanku Pandey JEE.

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On dividing the given number by 119, let k be the quotient and 19 as remainder. Then, number = 119k + 19

17*7k+17*1+217(7k+1)+2 Hence, the given number when divided by 17,

Solution :-

Given - Divisor = 53, Quotient = 33 and Remainder = 19

We know that

Dividend = (Divisor× Quotient) + Remainder

⇒ (53 × 33) + 19

⇒ 1749 + 19

= 1768

So, the number is 1768.

a=9d=17-9=8S=(n/2)(2a+(n-1)d)636*2=n(18+(n-1)8)1272=n(8n+10)1272=8n^2+10n8n^2+10n-1272=0n=12

answer for the first column element of each row is the sum of the 2nd column element and the 3rd column element of that row

for the 1st row:30 = 10+20

for the 3rd row:25=15+10

thus for the 2nd row:12=x+7x=5

Thanx bro!

1/√2+√3*√2-√3/√2-√3=√2-√3/2-3=√3-√2

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Please solve my other problems

If sum of first and third digit is twice of second digit then the three numbers are in AP.This is true here so the given three terms are consecutive terms of an AP

Let side of square be S

Then,Area of square = S^2

Hence,S^2 = x^2 + 4x + 4S^2 = (x + 2)^2S = (x + 2)

Diagonal of square = side*root(2)= (x + 2)*root(2)= root(2)(x + 2)

(d) is correct option