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Is this a complete question?

bhai ye he find zeros of polynomials

5x - 2--------- = 9 2

5x - 2 = 18

5x = 20

x = 4

Ans :- Theaddition rule statestheprobabilityoftwo eventsis thesumof theprobabilitythat either will happen minus theprobabilitythatbothwill happen.

Cost of 6 notebooks=₹270Cost of 1 notebook=₹270÷6=45

therefore cost of 15 notebooks=15×45=675Therefore cost of 15 notebooks is ₹675

10 cp = 8 sp

ratio of cp : sp

cp : sp

8 : 10

4 : 5

% gain = (5-4)/4×100= 1/4 × 100= 25 %

t

thank you

P(A)=0.5/2=0.25P(B)=0.5/3=0.17P(A INTERSECTION B)=0.5/5=0.1P(A U B)=0.25+0.17-0.1=0.32P(A|B)=P(A)/P(A INTERSECTION B) =0.25/0.32=25/32

loda koto javab che chodina

P(B) = 2/3

P(B) + P(B') = 1

P(B') = 1 - 2/3 = 1/3

P(B') = 1/3

thanks

BhI 15000 is correct answer according to your question

the correct answer will be -15000

-15000 is the correct answer.

-15000 is the correct answer of this question

15 *(-25)*(-4)*(-10)=-15000 is correct answer.

15000 is the right answer

-15000 is the best ans

-15000 is the correct answer

-15000 is the correct answer

15000 is the correct answer

-15000 is the best answer

15× (-25)×(-4)×(-10) =-15000

-15000 is the right answer according to your question

- 15,000 is a correct answer

15 × (-25) × (-4) × (-10) = (-15000) so is the correct answer.

assuming base is 10 solog (n-2) = 2log (n-2) = log 10²comparing we getn-2 = 10²n = 100+2n = 102

I think we multiply45×30and divide

coming from opposite relative speed = 45+30=75km/hto cross each other total distance= 450+550=1000m75000mt goes in 3600 sec1 Mt goes in 3600/75000 sec1000 mt goes in = 1000×3600/75000=48 sec

48 seconds is the correct answer

10 second s per km to the train

n = (2017)!

use the identity loga[b] = logb/loga

so, log2[n] = logn/log2 and 1/log2[n] = 1/(logn/log2)=> 1/log2[n] = log2/logn

similarly for all other terms 1/log3[n] = log3/logn1/log4[n] = log4/logn

now adding all we will get , denominator as same in all as logn

on adding= (log2+log3+log4 +......+log2017)/(logn)= log(2×3×4×....2017)/log(n)= log(2017!)/logn= logn/logn = 1

Log(m+n)= logm+ logn

log(m+n) = log(mn). [ loga+logb =log ab]

log on both sides geys cancelthen,

m+n = mn => m= mn -n=> m = n(m-1)=> n = m/m-1 [proved]

A pair of numbers used to locate a point on a coordinate plane is called an ordered pair. An ordered pair is written in the form (x, y) where x is the x-coordinate and y is the y-coordinate.

Sn = 3 n² + 4 nTn = Sn - S_n-1 = 3 n² + 4 n - 3 (n-1)² - 4 (n-1) = 6 n + 1

d = Tn - T_n-1 = 6 n + 1 - 6(n-1) - 1 = 6Since d = 6 = constant, The sequence is an AP

thank u

thanks sir

Let the height of the equal pillars be AB = CD = h

Given, width of the road is 150 m

Let BE = x, the DE = 150 - x

In right angle triangle ABE,

tan 60 = h/x

=> √3 = h/x

=> h = √3x ............1

In right angle triangle CDE,

tan 30 = h/(150 - x)

=> 1/√3 = h/(150 - x)

=> √3h = 150 - x

=> √3h = 150 - h/√3 {from equation 1}

=> √3h + h/√3 = 150

=> (3h + h)/√3 = 150

=> 4h = 150√3

=> h = 150√3/4

=> h = 37.5√3 m

So, the height of the equal pillers is 37.5√3 m

Let the height of the equal pillars be AB = CD = h

Given, width of the road is 150 m

Let BE = x, the DE = 150 - x

In right angle triangle ABE,

tan 60 = h/x

=> √3 = h/x

=> h = √3x ............1

In right angle triangle CDE,

tan 30 = h/(150 - x)

=> 1/√3 = h/(150 - x)

=> √3h = 150 - x

=> √3h = 150 - h/√3 {from equation 1}

=> √3h + h/√3 = 150

=> (3h + h)/√3 = 150

=> 4h = 150√3

=> h = 150√3/4

=> h = 37.5√3 m

So, the height of the equal pillers is 37.5√3 m

nahi aaya