Postgraduate Mathematics II task 78: Find the distance between (2,4) and (5,8).
The coordinate differences are 3 and 4, so the distance is √(3²+4²)=5.
Practice china-kaoyan-math-2 questions with answers and explanations.
The coordinate differences are 3 and 4, so the distance is √(3²+4²)=5.
f′(x)=15x²+8x. Substituting x=1 gives 23.
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The sum is n(n+1)/2=15, so the mean is 15/5.
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There are 5 favourable outcomes out of 9 equally likely outcomes, so P(red)=5/9.
The coordinate differences are 3 and 4, so the distance is √(3²+4²)=5.
f′(x)=12x²+6x. Substituting x=4 gives 216.
An antiderivative is 4/2 x²+4x. At x=5, the value is 70.0.
The dot product is 4×3+5×2=22.
Rewrite the value using base 5: 3125 = 5^5. Matching exponents gives x = 5.
aₙ=a₁+(n−1)d=9+(12−1)×2=31.
f′(x)=9x²+4x. Substituting x=3 gives 93.
An antiderivative is 3/2 x²+3x. At x=4, the value is 36.0.