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Mathematics

1 answer:

Step2247 [10]1 year ago

3 0

Answer:

linear is when the graph shows a straight line

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A . What is the slope in the line of the graph ?

vampirchik [111]

Answer:

0.2x+3

Step-by-step explanation:

5 0

2 years ago

For what values of θ on the polar curve r=θ, with 0≤θ≤2π , are the tangent lines horizontal? Vertical?

Bond [772]

Given that $r=\theta$ , then $r'=1$

The slope of a tangent line in the polar coordinate is given by:

$m= \frac{r'\sin\theta+r\cos\theta}{r'\cos\theta-r\sin\theta}$

Thus, we have:

$m= \frac{\sin\theta+\theta\cos\theta}{\cos\theta-\theta\sin\theta}$

Part A:

For horizontal tangent lines, m = 0.

Thus, we have:

$\sin\theta+\theta\cos\theta=0 \\ \\ \theta\cos\theta=-\sin\theta \\ \\ \theta=- \frac{\sin\theta}{\cos\theta} =-\tan\theta$

Therefore, the <span>values of θ on the polar curve r = θ, with 0 ≤ θ ≤ 2π, such that the tangent lines are horizontal are:

</span><span>θ = 0

</span>θ = <span>2.02875783811043
</span>
θ = <span>4.91318043943488

Part B:

For vertical tangent lines, $\frac{1}{m} =0$

Thus, we have:

$\cos\theta-\theta\sin\theta=0 \\ \\ \Rightarrow\theta\sin\theta=\cos\theta \\ \\ \Rightarrow\theta= \frac{\cos\theta}{\sin\theta} =\sec\theta$

</span>Therefore, the <span>values of θ on the polar curve r = θ, with 0 ≤ θ ≤ 2π, such that the tangent lines are vertical are:

</span>θ = <span>4.91718592528713</span>

3 0

2 years ago

Point A is located at negative 5 over 8 and point B is located at negative 2 over 8. What is the distance between points A and B

ki77a [65]

<h2>Answer:</h2>

Option: A is the correct answer.

A. negative 5 over 8 minus negative 2 over 8 = negative 3 over 8; therefore, the distance from A to B is absolute value of negative 3 over 8 equals 3 over 8 units.

<h2>Step-by-step explanation:</h2>

The location of Point A is : A(-5/8)

and Point B is located at : B(-2/8)

We know that when two points lie over a number line then the distance between the two points is the absolute difference between the two points.

If A is located at a.

and B is located at b.

Then the distance between A and B is given by:

|a-b|

Hence, here the distance between A and B is given by:

$|\dfrac{-5}{8}-(\dfrac{-2}{8})|=|\dfrac{-5}{8}+\dfrac{2}{8}|=|\dfrac{-3}{8}|=\dfrac{3}{8}\\\\i.e.\\\\\text{Distance between A and B is}=\dfrac{3}{8}\ \text{units}$