The line graphed by the equation, y=5/6x +1/2, has a slope of 1/2 and a y-intercept of 5/6 true or false?

Mathematics

1 answer:

vazorg [7]2 years ago

8 0

False. It has a slope of 5/6 and y intercept of 1/2.

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Solve the quadratic equation by factoring. 9x^2+24x+20=4

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D. -4/3

Let's solve your equation step-by-step.

9x2+24x+20=4

Step 1: Subtract 4 from both sides.

9x2+24x+20−4=4−4

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Step 2: Factor left side of equation.

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Step 3: Set factors equal to 0.

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Given the circle with the equation (x + 1)2 + y2 = 36, determine the location of each point with respect to the graph of the cir

kati45 [8]

$\bf \textit{equation of a circle}\\\\ 
(x- h)^2+(y- k)^2= r^2
\qquad 
center~~(\stackrel{}{ h},\stackrel{}{ k})\qquad \qquad 
radius=\stackrel{}{ r}\\\\
-------------------------------\\\\
(x+1)^2+y^2=36\implies [x-(\stackrel{h}{-1})]^2+[y-\stackrel{k}{0}]^2=\stackrel{r}{6^2}~~~~
\begin{cases}
\stackrel{center}{(-1,0)}\\
\stackrel{radius}{6}
\end{cases}$

so, that's the equation of the circle, and that's its center, any point "ON" the circle, namely on its circumference, will have a distance to the center of 6 units, since that's the radius.

$\bf ~~~~~~~~~~~~\textit{distance between 2 points}
\\\\
(\stackrel{x_1}{-1}~,~\stackrel{y_1}{0})\qquad 
A(\stackrel{x_2}{-1}~,~\stackrel{y_2}{1})\qquad \qquad 
% distance value
d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2}
\\\\\\
\stackrel{distance}{d}=\sqrt{[-1-(-1)]^2+(1-0)^2}\implies d=\sqrt{(-1+1)^2+1^2}
\\\\\\
d=\sqrt{0+1}\implies d=1$

well, the distance from the center to A is 1, namely is "inside the circle".

$\bf ~~~~~~~~~~~~\textit{distance between 2 points}
\\\\
(\stackrel{x_1}{-1}~,~\stackrel{y_1}{0})\qquad 
B(\stackrel{x_2}{-1}~,~\stackrel{y_2}{6})\\\\\\
\stackrel{distance}{d}=\sqrt{[-1-(-1)]^2+(6-0)^2}\implies d=\sqrt{(-1+1)^2+6^2}
\\\\\\
d=\sqrt{0+36}\implies d=6$

notice, the distance to B is exactly 6, and you know what that means.

$\bf ~~~~~~~~~~~~\textit{distance between 2 points}
\\\\
(\stackrel{x_1}{-1}~,~\stackrel{y_1}{0})\qquad 
C(\stackrel{x_2}{4}~,~\stackrel{y_2}{-8})
\\\\\\
\stackrel{distance}{d}=\sqrt{[4-(-1)]^2+[-8-0]^2}\implies d=\sqrt{(4+1)^2+(-8)^2}
\\\\\\
d=\sqrt{25+64}\implies d=\sqrt{89}\implies d\approx 9.43398$

notice, C is farther than the radius 6, meaning is outside the circle, hiking about on the plane.

3 0

2 years ago

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