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postnew [5]
3 years ago
13

Please help me, I'm suffering from all of these parallel lines ;9

Mathematics
1 answer:
Anarel [89]3 years ago
6 0

Answer:

1. y = 4x - 27

2. y = -4x - 15

Step-by-step explanation:

If two lines are parallel, then they have the same slope. So, the slope of the line we are looking for needs to be 4. We can start by writing a point-slope equation:

y - y1 = m(x - x1)

We can substitute the values we have, the point we are using is (8, 5) because it needs to be on the line:

y - 5 = 4(x - 8)

We can distribute:

y - 5 = 4x - 32

y = 4x - 27

We are not given the slope-intercept form, so we must divide both sides by two to get it:

y = 1/4 x + 8

A perpendicular line has the slope that is the negative reciprocal of the one that is given. So, the slope of the line would be - 4. We can start by writing a point-slope equation:

y - y1 = m(x - x1)

We can substitute the values we have, the point we are using is (-5, 5) because it needs to be on the line:

y - 5 = -4(x + 5)

We can distribute:

y - 5 = -4x - 20

y = -4x - 15

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Step-by-step explanation:

<u>Trigonometric Identities</u>

\cos(A \pm B)=\cos A \cos B \mp \sin A \sin B

<u>Trigonometric ratios</u>

\sf \sin(\theta)=\dfrac{O}{H}\quad\cos(\theta)=\dfrac{A}{H}\quad\tan(\theta)=\dfrac{O}{A}

where:

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Using the trig ratio formulas for cosine and sine:

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Therefore, using the trig identities and ratios:

\begin{aligned}\implies \cos(2 \cdot \angle ABC) & = \cos(\angle ABC + \angle ABC)\\\\& = \cos (\angle ABC) \cos (\angle ABC) - \sin(\angle ABC) \sin (\angle ABC)\\\\& = \cos^2(\angle ABC)-\sin^2(\angle ABC)\\\\& = \left(\dfrac{3}{5}\right)^2-\left(\dfrac{4}{5}\right)^2\\\\& = \dfrac{3^2}{5^2}-\dfrac{4^2}{5^2}\\\\& = \dfrac{9}{25}-\dfrac{16}{25}\\\\& = \dfrac{9-16}{25}\\\\& = -\dfrac{7}{25} \end{aligned}

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