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3 years ago

What line is parallel to y=7x-4

Mathematics

2 answers:

PtichkaEL [24]3 years ago

4 0

Y=7x-z
replace z with any number and it will be parallel

laiz [17]3 years ago

3 0

Answer:

y=7x+2

Step-by-step explanation:

The equation that i gave you is parallel to the one you mentioned

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3 years ago

Reparametrize the curve with respect to arc length measured from the point where t = 0 in the direction of increasing t. (Enter

jolli1 [7]

Answer:

the answer is incomplete, below is the complete question

"Reparametrize the curve with respect to arc length measured from the point where t = 0 in the direction of increasing t. (Enter your answer in terms of s.) r(t) = 3ti + (1 - 4t)j + (1 + 2t)k r(t(s)) ="

answer

$r(t(s)) = \frac{3s}{\sqrt{29} } i + (1 -\frac{4s}{\sqrt{29} }t)j + (1 + \frac{2s}{\sqrt{29} })k$

Step-by-step explanation:

The step by step procedure is to first determine the differentiate the given vector function

r(t) = 3ti + (1 - 4t)j + (1 + 2t)k

$\frac{d(r(t) = 3ti + (1 - 4t)j + (1 + 2t)k)}{dt} \\r'(t)=3i-4j+2k\\$

since s(t) is the arc length for r(t), which is define as

$s(t)=\int\limits^t_0 {||r'(t)||} \, dt$

if we substitute the value of r'(t) we arrive at

$s(t)=\int\limits^t_0 {||r'(t)||} \, dt\\s(t)=\int\limits^t_0 {\sqrt{3^{2} +4^{2}+2^{2}} \, dt\\s(t)=\int\limits^t_ 0{\sqrt{29} } \, dx\\$

$s(t)=\int\limits^t_ 0{\sqrt{29} } \, dx\\\\s(t)=\sqrt{29} t\\hence \\t(s)=\frac{s}{\sqrt{29} }$

substituting the value of t in to the given vector equation we have

$r(t(s)) = \frac{3s}{\sqrt{29} } i + (1 -\frac{4s}{\sqrt{29} }t)j + (1 + \frac{2s}{\sqrt{29} })k$

4 0

4 years ago

Find the equation of the line that is parallel to the given line and passes through the given point.

lapo4ka [179]

Answer:

y=0.2+5.8

Step-by-step explanation:

Parallel lines have the same slope. slope-intercept form is y=mx+b

The slope of y=0.2x-5 is 0.2, and you need to plug in the new point to find the equation.

5=0.2(-4)+b

5=-0.8+b --> b=5.8

final equation: y=0.2+5.8

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Answer:

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