The expression "p x p x p" represents the product of three identical variables, which can be simplified as p^3. This is equivalent to multiplying p by itself three times, resulting in p multiplied by p, then that product multiplied by p again. In mathematical terms, p^3 represents p raised to the power of 3, indicating the cube of the variable p.
Oh, dude, you're really testing my math skills here. So, when you multiply p by itself three times, you get p cubed. It's like you're making a little p sandwich with three layers. So, p x p x p = p^3. Math can be fun when you think of it in terms of Sandwiches, right?
p2 X p2 = p4or p X p X p X p = p4
A) x-5 is a factor of p(x) B) x-2 is a factor of p(x) C) x+2 is a factor of p(x) D) The remainder when p(x) is divided by x-3 is -2 I need help solving this question because I didn't really get the answer explanation that college board gave it. Thanks in advance!
The perimeter can be anything that's 138.6 or greater.The area doesn't tell you the perimeter.Here are a few examples:30 x 40 . . Area = 1200 . . Perimeter = 14025 x 48 . . . A = 1200 . . . P = 14624 x 50 . . . A = 1200 . . . P = 14820 x 60 . . . A = 1200 . . . P = 16015 x 80 . . . A = 1200 . . . P = 19012 x 100 . . . A = 1200 . . . P = 22410 x 120 . . . A = 1200 . . . P = 2608 x 150 . . . A = 1200 . . . P = 3166 x 200 . . . A = 1200 . . . P = 4125 x 240 . . . A = 1200 . . . P = 4903 x 400 . . . A = 1200 . . . P = 8062 x 600 . . . A = 1200 . . . P = 1,2041 x 1200 . . . A = 1200 . . . P = 2,402
nx - m = p so x = (m+p)/n
The sum of p and q
4p
p2 X p2 = p4or p X p X p X p = p4
p2 X p2 = p4or p X p X p X p = p4
f(x) = P(x)/Q(x) where P(x) and P(x) are polynomials and P(x) is not zero.
Well for one you cant buy Alex Rider the Gemini Project because it doesn't exist what your after is Alex Rider Point Blanc and as for that you can buy it in many places like amazon or got to the libary and get it :) :) :) :) :) :) :) :) :) :) :) :) :) :) :) :) :) :) :) :) :) :) :) :) :) :) :) :) :) :) :) :) :) :) :) :) :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :p :p :p :p :p :p :p :p :p :p :p :p :p :p :p :p :p :p :p :p :p :p :p :p :p :p :p :p :p :p :o :o :o :o :o :o :o :o :o :o :o :o :o :o :o :o :o :o :o :o :o :o :o :o :o :o :o :o :o :o :b :b :b :b :b :b :b :b :b :b :b :b :b :b :b :b :b :b :b :b :b :b :b :b :b :b :b :b :b :b :x :x :x :x :x :x :x :x :x :x :x :x :x :x :x :x :x :x :x :x :x :x :x :x :x :x :x :x :x :x :x :x :x :) :) :) :) :) :) :) :) :) :) :) :) :) :) :) :) :) :) :) :) :) :) :) :) :) :) :) :) :) :) :) :) :) :) :) :) :)
2 rows of planks going downwards vertically. |P|P|X| |P|P|X| |P|P|X| Where P=plank and X=nothing
x-a is a factor of the polynomial p(x),if p(a)=0.also,if x-a is a factor of p(x), p(a)=0.
P= positive N=negative P x N = N N x P = N P x P = P N x N = P Hope that helps!?!?!
We have to use the expression P(X=x) = nCx px (1--p)(n--x) Here n = n and p=p and x = 1 or x>1 P(X>/=1) = 1 -- P(X</=1) So, P(X<=1) = P(X=0) + P(X=1) This gives nC0 p0 (1--p)(n-0)+ nC1 p1 (1--p)(n--1) ie (1--p)n + n p (1--p)(n--1)
Yes.
Suppose p(x) is a polynomial in x. Then p(a) = 0 if and only if (x-a) is a factor of p(x).
#include <stdio.h> #include <stdlib.h> main(){ int x, *p; p = &x; /* initialise pointer */ *p = 0; /* set x to zero */ printf("x is %d\n", x); printf("*p is %d\n", *p); *p += 1; /* increment what p points to */ printf("x is %d\n", x); (*p)++; /* increment what p points to */ printf("x is %d\n", x); exit(EXIT_SUCCESS);