# We start with the equation

Please check my work and give me the correct answer. Thank you.

We start with the equation:

X1^2-6*X1+2=0

X1(1)=1-6+2=(-3)

X1(2)+9-6*(-3)+2=29

X1(3)=29^2-6*29+2=841-174+2=669

X1(4)=669^2-6*669+2=443549

X1=1/3+1/29+1/669+1/443549=

.3333333+.0344828+.0014947+.00000023=.369311

This is approximately equal to .36454.

Thus A my answer

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#### Solution Summary

The expert checks equations for the correct answers.

First-Order Differential Equations

Two Snowplows - Differential Equations (First-Order Differential Equations)

One day it began to snow exactly at noon at a heavy and steady rate. A snowplow left its garage at 1:00pm, and another one followed in its tracks at 2:00pm.

a)At what time did the second snowplow crash into the first? To answer this question, assume that the rate (in mph) at which a snowplow can clear the road is inversely proportional to the depth of the snow (and hence to the time elapsed since the road was clear of snow). [Hint: begin by writing differential equations for x(t) and y(t), the distance traveled by the first and second snowplows, respectively, at t hours past noon. To solve the differential equation involving y, let t rather than y be the dependent variable]

b)Could the crash have been avoided by dispatching the second snowplow at 3:00pm instead?

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