(t+h)(x+2)
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.
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.
(t+h)(x+2)
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.
3tx + 6t + 3hx +6h = 3(tx + 2t +hx +2h) = 3[t(x + 2) + h(x + 2)] = 3(x + 2)(t + h)
HX ---> H+ + X- Keq = [H+][X-]/[HX]
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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
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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
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HX Draw was created in 1911.