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1 year ago

Given the following 2 equations, state whether these equations are parallel, perpendicular, or neither.

Mathematics

1 answer:

stiks02 [169]1 year ago

3 0

Answer:

Neither

Explanation:

The slope-intercept form of the equation of a line is generally given as;

$y=mx+b$

where m = the slope of the line

b = the y-intercept of the line

Two lines are said to be parallel if they have the same slope. Looking at the two given equations below, we see that the slope of the 1st one, 5/7, is different from the slope of the 2nd one, 7/5;

$\begin{gathered} y=\frac{5}{7}x+14 \\ y=\frac{7}{5}x-20 \end{gathered}$

Two lines are said to be perpendicular if their slopes are negative reciprocals of each other, let's go ahead and compare their slopes;

$\begin{gathered} m_1=\frac{5}{7} \\ m_2=\frac{7}{5} \\ \therefore m_2=\frac{1}{m_1_{}} \end{gathered}$

So the two lines are not perpendicular to each other. It, therefore, means that the correct answer will be Neither.

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Answer:

The distance in one sweep is 100 cm approximately

Step-by-step explanation:

Here, we want to calculate the distance the wiper covers in one sweep

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Answer:

See below

Step-by-step explanation:

3. What are two ways that a vector can be represented?

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$\vec{v} = \langle a,b\rangle$

This is the component form. I don't like that way. It is probably used in high school, but

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For this question, I think you meant

vectors

$\vec{u_1} = (-8, 12)$

$\vec{u_2} = (13, 15)$

Once

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Considering that the dot product is

$\vec{u_1}\cdot \vec{u_2} = (-8)\cdot 13 + 12\cdot 15 = -104+180= 76$

and the norm of $\vec{u_1}$ is $||\vec{u_1}|| = \sqrt{(-8)^2 + 12^2} = \sqrt{64 + 144}= \sqrt{208}$

and the norm of $\vec{u_2}$ is $||\vec{u_2}|| = \sqrt{13^2 + 15^2} = \sqrt{169 + 225}= \sqrt{394}$

Thus,

$\cos(\theta)=\dfrac{76}{\sqrt{208} \sqrt{394}} = \dfrac{19}{\sqrt{13}\sqrt{394}}=\dfrac{19}{\sqrt{5122}}$

$\therefore \theta = \arccos \left(\dfrac{19}{\sqrt{5122}} \right)$

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2 years ago

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Answer:

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