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.
(t+h)(x+2)
(x + 2)(h + t)
3tx + 6t + 3hx +6h = 3(tx + 2t +hx +2h) = 3[t(x + 2) + h(x + 2)] = 3(x + 2)(t + h)
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In medical terms you can abbreviate it to "Tx" like you abbreviate History to "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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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.
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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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