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andriy [413]
3 years ago
14

The surface area of the prism is ______ square units. All measurements in the image below are in units. (Input whole number only

.)

Mathematics
2 answers:
pogonyaev3 years ago
7 0
It is probably 82 because you multiply 8 by 2 which is 16 and multiply 16 by 2. You then do the same thing do the 6 and 6.5. You then add all the numbers together and you get 82. Please review your steps.
Leni [432]3 years ago
6 0
The surface are of the shape is 251.5 units (sorry if that is wrong)

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Find the total surface area of the following cone. Leave your answer in terms of Pi. 4cm 3cm
castortr0y [4]

The triangle formed from the height of the cone and the radius of its base is a special 3,4,5 triangle. The slant height is 5cm.

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2 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 \\&#10;\begin{array}{lllll}&#10;&x_1&y_1&x_2&y_2\\&#10;%  (a,b)&#10;R&({{ -1}}\quad ,&{{ 6}})\quad &#10;%  (c,d)&#10;S&({{ 5}}\quad ,&{{ 5}})&#10;\end{array}\qquad&#10;%   coordinates of midpoint &#10;\left(\cfrac{{{ x_2}} + {{ x_1}}}{2}\quad ,\quad \cfrac{{{ y_2}} + {{ y_1}}}{2} \right)&#10;\\\\\\&#10;\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}&#10;&x_1&y_1&x_2&y_2\\&#10;%   (a,b)&#10;&({{ -1}}\quad ,&{{ 6}})\quad &#10;%   (c,d)&#10;&({{ 5}}\quad ,&{{ 5}})&#10;\end{array}&#10;\\\\\\&#10;% slope  = m&#10;slope = {{ m}}= \cfrac{rise}{run} \implies &#10;\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}

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\bf \textit{perpendicular, negative-reciprocal slope for slope}\quad -\cfrac{1}{6}\\\\&#10;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}&#10;&x_1&y_1\\&#10;%   (a,b)&#10;&({{ 2}}\quad ,&{{ \frac{11}{2}}})&#10;\end{array}&#10;\\\\\\&#10;% slope  = m&#10;slope = {{ m}}= \cfrac{rise}{run} \implies 6&#10;\\\\\\&#10;% point-slope intercept&#10;\stackrel{\textit{point-slope form}}{y-{{ y_1}}={{ m}}(x-{{ x_1}})}\implies y-\cfrac{11}{2}=6(x-2)&#10;\\\\\\&#10;y-\cfrac{11}{2}=6x-12\implies -6x+y=-12+\cfrac{11}{2}\implies \stackrel{\textit{standard form}}{-6x+y=-\cfrac{13}{2}}
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Step-by-step explanation:

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