
Solve the Equation 6x-8=8 - Answer | Math Problem Solver - Cymath
Solve the Equation 6x-8=8 - Answer | Math Problem Solver - Cymath ... \\"Get
How do you solve 6x-8=20? - Socratic
2017年4月5日 · It 6x=28 or x=14/3 You can arrange the equation as 6x=20+8 then solve it for x: x=28/6 or by simplification x=14/3.
How do you write f(x) = -x^2 + 6x + 8 in vertex form? | Socratic
2016年6月20日 · In that #f(x)=-x^2+6x+8+k# at this stage #k=0# Write as: #y=( -1x^2+6x)+8+k# Factor out the #-1# (Sometimes this value is not 1) #y=-1(x^2-6x)+8+k# take the exponent …
Resolver la Ecuación 6x-8=4 - Respuesta - Cymath
Resolver la Ecuación 6x-8=4 - Respuesta - Cymath ... \\"Obtén
What is the equation that best represents "six times a ... - Socratic
2016年10月31日 · 6x-8<5x+2 Six times a number: 6x Decreased by 8: -8 Is less than: < Five times a number: 5x Plus two: +2 Put it all together to get: 6x-8<5x+2
How do you find the x and y intercepts for - 6x + 8? | Socratic
2017年4月26日 · Given: #" "f(x)=-x^2+6x+8# The use of the term #f(x)# is that of giving a label (name) to a particular mathematical construct. The #f# is the name and the #(x)# bit means it …
How do you find the vertex and the intercepts for y = x^2-6x+8?
2016年7月9日 · Vertex is at (3, -1) ; y-intercept is at (0,8) and x-intercepts are at (2,0) and (4,0) We know the equation of parabola in vertex form isy=a(x-h)^2+k where vertex is at (h,k).Here …
What is the slope of a line perpendicular to 2y=-6x+8? - Socratic
2014年9月13日 · First we need to solve the linear equation for y because we need to get the slope. Once we have the slope we need to convert it to its negative reciprocal, this means to …
How do you solve the equation x^2+8=-6x by graphing? - Socratic
2017年10月2日 · See below. To graph x^2+8=-6x First add -6x to both sides to get: x^2+6x+8=0 We can now write this in function form by replacing the 0 with a y: x^2 +6x+8= y or. y=x^2 …
How do you simplify 4x-2+(-6x)+8? - Socratic
2017年1月21日 · First, subtract #color(red)(4x)# and add #color(blue)(8)# to each side of the equation to isolate the #x# term while keeping the equation balanced: