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Tx plus 2t plus hx plus 2h equals?
(t+h)(x+2)
Tx plus 2t plus hx plus 2h?
What does the question ask? Does it ask you to simplify the above, or is it part of a larger question where values for t, x and h are given? Is it definitely Tx rather than tx? If asked to simplify it, there are numerous ways to do so (but none really make it 'simpler' as such, so it is a bit of a strange one). For example, by factorising: tx + 2t + hx + 2h = t(x + 2) + h(x+2) or we could write (again by factorising) tx + 2t + hx + 2h = x(t + h) + 2(t + h) Which is useful depends on what the question asks, really.
How do you cheat on Aleks math?
If the graph of the function hdefined by =hx+−2x29is translated verticallydownward by 6units, it becomes the graph of a function g.Find the expression for gx.
Tx plus 2t plus hx plus 2h equals?
(t+h)(x+2)
Tx plus 2t plus hx plus 2h?
What does the question ask? Does it ask you to simplify the above, or is it part of a larger question where values for t, x and h are given? Is it definitely Tx rather than tx? If asked to simplify it, there are numerous ways to do so (but none really make it 'simpler' as such, so it is a bit of a strange one). For example, by factorising: tx + 2t + hx + 2h = t(x + 2) + h(x+2) or we could write (again by factorising) tx + 2t + hx + 2h = x(t + h) + 2(t + h) Which is useful depends on what the question asks, really.
Factor 3tx plus 6t plus 3hx plus 6h?
3tx + 6t + 3hx +6h = 3(tx + 2t +hx +2h) = 3[t(x + 2) + h(x + 2)] = 3(x + 2)(t + h)
What is the perpendicular bisector equation of the line joined by the points of h k and 3h -5k on the coordinated grid?
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What is the perpendicular bisector equation of the line whose coordinates are at h k and 3h -5k on the Cartesian plane showing work?
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What is the perpedicular bisector equation of a line that cuts through the line of h plus k and 3h minus 5k showing key stages of work?
Points: (h, k) and (3h, -5k) Slope: -3k/h Perpendicular slope: h/3k Midpoint: (2h, -2k) Perpendicular equation: y--2k = h/3k(x-2h) Multiply all terms by 3k: 3ky--6k2 = h(x-2h) Equation in terms of 3ky = hx-2h2-6k2 Perpendicular bisector equation in its general form: hx-3ky-2h2-6k2 = 0
What is the perpendicular bisector equation of the line joined by the points h k and 3h -5k showing work?
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What is the acid dissociation constant for an acid at equilibrium HX ----- H (spuare Plus) plus x -?
The acid dissociation constant, Ka, is a measure of how well an acid donates a proton in a chemical reaction. For the reaction HX ⇌ H+ + X-, the expression for Ka is [H+][X-]/[HX]. The value of Ka indicates the strength of the acid - higher Ka values indicate stronger acids.
What is the perpendicular bisector equation of the line spanning the points of h k and 3h -5k on the Cartesian plane showing key stages of work?
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What is the acid dissociation constant for an acid at equilibrium hx h plus plus x-?
The acid dissociation constant (Ka) for a weak acid (HX) at equilibrium is defined by the equation: ( Ka = \frac{[H^+][X^-]}{[HX]} ). Here, ([H^+]) and ([X^-]) are the concentrations of the hydrogen ions and the conjugate base at equilibrium, respectively, while ([HX]) is the concentration of the undissociated acid. A higher Ka value indicates a stronger acid, as it signifies a greater tendency to dissociate into its ions.
What is the straight line equation in its generals form whose end points are at k 3h and 3k h showing work?
Points: (k, 3h) and (3k, h) Slope: (h-3h)/3k-k) = -2h/2k => -h/k Equation: y-3h = -h/k(x-k) => ky-3hk = -hx+hk => ky = -hx+4hk Equation in its general form: hx+ky-4hk = 0
What in its general form is the perpendicular bisector equation passing through the line segment of h k and 3h -5k showing key stages of your work?
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