ALGEBRA 2 ; I need help with solving systems of equations, and solving inequalities!?
Standard form: 3x-4y=5; 6x-8y=-5; substitute/eliminate x = +4/3y+5/3 inconsistent system - no solutions. Standard form: 2x-7y=14; x+3y=6; substitute/eliminate x = +7/2y+7 substitute/eliminate y = -2/13 Solution: x = 84/13; y = -2/13; (x, y) = (6.461538462, -0.153846154) Standard form: 4x-y=10; -3x+y=-6; substitute/eliminate x = +1/4y+5/2 substitute/eliminate y = +6 Solution: x = 4; y = 6; (x, y) = (4, 6) Standard form: x-y=6; 3x+2y=-22; substitute/eliminate x = +1y+6 substitute/eliminate y = -8 Solution: x = -2; y = -8; (x, y) = (-2, -8) Standard form: 5x+2y=1; 2x+3y=7; substitute/eliminate x = -2/5y+1/5 substitute/eliminate y = +3 Solution: x = -1; y = 3; (x, y) = (-1, 3) Standard form: 5x-3y=16; 2x+7y=-10; substitute/eliminate x = +3/5y+16/5 substitute/eliminate y = -2 Solution: x = 2; y = -2; (x, y) = (2, -2)
Algebra 2 Honors:solving y=3x^2-6x+8 and -4x^2+6x-3=0 (URGENT!)?
Hi my answer is coming up so as you know standard form is ......y = ax 2 + bx + c and vertex is .....y= a(x-h)2+k where h and k are the vertex coordinates of the parabola y=3x^2-6x+8 ....for vertex form you need the coordinate points of the vertex which is this case is (1,5)=(h,k) so using the equation for the vertex form we plugin the 1,5 where they belong: ......y= a(x-1)2+5 <--- now we just plugin a which according to the other equation is the 3 next to the x squared ......y= 3(x-1)2+5 <---SOLUTION PART 2: -4x^2+6x-3=0 4x^2-6x+3=0 ...... multiplied by negative one on both sides not you have to use the quadratic equation... so plugin the numbers... x= [ 6 +- sqrt( (-6^2) -4(4)(3) ] / 2(4) x= [ 6 +- sqrt( 36 -48 ) ] / 8 x= [ 6 +- sqrt( -12 ) ] / 8 <---the roots are "imaginary" since the discriminant is negative x= [ 6 +- sqrt( 12 ) * sqrt( -1 ) ] / 8 <---sqrt of -1 is i which is an imaginary number: i=sqrt(-1) x= [ 6 +- (3.46410162)*i ] / 8 so not we have two answers x1= [ 6 + (3.46410162)*i ] / 8 x2= [ 6 - (3.46410162)*i ] / 8 for x1... x1= 6/8 + (3.46410162)*i / 8 <---simple algebra concept... ex. (x-y)/2= (x/2) - (y/2) x1= 3/4 + 0.433012702i <---- if you dont like the decimal number you can replace it for the exact answer by simply replacing the sqrt of 12 to what it was before which is x1= 3/4 + sqrt(-12)/8 do the same for x2...
Solve algebraically... log[4](3x-1)=2... help!?
1) log[4]( 3x - 1 ) = log[4]16, since 2 is equal to log[4]16. Now you know that 3x - 1 must also equal 16 for the equation to be true, so 3x = 17 x = 17/3 2) log[3]( x + 1 ) - log[3] ( x - 4 ) = log[3]9 Using log laws, you know that log[3] ( x +1 / x - 4) = log[3]9 x + 1 / x - 4 = 9 x + 1 = 9 ( x - 4 ) = 9x - 36 37 = 8x 37/8 = x
How do I algebraically solve this type of equation for X (minimal calculators please): 1.3*10^-6 = 0.4x^2 + 4x^3?
From the ratios of the coefficients, the constant on the left-hand side is quite small, so you can solve the equation by iteration:First, recognize that the cubic term is going to be much smaller than the quadratic term, so to lowest order, we just haveC = 4.0E-1 X**2, with C = 1.3*E-6X = sqrt(C/0.4).We can now calculate the presumably small correction to it by noting that the full solution “x” will be “X+d,” where “d” is the small correction. We are now interested in solving for “d,” since we already have X.C = 0.4 * (X + d)**2 + 4 * (X+d)**3,C = 0.4 * (X**2 + 2*X*d +d**2) + 4 * (X**3 + 3*X**2 d + 3* X *d**2 + d**3 ).So far we’ve just rewritten the original equation, but if we take d to be sufficiently small, we then disregard terms involving d**2 or d**3:C = 0.4 * (X**2 + 2*X*d) + 4 * (X**3 + 3 * X**2 * d),or d = (C-0.4*X**2–4.0*X**3) / (0.8*X + 1.2E1 X**2).Note that X**2 = C/0.4, so we haved = -X**2/(0.2 + 0.3*X**2). (Since this solution is higher-order in X, our assumption that d is sufficiently small is retroactively justified.)We can then construct a new “X” from the old values of X+d, and iterate again if we find our first-order answer X+d to have insufficient precision.The zero-order solution will be Order(10E-2), and the first order-correction will be Order(10E-2)+Order(10E-4), which just corrects the zero-order solution by two digits. A second-order correction will give a better correction to Order(10E-4), and at some point, we will start correcting the terms Order(10E-6), and so on.
Algebra Questions.?
I have several math problems that have been troubling me. I have answers for some but Im 100% sure that they are wrong. I've tried tutors but they had trouble with them and their answers were so complex I dont trust them. Any help would be appriciated. ^ = Exponent. 1) Factor y^2 + 8y + 16 - 169t^2 2) Multiply (x^2 + x - 1)(-x^3 + x^2 + 1) 3) The product of the reciprocals of two consecutive integers is 1/90. Find the integers. 4) __6__ - __2__ x^3 - 125 - x^2 - 25 5) One summer, Oscar mowed 4 lawns for every 5 lawns mowed by Luis. Together they mowed 144 lawns. How many did each mow? 6)David can paint a bedroom in 1.75 hr. Jay can paint the same room in 2.5 hr. How long will it take them, working together, to paint the bedroom? 7) Solve 9x^2 + 3x = 0 8) Solve 3x^3 + 4x = -8x^2 9)Left f(x) = 3x2 - 21x + 286. Find a such that f(a) = 286. 10) Find the domain of the function f given by f(x) = __8-x__ x^2 + 6
Solve the systems with linear equation graphic, put in form Y=MX+B?
A) x-y= 1 → y = x - 1 .................. (1) 3x+2y = -12 → y = -3x/2 - 6 .... (2) You should now graph the two lines and see where they intersect. I am unable to do the graphics here so I will solve them algebraically Substitute equation (2) in equation (1) x - 1 = -3x/2 - 6 So 5x/2 = -5 → x = -2 So y = -3 So the lines should intersect at (-2, -3) Check x - y = -2 - -3 = 1 3x + 2y = -6 + -6 = -12 YESSSSSSSSSSSSSSS!!!!! So the lines should intersect at (-2, -3) So the solution is x = -2 y = -3 B) m+ 5n+ 9= 0 → n = -m/5 - 9/5 ....... (1) 3m - n -5 = 0 →n = 3m - 5 ..............(2) Graph the two lines (n is the equivalent of Y and m is the equivalent of X in Y = MX + B) and find where they intersect I am unable to do the graphics here so I will solve them algebraically Substitute equation (2) in equation (1) So 3m - 5 = -m/5 - 9/5 ie 16m/5 = 16m/5 → m = 1 so n = -2 Check m+ 5n+ 9 = 1 - 10 + 9 = 0 3m - n -5 = 3 - -2 - 5 = 0 YESSSSSSSSSSSSSS!!!!!!!! So lines should intersect at (1, -2) So the solution is m = 1, n = -2
How can I solve (x^4 -2x^3 +3x^2 -2x +1=0) manually?
This equation has a symmetric form, and what you do with such equations is that you take the last and first terms together, try to apply some algebraic identities to them, and then take the second terms from the beginning and the end, and bring out something common from the previous factorisation. Here we can take the following approach,x^4–2x^3+3x^2–2x+1=0x^4+1–2x(x^2+1)+3x^2=0(x^2+1)^2–2x(x^2+1)+x^2=0(x^2-x+1)^2=0((x-1/2)^2+3/4)^2=0which has no real roots because it is always positive. However the two repeated complex roots can be found out by using SreedharAcharya’s formula.
Nova net algebra 2 linear equation 3x + 2(x + 3) = -2 + 4(x - 4)?
3x + 2(x + 3) = -2 + 4(x - 4) 3x + 2x + 6 = -2 + 4x - 16 5x + 6 = 4x - 18 x + 6 = -18 x = -24