Find HCF on 285 and 1294 whose remainders are9 and 7

Notice: Undefined variable: position6 in /home/domains2/public_html/painphr.com/show.php on line 143

Guide :

# Find HCF on 285 and 1294 whose remainders are9 and 7

find HCF  of  285and1249 whose remainders are 9 and 7

## Research, Knowledge and Information :

Notice: Undefined variable: go_page in /home/domains2/public_html/painphr.com/show.php on line 211

Notice: Undefined variable: go_page in /home/domains2/public_html/painphr.com/show.php on line 242

We did not find results for: find hcf on 285 and 1294 whose remainders are9 and 7.

## Suggested Questions And Answer :

Notice: Undefined variable: position4 in /home/domains2/public_html/painphr.com/show.php on line 277

### Find HCF on 285 and 1294 whose remainders are9 and 7

me dont understand the quesshun "HCF" normally meen hiest konnom fakter, but 285 & 1294 dont hav NE kommon fakters Then yu talk bout "remaenders".  Fakters defined==number that divide intu start number with leftover=0.

Notice: Undefined variable: position4 in /home/domains2/public_html/painphr.com/show.php on line 277

### find all reconstuctions of the sum: ABC+DEF+GHI=2014 if all letters are single digits

There is no solution if A to I each uniquely represent a digit 1 to 9, because the 9's remainder of ABC+DEF+GHI (0) cannot equal the 9's remainder of 2014 (=7). Let me explain. The 9's remainder, or digital root (DR), is obtained by adding the digits of a number, adding the digits of the result, an

Notice: Undefined variable: position4 in /home/domains2/public_html/painphr.com/show.php on line 277

### 10 vedic maths rules for class IX

2 instead of 5: 34/5 can be calculated by multiplying 34 by 2 instead of dividing by 5. 34*2=68. We just move the decimal point one place to the left: 34/5=6.8. 124/5=24.8 because 124*2=248. Move the decimal point: 248 becomes 24.8. 34*5 is the same as 34/2=17 but we add a zero to make 17 into

Notice: Undefined variable: position4 in /home/domains2/public_html/painphr.com/show.php on line 277

### find HCF of3x^4+17x^3+27x^2+7x-6;6x^4+7x^3-27x^2+17x-3 by division method.

3x^4+17x^3+27x^2+7x-6=(x+1)(3x^3+14x^2+13x-6)=(x+1)(x+3)(3x^2+5x-2)= (x+1)(x+3)(3x-1)(x+2) CLUE: Write coefficients: 3 17 27 7 -6. If we add them together, do we get zero? No, so 1 is not a zero and x-1 is not a factor. Look at the odd powers of x (17x^3 and 7x). Change the sign in fro

Notice: Undefined variable: position4 in /home/domains2/public_html/painphr.com/show.php on line 277

### find HCF and LCM of3x^4+17x^3+27x^2+7x-6;6x^4+7x^3-27x^2+17x-3 by division method.

3x^4+17x^3+27x^2+7x-6=(x+1)(3x^3+14x^2+13x-6)=(x+1)(x+3)(3x^2+5x-2)= (x+1)(x+3)(3x-1)(x+2) CLUE: Write coefficients: 3 17 27 7 -6. If we add them together, do we get zero? No, so 1 is not a zero and x-1 is not a factor. Look at the odd powers of x (17x^3 and 7x). Change the sign in fro

Notice: Undefined variable: position4 in /home/domains2/public_html/painphr.com/show.php on line 277

### find a geometric progression whose sum is 5/2. Is it possible to find another one?

In an infinite GP converging to a finite limit, S=a/(1-r) because Sn=a(r^n-1)/(r-1)=a(1-r^n)/(1-r). When r is small (<1) and n is large r^n approaches zero and is zero when n=infinity. Therefore, 5/2=a/(1-r) so 5(1-r)=2a. This shows a relationship between a and r for 0 Read More: ...

Notice: Undefined variable: position4 in /home/domains2/public_html/painphr.com/show.php on line 277

### Find the volume of the solid whose base is the region bounded by y=x^7, y=1,

The radius of each semicircle is (1-y)/2 which equals (1-x^7)/2. The area of such a semicircle is π((1-y)/2)^2/2=π(1-y)^2/8. If the semicircle has a thickness dx the volume of the semicircular section is π(1-y)^2dx/8. When y=1, x=1, and (0,0) is the starting point. The volume of the first semicir

Notice: Undefined variable: position4 in /home/domains2/public_html/painphr.com/show.php on line 277

### i need the answer for these questions

Part 1 Newton’s Method for Vector-Valued Functions Our system of equations is, f1(x,y,z) = 0 f2(x,y,z) = 0 f3(x,y,z) = 0 with, f1(x,y,z) = xyz – x^2 + y^2 – 1.34 f2(x,y,z) = xy –z^2 – 0.09 f3(x,y,z) = e^x + e^y + z – 0.41 we can think of (x,y,z) as a vector

Notice: Undefined variable: position4 in /home/domains2/public_html/painphr.com/show.php on line 277

### the polynomial x^4-2x^3 +3x^2-ax+3a-7 when divided by x+1 leaves remainder 19 .find [a]

Subtract 19: x^4-2x^3+3x^2-ax+3a-26. Because we removed the remainder, x+1 must be a factor of the revised polynomial. We can use synthetic division to divide by x+1 using -1 as the root. When we do this we end up with 4a-20. This must be equal to zero for there to be no remainder, so 4a-20=0 and a=

Notice: Undefined variable: position4 in /home/domains2/public_html/painphr.com/show.php on line 277

### p(x)=kx^3+15x^2+9x-40 mod (x+4)=-28. find the value of 'k'.

Try different values of x to observe the behaviour of p(x). x=0: p(0)=0, because 40 mod (x+4)=40 mod 4=0. x=1: p(1)=k+24, because 40 mod 5=0. If k+24=-28, then k=-52. x=-1: p(-1)=5-k, because 40 mod 3=1. If 5-k=-28, then k=33. x=2: p(2)=8k+74, because 40 mod 6=4. If 8k+74=-28, then k=-10

#### Tips for a great answer:

- Provide details, support with references or personal experience .
- If you need clarification, ask it in the comment box .
- It's 100% free, no registration required.
next Question || Previos Question

Notice: Undefined variable: position5 in /home/domains2/public_html/painphr.com/show.php on line 357
• Start your question with What, Why, How, When, etc. and end with a "?"
• Be clear and specific
• Use proper spelling and grammar