Find the equation of the line tangent to the function at the given point f(x)=5x^2 at x=10
1 answer:
f(x) = 5x²
f(10) = 5(10)² = 5(100) = 500
f'(x) = 10x
f'(10) = 10(10) = 100
Now, find the line that passes through (10, 500) and has a slope of 100
y - y₁ = m(x - x₁)
y - 500 = 100(x - 10)
y - 500 = 100x - 1000
y = 100x - 500
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40. Divide 81 by 41. 41 goes into 81 once with a remainder of 40
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Step-by-step explanation:
Step-by-step explanation:
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