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sertanlavr [38]
3 years ago
7

Can somebody please help me with this equation

Mathematics
1 answer:
Simora [160]3 years ago
5 0
The answer is c your welcome
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Which logarithmic graph can be used to approximate the value of y in the equation 3^y = 4?
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Step-by-step explanation:

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Given m ZLMN = 145°, what is m ZXMN?<br> Show all relevant work.
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80 degrees. 145 = (4x+5) + (6x-10) = 10x-5... so 10x = 150, x = 15. LMN = 6x-10 = 6(15) - 10 = 80.
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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}\\\\ &#10;(x- h)^2+(y- k)^2= r^2&#10;\qquad &#10;center~~(\stackrel{}{ h},\stackrel{}{ k})\qquad \qquad &#10;radius=\stackrel{}{ r}\\\\&#10;-------------------------------\\\\&#10;(x+1)^2+y^2=36\implies [x-(\stackrel{h}{-1})]^2+[y-\stackrel{k}{0}]^2=\stackrel{r}{6^2}~~~~&#10;\begin{cases}&#10;\stackrel{center}{(-1,0)}\\&#10;\stackrel{radius}{6}&#10;\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}&#10;\\\\&#10;(\stackrel{x_1}{-1}~,~\stackrel{y_1}{0})\qquad &#10;A(\stackrel{x_2}{-1}~,~\stackrel{y_2}{1})\qquad \qquad &#10;%  distance value&#10;d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2}&#10;\\\\\\&#10;\stackrel{distance}{d}=\sqrt{[-1-(-1)]^2+(1-0)^2}\implies d=\sqrt{(-1+1)^2+1^2}&#10;\\\\\\&#10;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}&#10;\\\\&#10;(\stackrel{x_1}{-1}~,~\stackrel{y_1}{0})\qquad &#10;B(\stackrel{x_2}{-1}~,~\stackrel{y_2}{6})\\\\\\&#10;\stackrel{distance}{d}=\sqrt{[-1-(-1)]^2+(6-0)^2}\implies d=\sqrt{(-1+1)^2+6^2}&#10;\\\\\\&#10;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}&#10;\\\\&#10;(\stackrel{x_1}{-1}~,~\stackrel{y_1}{0})\qquad &#10;C(\stackrel{x_2}{4}~,~\stackrel{y_2}{-8})&#10;\\\\\\&#10;\stackrel{distance}{d}=\sqrt{[4-(-1)]^2+[-8-0]^2}\implies d=\sqrt{(4+1)^2+(-8)^2}&#10;\\\\\\&#10;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
3 years ago
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Write the equation in slope-intercept form. y + 5 =- 6(x + 7)​
Dmitry [639]
The answer is : y=-6-47
:)
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3 years ago
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