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3 years ago

TRUE OR FALSE: a rectangular equation that is not a function may be a function in the polar coordinate system.

Mathematics

2 answers:

Sindrei [870]3 years ago

5 0

It is false. The website i am on says that the statement uptop is false

Crank3 years ago

4 0

The statement above is true. Polar equations indeed can describe graphs as functions, even if when the equations in the rectangular coordinate system are not one of the functions. Polar equations can be graphed accurately using hands by using the Polar Coordinate System.

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Find m (2x+14)<br> (X+7)

LekaFEV [45]

Answer:

Step-by-step explanation:

Ⓗⓘ ⓣⓗⓔⓡⓔ

Well, assuming m of angle ABD is 90 degrees, then you would add and then solve.

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3 years ago

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shusha [124]

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Answer:

C. (-4, -3)

Step-by-step explanation:

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3 years ago

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RideAnS [48]

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3 years ago

Read 2 more answers

For what x value(s) does f(x)=2? HELP ANSWER FAST!!!!!!

GuDViN [60]

Answer:

x = 2

Step-by-step explanation:

Given

See attachment

Required

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The vertical axis represents f(x) while the horizontal represents x.

From the attachment:

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The circled point on attachment 2 represents the required point

<em>Assume that each box is 1 unit.</em>

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3 years ago

I need help please help me thanks please

Vitek1552 [10]

so we have a table of values, with x,y coordinates, so let's use any two of those points to get the slope of the table and use the point-slope form to get its equation

$~\hspace{2.7em}\stackrel{\textit{let's use}}{\downarrow }\qquad \stackrel{\textit{and this}}{\downarrow }\\\begin{array}{|lr|r|r|r|r|}\cline{1-6}x&0&1&2&3&4\\y&-1&3&7&11&15\\\cline{1-6}\end{array}\\\\[-0.35em]\rule{34em}{0.25pt}\\\\(\stackrel{x_1}{1}~,~\stackrel{y_1}{3})\qquad(\stackrel{x_2}{4}~,~\stackrel{y_2}{15})$

$\stackrel{slope}{m}\implies\cfrac{\stackrel{rise}{\stackrel{y_2}{15}-\stackrel{y1}{3}}}{\underset{run}{\underset{x_2}{4}-\underset{x_1}{1}}}\implies \cfrac{12}{3}\implies 4\\\\\\% point-slope intercept\begin{array}{|c|ll}\cline{1-1}\textit{point-slope form}\\\cline{1-1}\\y-y_1=m(x-x_1)\\\\\cline{1-1}\end{array}\implies y-\stackrel{y_1}{3}=\stackrel{m}{4}(x-\stackrel{x_1}{1})\\\\\\y-3=4x-4\implies y = 4x-1\implies \blacktriangleright y = 4x+(-1)\blacktriangleleft$

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3 years ago