t=t0+(frac{k}{2})w, for w
\(\displaystyle{t}={t}{0}+{\left({\frac{{{k}}}{{{2}}}}\right)}{w}\), for w
We are given:  

\(\displaystyle{t}={t}{0}+{\left({\frac{{{k}}}{{{2}}}}\right)}{w}\)  

Subtract t0 from both sides:  

\(\displaystyle{t}-{t}{0}={t}{0}+{\left({\frac{{{k}}}{{{2}}}}\right)}{w}-{t}{0}\)  

\(\displaystyle{t}-{t}{0}={\left({\frac{{{k}}}{{{2}}}}\right)}{w}\)  

Multiply both sides by \(\frac{2}{k}\)  

\(\displaystyle{\left({t}-{t}{0}\right)}{\frac{{{2}}}{{{k}}}}={\left({\frac{{{k}}}{{{2}}}}\right)}{w}\cdot{\left({\frac{{{2}}}{{{k}}}}\right)}\)  

\(\displaystyle{\frac{{{2}{\left({t}-{t}{0}\right)}}}{{{k}}}}={w}\)  

\(\displaystyle{w}={2}\frac{{{t}-{t}{0}}}{{k}}\)
f = ma(t) = kte^(-at) => a(t) = k/m * te^(-at)

v(t) = ∫a(t) dt = k/(ma^2) * (-ate^(-at) - e^(-at) + c)
= -k/(ma^2) * e^(-at)*(at + 1 + ce^(at)) [readmore]typed answer2no + o2 → 2no2rate=knomo2n2nd order in no → m = 23rd order overall → m + n = 3so 2 + n = 3n = 1rate=k[no]2[o2] in the given expressionwe know that(a + b)2 = a2 + 2ab + b2(a-b)2 = a2-2ab + b2hence4[a2 + 2ab + b2] – 9[a2-2ab + b2]4a2 + 8ab + 4b2 - 9a2 + 18ab - 9b226ab – 5a2 - 5b225ab + ab – 5a2 – 5b2[25ab – 5a2] + [ab – 5b2]5a[5b – a] – b[5b – a][5a – b][5b – a]hence the factors 4 (a + b)2 – 9 (a – b)2 are [5a – b] and [5b – a] $$s\_{m+n}=\dfrac{(s\_m-s\_n)(m+n)}{m-n}$$

$$s\_{m}-s\_{n} = a\_1m+\dfrac{m(m-1)d}{2}-a\_1n-\dfrac{n(n-1)d}{2}$$
$$=a\_1(m-n)+\dfrac{(m^2-n^2-m+n)d}{2}$$
$$\dfrac{s\_{m}-s\_{n}}{m-n} = a\_1+\dfrac{(m+n-1)d}{2} $$
$$\dfrac{(s\_{m}-s\_{n})(m+n)}{m-n}$$
$$ =a\_1(m+n)+\dfrac{(m+n)(m+n-1)d}{2} $$
$$= s\_{m+n}$$
therefore,
$$s\_{m+n}=\dfrac{(s\_m-s\_n)(m+n)}{m-n}$$
if $$s\_{m}=s\_{n}$$, $$s\_{m+n} = 0$$ hint: mathematics includes the study of topics which are related to quantity, structure, space and change. to solve this problem, we first try to simplify the expression. for doing so we let ${{x}^{2}}-4x=t$ and solve by using this consideration. in this way we can easily factorise the expression.complete step-by-step answer:according to a question, we are given the expression $\left( {{x}^{2}}-4x \right)\left( {{x}^{2}}-4x-1 \right)-20$. to simplify this complex expression, we let ${{x}^{2}}-4x=t$.this can be mathematically expressed as: $t(t-1)-20$.since, these types of factorization are familiar to us, so we try to factorise the equation in terms of t:${{t}^{2}}-t-20$now, by using the middle term splitting method of separating the variables, we get$\begin{align}  & {{t}^{2}}+4t-5t-20=0 \\  & t(t+4)-5(t+4)=0 \\  & (t-5)(t+4)=0 \\ \end{align}$ so, the factors of the expression in terms of t are (t-5) and (t+4).now, finally replacing t as ${{x}^{2}}-4x$,we get$\left( {{x}^{2}}-4x-5 \right)\left( {{x}^{2}}-4x+4 \right)=0$ on further simplifying this expression, by using middle term splitting method again in part one:$\begin{align}  & {{x}^{2}}-4x-5={{x}^{2}}+x-5x-5 \\  & \rightarrow x(x+1)-5(x+1)=0 \\  & (x-5)(x+1)=0 \\ \end{align}$ now, for the second part, using the identity of ${{(a-b)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}$, we get${{x}^{2}}-4x+4={{(x-2)}^{2}}$$\therefore (x-5)(x+1){{(x-2)}^{2}}=0$ hence, the factors of the required equation as per problem statement are: $(x-5)(x+1){{(x-2)}^{2}}=0$ note: this question can be alternatively solved by expanding the powers of x in the given expression and then letting ${{x}^{2}}$ as another variable. but the complexity of the problem would increase by doing so. hence, for simplification we must assume some other variable which can be finally replaced in the original expression to obtain the answer. tw - st < sw < st + tw

4 < sw < 34 [readmore]typed answerh2 + i2 ⇌ 2hi      kc = 4.6i            0.5   0.5     0.5c          -x      -x       +2xe        0.5-x 0.5-x   0.5+2xkc=hi2hii2kc=0.5+2x20.5-x0.5-xkc=0.5+2x20.5-x24.6=0.5+2x0.5-xx=0.138hi=0.5+2x=0.776 m on sapling its

oxidation:
cu --> cu^2+ + 2e^-

reduction:
o\_2 + 4e^- -->2o^2- to find: limitsnote: first check the form of imit. used this method if the limit is satisfying any one from 7 indeterminate form.in this case, indeterminate form is formula used: = 1so, by using the above formula, we have = = therefore, = equating the expression with 0,3x2 + 5x + 2 = 0on factorising it further,3x2 + 3x + 2x + 2 = 03x(x + 1) + 2(x + 1) = 0(3x + 2) (x + 1) = 0 x = - or x = -1 the zeroes of 3x2 + 5x + 2 are – and -1.