Sh(2x) = (e^2x + e^-2x)/2
<span>Thus the integral becomes </span>
<span>Int[e^3x*(e^2x + e^-2x)/2] = Int[(e^5x + e^x)/2] </span>
<span>= e^5x/10 + e^x/2 + C
</span>=(1/10)(e^5x) + (1/2)(e^x) + C
Answer:
The correct options are: Interquartile ranges are not significantly impacted by outliers. Lower and upper quartiles are needed to find the interquartile range. The data values should be listed in order before trying to find the interquartile range. The option Subtract the lowest and highest values to find the interquartile range is incorrect because the difference between lowest and highest values will give us range. The option A small interquartile range means the data is spread far away from the median is incorrect because a small interquartile means data is nor spread far away from the median
We know that the law of sines states that:
For simplicity, let:
In triangle A1BC this leads to:
Therefore:
Now, triangle A2BC is isosceles which means that both the base angles are equal, since angle CA1B and A2A1B are supplementary we have:
Answer:
Step-by-step explanation:
distributive
Answer:
the answer is C
Step-by-step explanation:
BABYSHARK