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Rewrite the expression in the form 9n9^n. \newline97/95=9^7 / 9^5 =

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Marco's class is painting a mural on the side of their school. The mural covers a 312 m 3 \frac{1}{2} \mathrm{~m} high, rectangular wall. It has an area of 28 m2 28 \mathrm{~m}^{2} .\newlineWhat is the length of the mural?\newlinem
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A cup of hot coffee has been left to cool in a room with an ambient temperature of 21∘C 21^{\circ} \mathrm{C} .\newlineThe relationship between the elapsed time, m m , in minutes, since the coffee was left to cool, and the temperature of the coffee, T T , measured in ∘C { }^{\circ} \mathrm{C} , is modeled by the following function.\newlineT(m)=21+74⋅10−0.03m T(m)=21+74 \cdot 10^{-0.03 m} \newlineWhat will the temperature of the coffee be after 1010 minutes?\newlineRound your answer, if necessary, to the nearest hundredth.\newline∘C { }^{\circ} C
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Takumi plants a tree in his backyard and studies how the number of branches grows over time.\newlineHe predicts that the relationship between N N , the number of branches on the tree, and t t , the elapsed time, in years, since the tree was planted can be modeled by the following equation.\newlineN=5⋅100.3t N=5 \cdot 10^{0.3 t} \newlineAccording to Takumi's model, in how many years will the tree have 100100 branches?\newlineGive an exact answer expressed as a base−10-10 logarithm.\newlineyears
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Let g(x)=2x g(x)=2^{x} .\newlineCan we use the mean value theorem to say the equation g′(x)=16 g^{\prime}(x)=16 has a solution where 3<x<5 3<x<5 ?\newlineChoose 11 answer:\newline(A) No, since the function is not differentiable on that interval.\newline(B) No, since the average rate of change of g g over the interval 3≤x≤5 3 \leq x \leq 5 isn't equal to 16‾ 1 \overline{6} .\newline(C) Yes, both conditions for using the mean value theorem have been met.
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ddx(3x2−1x−2)= \frac{d}{d x}\left(\frac{3 x^{2}-1}{x-2}\right)=
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dydt=2t+3 and y(1)=6 \frac{d y}{d t}=2 t+3 \text { and } y(1)=6 \newlineWhat is t t when y=0 y=0 ?\newlineChoose all answers that apply:\newline(A) t=−1 t=-1 \newline(B) t=−3 t=-3 \newline(C) t=0 t=0 \newline(D) t=−4 t=-4 \newline(E) t=1 t=1 \newline(F) t=−2 t=-2
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y=x13 y=x^{13} \newlinedydx= \frac{d y}{d x}=
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ddx[x6]= \frac{d}{d x}\left[x^{6}\right]=
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