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3 years ago

Find the equation of the line whose x-intercept is 8 and y-intercept is -2. write the equation in slope-intercept form.

2 answers:

7 0

Find the slope, m = (change in y) / (change in x)

0-(-2)
Here, m = ------------- = 1/4
8 - 0
y-k = m(x-h) becomes y+2 = (1/4)(x), or y = (1/4)x - 2

Then the desired equation is y = (1/4)x - 2 (slope-intercept form)

Gre4nikov [31]3 years ago

3 0

Answer:

The equation of the line is y = 1/4x - 2

Step-by-step explanation:

In order to find this, we must first find the slope. We can do that by using the slope formula. For this, we will use the points given of (8, 0) and (0, -2)

m(slope) = (y2 - y1)/(x2 - x1)

m = (0 - -2)/(8 - 0)

m = 2/8

m = 1/4

Now that we have the slope, we can use that and either point in point-slope form to find the equation.

y - y1 = m(x - x1)

y - -2 = 1/4(x - 0)

y + 2 = 1/4x

y = 1/4x - 2

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3 years ago

A paintball court charges an initial entrance fee plus a fixed price per ball.

m_a_m_a [10]

Answer : The cost of 10 balls will be, 13.5 dollars

Step-by-step explanation :

As we are given the following expression:

$P=0.80\times n+5.50$

where,

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n = number of balls used

Given:

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Now we have to determine the cost of 10 balls.

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3 years ago

HELP! Find the value of sin 0 if tan 0 = 4; 180 < 0< 270

BabaBlast [244]

Hi there! Use the following identities below to help with your problem.

$\large \boxed{sin \theta = tan \theta cos \theta} \\ \large \boxed{tan^{2} \theta + 1 = {sec}^{2} \theta}$

What we know is our tangent value. We are going to use the tan²θ+1 = sec²θ to find the value of cosθ. Substitute tanθ = 4 in the second identity.

$\large{ {4}^{2} + 1 = {sec}^{2} \theta } \\ \large{16 + 1 = {sec}^{2} \theta } \\ \large{ {sec}^{2} \theta = 17}$

As we know, sec²θ = 1/cos²θ.

$\large \boxed{sec \theta = \frac{1}{cos \theta} } \\ \large \boxed{ {sec}^{2} \theta = \frac{1}{ {cos}^{2} \theta} }$

And thus,

$\large{ {cos}^{2} \theta = \frac{1}{17}} \\ \large{cos \theta = \frac{ \sqrt{1} }{ \sqrt{17} } } \\ \large{cos \theta = \frac{1}{ \sqrt{17} } \longrightarrow \frac{ \sqrt{17} }{17} }$