Slope=4/3 y-intercept=-5

Mathematics

2 answers:

Flura [38]3 years ago

4 0

Answer:

The equation of the line in point-slope form is:

$y+5=\:\frac{4}{3}\left(x-0\right)$

The graph is also attached below.

Step-by-step explanation:

Given

Slope m = 4/3

y-intercept = -5

We know that the value of the y-intercept can be determined by setting x = 0 and determining the corresponding value of y.

As we are given the y-intercept = -5

Thus, the point (0, -5) represents the y-intercept.

Point slope form:

$y-y_1=m\left(x-x_1\right)$

where

m is the slope of the line

(x₁, y₁) is the point

In our case,

m = 4/3

(x₁, y₁) = (0, -5)

substituting m = 4/3 and the point (0, -5) in the point-slope form

$y-y_1=m\left(x-x_1\right)$

$y\:-\:\left(-5\right)\:=\:\frac{4}{3}\:\left(x\:-\:0\right)$

$y+5=\:\frac{4}{3}\left(x-0\right)$

Therefore, the equation of the line in point-slope form is:

$y+5=\:\frac{4}{3}\left(x-0\right)$

The graph is also attached below.

MrRissso [65]3 years ago

3 0

Answer: 4/3 - 5

I assume you want to know the numbers in slope intercept form so... yeah

Step-by-step explanation

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xxTIMURxx [149]

If point $\left(x,\ \frac{\sqrt{7}}{3}\right)$ is on the unit ircle, then:

$x^2+\left( \frac{ \sqrt{7} }{3} \right)^2=1 \\ \\ \Rightarrow x^2+ \frac{7}{9} =1 \\ \\ \Rightarrow x^2=1- \frac{7}{9} = \frac{2}{9} \\ \\ \Rightarrow x= \sqrt{ \frac{2}{9} } = \frac{ \sqrt{2} }{3}$

Since, the point is in the second quadrant, x is negative.

Thus, $(x,\ y)=\left(x,\ \frac{\sqrt{7}}{3}\right)=\left(-\frac{ \sqrt{2} }{3},\ \frac{\sqrt{7}}{3}\right)$

Part A:

$3 \sqrt{7} =6\left( \frac{ \sqrt{7} }{2} \right) \\ \\ =6\left( \frac{ \sqrt{7} }{3} \cdot \frac{3}{ \sqrt{2} } \right) \\ \\ =6\left( \frac{ \frac{ \sqrt{7} }{3} }{ \frac{ \sqrt{2} }{3} } \right)=-6\left( \frac{ \frac{ \sqrt{7} }{3} }{ -\frac{ \sqrt{2} }{3} } \right) \\ \\ =-6\left( \frac{y}{x} \right)=-6\tan\theta$

Therefore, $-6\tan\theta=3\sqrt{7}$ .

Part B:

$- \frac{ \sqrt{7} }{2} =- \frac{ \sqrt{7} }{3} \cdot \frac{3}{ \sqrt{2} } \\ \\ =- \frac{ \frac{ \sqrt{7} }{3} }{ \frac{ \sqrt{2} }{3} } = \frac{ \frac{ \sqrt{7} }{3} }{ -\frac{ \sqrt{2} }{3} } = \frac{y}{x} \\ \\ =\tan\theta$

Therefore, $\tan\theta=- \frac{ \sqrt{7} }{2}$