What is Pi equal to (First 20 digits)

Please help me with this geometry question. It’s over similar triangles all you have to do is sent up the other half of the prob

Brilliant_brown [7]

Set up a proportion. y=6 should be your answer!

7 0

3 years ago

Indicate the equation of the given line in standard form. The line that is the perpendicular bisector of the segment whose endpo

noname [10]

Well the line that bisects RS, will cut RS in two equal halves, therefore, that line will cut RS perpendicularly at the midpoint of RS.

now, what the dickens is the midpoint of RS anyway?

$\bf \textit{middle point of 2 points }\\ \quad \\
\begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
% (a,b)
R&({{ -1}}\quad ,&{{ 6}})\quad 
% (c,d)
S&({{ 5}}\quad ,&{{ 5}})
\end{array}\qquad
% coordinates of midpoint 
\left(\cfrac{{{ x_2}} + {{ x_1}}}{2}\quad ,\quad \cfrac{{{ y_2}} + {{ y_1}}}{2} \right)
\\\\\\
\left( \cfrac{5-1}{2}~~,~~\cfrac{5+6}{2} \right)\implies \stackrel{midpoint}{\left(2~~,~~\frac{11}{2} \right)}$

so, we know that perpendicular line, will have to go through (2, 11/2)

now, a perpendicular line to RS, will have a negative reciprocal slope to it. Well, what is the slope of RS anyway?

$\bf \begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
% (a,b)
&({{ -1}}\quad ,&{{ 6}})\quad 
% (c,d)
&({{ 5}}\quad ,&{{ 5}})
\end{array}
\\\\\\
% slope = m
slope = {{ m}}= \cfrac{rise}{run} \implies 
\cfrac{{{ y_2}}-{{ y_1}}}{{{ x_2}}-{{ x_1}}}\implies \cfrac{5-6}{5-(-1)}\implies \cfrac{5-6}{5+1}\implies -\cfrac{1}{6}$

and let's check the reciprocal negative of that,

$\bf \textit{perpendicular, negative-reciprocal slope for slope}\quad -\cfrac{1}{6}\\\\
slope=-\cfrac{1}{{{ 6}}}\qquad negative\implies +\cfrac{1}{{{ 6}}}\qquad reciprocal\implies + \cfrac{{{ 6}}}{1}\implies 6$

so, then, what's is the equation of a line whose slope is 6, and goes through 2, 11/2?

$\bf \begin{array}{lllll}
&x_1&y_1\\
% (a,b)
&({{ 2}}\quad ,&{{ \frac{11}{2}}})
\end{array}
\\\\\\
% slope = m
slope = {{ m}}= \cfrac{rise}{run} \implies 6
\\\\\\
% point-slope intercept
\stackrel{\textit{point-slope form}}{y-{{ y_1}}={{ m}}(x-{{ x_1}})}\implies y-\cfrac{11}{2}=6(x-2)
\\\\\\
y-\cfrac{11}{2}=6x-12\implies -6x+y=-12+\cfrac{11}{2}\implies \stackrel{\textit{standard form}}{-6x+y=-\cfrac{13}{2}}$

6 0

3 years ago

Show your work, type the answer 6y + 5(6 + 4у + Зу)

Andreas93 [3]

6y + 5*(6 + 7y)
6y + 30 + 35y
41y + 30

6 0

3 years ago

What is the area of this polygon?

ra1l [238]

Step 1) Draw a line from point F to point S. A rectangle forms (rectangle FSCW)

Step 2) Find the area of this rectangle. The area is 18 square units because it is 2 units high and 9 units across (9*2 = 18). You can count out the spaces or you can note how we go from x = -4 to x = 5 so subtracting the values gives -4-5 = -9 which has an absolute value of 9.

Step 3) Find the area of triangle FSN. The base is 9 units and the height is 6 units (count out the spaces or subtract y values). So the area is A = b*h/2 = 9*6/2 = 54/2 = 27

Step 4) Add up the area of the rectangle to the area of the triangle: 18+27 = 45

Final Answer: 45 square units

note: another way to find the answer is to find the area of rectangle WABC where point A is at (-4,4) and point B is at (5,4). Then subtract off the triangular areas of AFN and BNS

6 0

3 years ago

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Plz can any1 answer this question asap

vfiekz [6]

Answer:

the answer is

17+19+11=47

55-47=8

3 0

3 years ago

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