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Given a circle has the equation (x-4)^2 + (y)^2 = 25 what are the ... 30 Nov 2015 · O = (4, 0), r = 5 (x - a)^2 + (y - b)^2 = r^2 Center is (a, b), radius is r. Given a circle has the equation #(x-4)^2 + (y)^2 = 25# what are the coordinates of the center and the length of its radius?
How do you use the Binomial Theorem to expand #(x + 2)^4#? 27 Jan 2017 · (x+2)^4" "=" "x^4+8x^3+24x^2+32x+16. The Binomial Theorem says that, for a positive integer n, (x+b)^n= ""_nC_0 x^n + ""_nC_1 x^(n-1)b+""_nC_2 x^(n-2)b^2 + ... + ""_nC_n b^n This can be succinctly written as the sum (x+b)^n= sum_(k=0)^n ""_nC_k x^(n-k)b^k To see why this formula works, let's use it on the binomial for this question, (x+2)^4. If we were to …
How do you evaluate 2x+3 when x=4? - Socratic 27 Jul 2016 · Substitute 4 for x. Just substitute 4 for the variable x: 2(4)+3=8+3=11. 6463 views around the world
How do you solve #2 log x = log 2 + log (3x - 4)#? - Socratic 10 Dec 2015 · x= 2 ; x= 4 > Given 2log x= log 2 + log(3x-4) Step 1: Rewrite the equation as a single logarithm on the right hand side, using the sum to product rule, like this logx^2 = log(2*(3x-4)) logx^2 = log(6x-8) Step 2: Transform into the exponential form with the base of 10 like this (or most simple way to put it, if there are log with same base on each side of equation then we …
How do you solve #x^2 +6x +8 =0# using the quadratic formula? 8 May 2018 · The answers are x=-2 and x=-4. To start, the quadratic formula is x=(-bpmsqrt(b^2-4ac))/(2a) In this problem, a = 1 (as the x^2 term has no coefficient), b=6, and c=8. Plug those values into the quadratic equation to get: x=(-6pmsqrt(6^2-4(1)(8)))/(2(1)) Multiply 2*1 on the bottom of the fraction: x=(-6pmsqrt(6^2-4(1)(8)))/(2) Square 6 and multiply 4*1*8 within the …
How do you factor x^4+x^2+1? - Socratic 10 May 2015 · x^4 + x^2 + 1 = (x^2 + x + 1)(x^2 - x + 1) To find this, first notice that x^4 + x^2 + 1 > 0 for all (real) values of x. So there are no linear factors, only quadratic ones. x^4 + x^2 + 1 = (ax^2 + bx + c)(dx^2 + ex + f) Without bothering to multiply this out fully just yet, notice that the coefficient of x^4 gives us ad = 1. We might as well let a = 1 and d = 1. ... = (x^2 + bx + c)(x^2 …
Factorise? a) x^4+2x^3+3x^2+2x+1 - Socratic x^4+2x^3+3x^2+2x+1=(x^2+x+1)^2. When you have a "symmetric polynomial", one where the coefficients are the same when you read them forwards and backwards, like x^4+2x^3+3x^2+2x+1, then it will have symmetric factors too. Let us assume that the given fourth degree polynomial has a pair of symmetric, quadratic factors. The natural choice, since the …
How do you graph #y= (x-4)^2 +3 - Socratic 24 Aug 2015 · Determine the vertex and several points, preferably on mirror images of the parabola. Plot points and sketch a curve through the points. Do not connect the dots. y=(x-4)^2+3 The equation is in vertex form, y=a(x-h)^2=k, where (h,k) is the vertex, and a=1, h=4, and k=3. The vertex (h,k)=(4,3). Determine several points on the parabola, substituting both positive …
How do you find the limit of (x^2-4)/(x-2) as x approaches 2? 3 Nov 2016 · If we look at the graph of #y=(x^2-4)/(x-2) # we can see that it is clear that the limit exists, and is approximately #4# graph{(x^2-4)/(x-2) [-10, 10, -5, 5]} The numerator is the difference of two squares, and as such we can factorise using it as # A^2-B^2 -= (A+B)(A-B) # Se we can factorise as follows:
How do you solve |x - 4|< 2? - Socratic 29 May 2018 · Solution is set of values: 2< x< 6 Remove the absolute value term. This creates a +- on the right side of the equation because |x|= +-x x−4<+-2 Step 1: Set up the positive portion of the +- solution. x−4<2 Add 4 both sides: x-4+4<2+4 x<6 Step 2: Set up the negative portion of the +- solution. When solving the negative portion of an inequality, flip the direction of the …