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Hoochie [10]
4 years ago
7

What is the equation of the graphed line in point-slope form?

Mathematics
2 answers:
katrin2010 [14]4 years ago
7 0

Answer:

the first option

Step-by-step explanation:

the actual equation is y=2x+3 fully simplified and y + 3=2(x + 3) simplifies to y=2x+3 take y + 3=2(x + 3) distribute the 2 to make y + 3=2x+6 then subtract 3 from both sides to make y=2x+3

daser333 [38]4 years ago
4 0

Answer:

Its the first one :)

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Read 2 more answers
Consider the following vector function. R(t) = 9 2 t, e9t, e−9t (a) find the unit tangent and unit normal vectors t(t) and n(t)
garik1379 [7]

The unit tangent vector is T(u) and the unit normal vector is N(t) if the  vector function. R(t) is equal to 9 2 t, e9t, e−9t.

<h3>What is vector?</h3>

It is defined as the quantity that has magnitude as well as direction also the vector always follows the sum triangle law.

We have vectored function:

\rm R(t) = (9\sqrt{2t}, e^{9t}, e^{-9t})

Find its derivative:

\rm R'(t) = (9\sqrt{2}, 9e^{9t}, -9e^{-9t})

Now its magnitude:

\rm |R'(t) |= \sqrt{(9\sqrt{2})^2+ (9e^{9t})^2+ (-9e^{-9t})^2}

After simplifying:

\rm R'(t) = 9 \dfrac{e^{18t}+1}{e^{9t}}

Now the unit tangent is:

\rm T(u) = \dfrac{R'(t)}{|R'(t)|}

After dividing and simplifying, we get:

\rm T(u) = \dfrac{1}{e^{18t}+1} (\sqrt{2}e^{9t}, e^{18t}, -1)

Now, finding the derivative of T(u), we get:

\rm T'(u) = \dfrac{1}{(e^{18t}+1)^2} (9\sqrt{2}e^{9t}(1-e^{18t}), 18e^{18t}, 18e^{18t})

Now finding its magnitude:

\rm |T'(u) |= \dfrac{1}{(e^{18t}+1)^2} (9\sqrt{2}e^{9t}(1-e^{18t})^2+ (18e^{18t})^2+( 18e^{18t})^2)

After simplifying, we get:

\rm |T'(u)|= \dfrac{9\sqrt{2}e^{9t}}{e^{18t}+1}

Now for the normal vector:

Divide T'(u) and |T'(u)|

We get:

\rm N(t) = \dfrac{1}{e^{18t}+1} ( 1-e^{18t},          \sqrt{2}e^{9t},  \sqrt{2}e^{9t})

Thus, the unit tangent vector is T(u) and the unit normal vector is N(t) if the  vector function. R(t) is equal to 9 2 t, e9t, e−9t.

Learn more about the vector here:

brainly.com/question/8607618

#SPJ4

3 0
2 years ago
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