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kupik [55]
3 years ago
5

The equation of the line of best fit of a scatter plot is y = −5x − 2. What is the slope of the equation?

Mathematics
2 answers:
Brrunno [24]3 years ago
7 0

Answer:

5

Step-by-step explanation:

The response is 5. The slope is always the coefficient of x in a slope-intercept equation.

Rom4ik [11]3 years ago
5 0

Answer:

The correct answer for this equation is -5 I know this because when I went to my previous middle school my old teacher had me grade this exact question.

Step-by-step explanation:

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Find three consecutive integers such that three times the third is the same as 2 les than the sum of the first and second
harina [27]
Let the integers be x, x+1 and x+2.

Thus, 3(x + 2) = (x + x + 1) - 2

Then 3x + 6 = 2x + 1 - 2, or       3x + 6 = 2x - 1

Solving for x, x = - 7.

First number is -7
Second one is -6
Third one is -5
5 0
3 years ago
Use the grouping method to factor the polynomial below completely.<br> x - 3x² + 5x - 157
Levart [38]

Answer:

-3x^2+6x-157

Step-by-step explanation:

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4 years ago
There are___puppies that weigh less than 102 g
Nataly [62]
Need more info I can help just do not know what the question is
5 0
4 years ago
Steven recorded the growth of a plant over 10 weeks for his science project. He made a graph that shows how much the plant grew
Natasha2012 [34]

Answer:

  • D. y = 6/5x + 2

Step-by-step explanation:

  • Initial height is 2 in
  • Height after 5 weeks is 8 in

<u>Rate of change is</u>

  • (8 - 2)/5 = 6/5

<u>So the equation is:</u>

  • y = 6/5x + 2

Correct option is D

6 0
3 years ago
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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