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solmaris [256]
3 years ago
10

I need some help with this calculus question

Mathematics
1 answer:
Over [174]3 years ago
3 0

The tangent to y at the point (0, 5) has a slope equal to the derivative evaluated at x=0.

y=x^4+5e^x\implies y'=4x^3+5e^x\implies y'(0)=5

This tangent passes through the point (0, 5), so its equation is

y-5=5(x-0)\implies y=5x+5

The normal line passes through the same point, but is perpendicular to the tangent line, so its slope is the negative reciprocal of the slope of the tangent. So the equation for the normal line is

y-5=-\dfrac15(x-0)\implies y=5-\dfrac x5

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Answer:

x=5

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Step-by-step explanation:

the equation is

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3 years ago
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Answer:

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Step-by-step explanation:

Since the system of equations are d=18t+34 and d=22t, you first rewrite d=18t+34 as an equation. 18t+34=22 and that equals to 8.5=t. Then you solve the next one d=22t and the answer you got from the other equation, you substitute d=22(8.5) and that equals to 187.

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82/9= 9  1/9

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