Loading [MathJax]/jax/output/SVG/jax.js
 
  Questions   
Sort: 
 #3
avatar+484 
+3
Feb 16, 2018
 #2
avatar+149 
+3
Feb 16, 2018
 #1
avatar+118696 
+3

1.A telegraph has x arms and each arm is capable of (x-1) distinct positions , including the position of rest. The total no. of signals that can be made is?

Rosala, This is just      x(x-1)

 

 

2. How many natural numbers are there fro 1 to 1000 which have none of their digits repeated?

i've tried this question a no. of times but it just doesnt see to be clicked in my mind. i would be really appreciate if anyone can explain it to me in detail.

There are 9 one digit ones

There are no four digit ones

So how many 2 digit ones are there.

The tens digit can be 1 to 9 that is 9 choices, now you used one digit but you can use the zeros so there are 9 possible digits left for the units digit

So that is 9*9=81

AND how many three digit ones are then  9*9*8 = 648

Altogether there are  9+81+648 = 738

 

3. Number of different natural numbers which are smaller than two hundred million and using only the digits 1 or 2 is :

 

less than      200,000,000   only containing the digits 1 and 2

 

1 digit     2

2 digit     2*2=2^2

3 digit      2^3

4 digit       2^4

5 digit      2^5

6 digit      2^6

7 digit      2^7

8 digit      2^8

9 digit      2^8       The biggest value digit must be 1

 

So what do we get when we add these up

 

2+22+23+24+25+26+27+28+28=(2+22+23+24+25+26+27+28)+28The brackets is the sum of a GP, a=2 and r=2\;\;n=8=a(rn1)r1+28=2(281)21+28=292+28=29+282=28(2+1)2=3×282

 

 

i. (3).2^8 - 2 

 

ii. (3).2^8 - 1 

 

iii. 2 (2^9 - 1) 

 

iv) None 

 

I have a few more question like the questions above, and i just dont seem to get the correct answers. 

 

i have tried the questions above many times but i just cant understand a few things , i would be glad if anyone could explain to me in detail. Thank you very much. 

 

i really appreciate the time you'd take out for my answers! 

Feb 16, 2018
 #2
 #4
avatar+26396 
+1

In the diagram below, we have DE = 2EC and AB = DC = 20. Find the length of FG.

 

 

Let DC=20 Let DE=2320 Let EC=1320 Let FG=x+EC or x=FGEC Let BG=p Let GC=q 

 

1. intercept theorem

qx=pDExDE=2320qx=p2320xpq=403x3x(1)

 

2. intercept theorem

qDEx=px+ECDE=2320EC=1320q2320x=px+1320pq=3x+20403x(2)

 

(1)=(2):pq=403x3x=3x+20403x403x3x=3x+20403x(403x)2=3x(3x+20)1600240x+9x2=9x2+60x1600240x=60x300x=1600|:300x=1600300x=163FG=x+ECEC=203=163+203=363FG=12

 

 

laugh

Feb 16, 2018
 #1

2 Online Users

avatar