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21–12)*1=9 ……………sequence 1

(41–14)*2/3=18 ……………sequence 2

(53–35)*1=18 ……………sequence 1

(63–36)*2/3=18 ……………sequence 2

hence,(94–49)*1=45 ……………sequence 1



thus,(9,4=45)

Just REVERSE the NUMBER and subtract from ORIGINAL number and MULTIPLY with 1

Heya JAY...understand dis pis...rtta chda LO iska....QUES SHI HOGE hmesha.....

@skb

Any odd degree polynomial with real coefficients has at least one real root. In what FOLLOWS, I ..... OOPS - nice answer to the wrong QUESTION. 1 .... So necessary conditions are.

In TRIANGLE RBT
Line AS ||lineBT
So ANGLE 2 =angle5 AA test 1
Angle 3 = angle 2 given 2
Angle 5= angle 4 given 3
Angle 1 =angle 4 CA test 4
From 1 2 3 4
Angle 3=angle 5 = angle 4
So by equilateral triangle test
Angle 1= 60*

GIVEN.

= (16x2^n+1)-4x2^n / (16x2^n+2)-2x2^n+2

Take 2^n out SINCE it is common on both SIDES in NUMERATOR i.e

= 2^n(16*2-4)/2^n + 2(16-2)

= 28/56

= 1/2

Her m

2x+12-30=2
2x-18=2
2x=2+18
2x=20
x=20/2
x=10

19, 10, 3, -2, -5, ?

19 - 9 = 10
10 - 7 = 3
3 - 5 = -2
-2 - 3 = -5

Here we can observe subtraction of consecutive odd numbers (9, 7, 5, 3)

So, next term
=> -5 -1 = -6

It COMES after 99,99,999
Hope it HELPS.

A+8d=19........(1)
a+18d=3(a+5d)
a+18d=3a+15d
a+18d-3a-15d=0
-2a+3D=0
2a-3d=0..........(2)

a+8d=19 ×(2)
2a-3d=0 ×(1)

2a+16d=38
2a- 3d =0
- + -
__________
19d=38
d=2
2a-3d=0
2a-3(2)=0
2a-6=0
2a=6
a=3

3,5,7.........A.P.

This EQUATION is RELATED to the BOOK of CLASS 7

I FOUND the RATIO of both QUESTION in above PICTURE

Here is your answer

HOPE this HELPS you
PLZ mark as brainliest answer PLZZZZZ PLZZZZZZZZZZZZZ

12500+9280+10445 =32225

GST = 32225× 12%
=3867 Rs
Mr KUMAR has to PAID = (32225+3867)
=36092Rs

2.

we know
180=2×2×3×3×5
so ROOT 180= 6ROOT 5

hope helps
☆☆☆☆☆
pls MARK it as BRAINLIEST.
▪▪▪▪▪▪

The ANSWER is 3-root5

The above image CONTAINS the ANSWER for the QUESTION

#MARK as brainliest

Solution :

Let the numbers be x and 3X

sum = x + 3x = 4x

But actual sum = 145

∴ ACQ

4x = 145

x = 145/4

x = 36.25

so, the numbers are

x = 36.25

3x = 3 x 36.25 = 108.75

Check: 36.25 + 108 .75 = 145

__________________________________

Still any problem in UNDERSTANDING the solution, please feel free to ask in the COMMENTS box.

The ANSWER of this QUESTION is 8.5276

BE is PARALLEL to CD [ because Angle E = Angle D ( By CO INTERIOR Angle )

Also, BD ACTS as a transversal line..

THEREFORE, measure of angle 1 = measure of angle 3 ( By Alt Interior Angle )

Hence Proved.

Average weight of 10 kids = 35

Weight of 10 kids = 35*10 = 350

After sonu joined there will be 11 kids and their average weight is 36 kgs,

So actual weight of 11 kids = 11*36 = 396

Weight of sonu = 396 - 350 = 46

The ANSWER will be 15/22

The ANSWER is in PICTURE

The CUBE ROOT of +5

Angle BCA = 180 - (105+24) = 51

Since BC || DE, Angle CED = Angle BCA
i.e., y = 51°

Since AB || CD, Angle BAC = Angle ECD = 24

From ∆ABC, 51 + 3x + 24 = 180
x = 35°

GOOD QUESTION...........................

I am trying to calculate the multiplicative INVERSE of a large number MOD another large number. For example, I want to calculate the multiplicative inverse of 6003722857 mod 77695236973. I wrote some C++ code to do this. It works fine for small numbers like a = 1891 and n = 3797 but once I try very large numbers, the program doesn't work. There is no error. It looks like it is calculating something but then the program just ends.

Any advice would be appreciated. Thanks!

#include

USING namespace std;

int main ()

{

long long int a = 0, n = 0;

cout << "ENTER a: ";

cin >> a;

cout << "Enter n: ";

cin >> n;

for (long long int c = 1; c < n; c++) {

long long int num = a*c - 1;

if (num%n == 0) {

cout << "The multiplicative inverse is " << c << "." << endl;

break;

}

}

return 0;

5

You're probably overflowing the variables. Try using a "bignum" library such as GMP. – Some programmer dude May 19 '14 at 19:21

How are you running the program? (and what are you compiling it with?) "just ends" seems like a surprising outcome. For me (using VS2013), it just hangs, as might be expected when crawling over such a huge search space. – Rook May 19 '14 at 19:21

If the program "just ends", could it be that c >= n? If that wasn't expected to happen, it may be as @JoachimPileborg suggests. – Scott Hunter May 19 '14 at 19:23

It works, but it is very slow. Is there no way to run the problem in multiple threads? Which input did you use which caused the problem? – Flovdis May 19 '14 at 19:25

Works so far for 56757656857555855678 as input! Define 'very large numbers' more clearly please! – πάντα ῥεῖ May 19 '14 at 19:25

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2 Answers

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1

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accepted

And here it is in super-fast c++:

#include

#include

#include

#include

using namespace std;

using temp_t = std::tuple;

long long calcInverse(long long a, long long n)

{

long long t = 0, newt = 1;

long long r = n, newr = a;

while (newr != 0) {

auto quotient = r /newr;

TIE(t, newt) = make_tuple(newt, t- quotient * newt);

tie(r, newr) = make_tuple(newr, r - quotient * newr);

}

if (r > 1)

throw runtime_error("a is not invertible");

if (t < 0)

t += n;

return t;

}

int main ()

{

long long int a = 6003722857 , n = 77695236973;

/*

cout << "Enter a: ";

cin >> a;

cout << "Enter n: ";

cin >> n;

*/

try {

auto inverse = calcInverse(a, n);

cout << "The multiplicative inverse is " << inverse << endl;

}

catch(exception& e) {

cout << e.what() << endl;

}

return 0;

}

HOPE THIS HELPS..................

PLS MARK IT AS BRAINLIEST!!!!

Then what to FIND...

rhombus is a QUADRILATERAL with all sides EQUAL in length. A square is a quadrilateral with all sides equal in length and all interior angles right angles. Thus a rhombus is not a square unless the angles are all right angles. A square HOWEVER is a rhombussince all FOUR of its sides are of the same length.

1) An isosceles RIGHT triangle has an equal base and height

so, height = base
let base and height be x

its area is 8 cm²

so, area of triangle = 1/2(base * height)
8 = (x*x)/2
x² = 16
x = 4

so base of triangle is 4cm and height of triangle is 4cm

2)The area of the LARGER CIRCLE = 100cm²

so, area of circle = πr²

thus area of smaller circle not need for this calculation

so radius of larger circle = 10√(0.31818) cm

<P>A = Rs. 9261

P = Rs. 8000

R% = 5%

A = P * (1 + (R/100))^T

9261 = 8000 (1 + (5/100))^T

9261/8000 = (105/100)^T

9261/8000 = (21/20)^T

(21^3)/(20^3) = (21/20)^T

(21/20)^3 = (21/20)^T

Thus, T = 3 Years