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rodikova [14]
3 years ago
6

Shaun’s tent (shown below) is a triangular prism. What is the the surface area, including the floor, of his tent?

Mathematics
2 answers:
Arturiano [62]3 years ago
4 0

Answer:

52.8

Step-by-step explanation:

blondinia [14]3 years ago
3 0

Answer:

The surface area of the ten t including the floor is 52.8 m².

Step-by-step explanation:

We are given that,

Shaun's tent consists of 3 rectangles and 2 triangles.

1. The dimensions of the rectangles are 5 × 3 meters.

<h3>Since, Area of the rectangle = Length × Width </h3>

i.e. Area of 3 rectangles = 3 × Length × Width

i.e. Area of 3 rectangles = 3 × 5 × 3 = 45 m²

<em>Thus, area of the 3 rectangles is 45 m².</em>

2. The dimensions of the triangles are base= 3 meter and height= 2.6 meter

<h3>As, Area of the triangle = \frac{1}{2}\times Base\times Height</h3>

i.e. Area of 2 triangles = 2\times \frac{1}{2}\times Base\times Height

i.e. Area of 2 triangles = Base\times Height

i.e. Area of 2 triangles = 3\times 2.6

i.e. Area of 2 triangles = 7.8 m²

Thus, area of 2 triangles is 7.8 m².

<em>So, the total surface area of the prism = Area of rectangles + Area of triangles</em>

i.e. Total surface area = 45 + 7.8 = 52.8 m²

Hence, the surface area of the ten t including the floor is 52.8 m².

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Tom rolls 2 fair dice and adds the results from each. Work out the probability of getting a total that is a factor of 21.
Ber [7]

The probability of getting a total that is a factor of 21 is 0.22

<u>Explanation:</u>

Factor of 21 is = 1, 3, 7, 21

Out of all the factors of 21, we cannot get 21 when the two die are rolled as the maximum sum that can be obtained is 12

The possible combinations are:

(1,2) (1,6)

(2,1) (2,5)

(3,4)

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Total outcome = 6 X 6

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Therefore, the probability of getting a total that is a factor of 21 is 0.22

5 0
3 years ago
5 Exam-style ABCD is a kite.
Mnenie [13.5K]

let's recall that in a Kite the diagonals meet each other at 90° angles, Check the picture below, so we're looking for the equation of a line that's perpendicular to BD and that passes through (-1 , 3).

keeping in mind that perpendicular lines have negative reciprocal slopes, let's check for the slope of BD

y = \stackrel{\stackrel{m}{\downarrow }}{3}x-1\impliedby \begin{array}{|c|ll} \cline{1-1} slope-intercept~form\\ \cline{1-1} \\ y=\underset{y-intercept}{\stackrel{slope\qquad }{\stackrel{\downarrow }{m}x+\underset{\uparrow }{b}}} \\\\ \cline{1-1} \end{array}

\stackrel{\textit{perpendicular lines have \underline{negative reciprocal} slopes}} {\stackrel{slope}{3\implies \cfrac{3}{1}} ~\hfill \stackrel{reciprocal}{\cfrac{1}{3}} ~\hfill \stackrel{negative~reciprocal}{-\cfrac{1}{3}}}

so we're really looking for the equation of a line whose slope is -1/3 and passes through point A

(\stackrel{x_1}{-1}~,~\stackrel{y_1}{3})\qquad \qquad \stackrel{slope}{m}\implies -\cfrac{1}{3} \\\\\\ \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}{-\cfrac{1}{3}}[x-\stackrel{x_1}{(-1)}]\implies y-3=-\cfrac{1}{3}(x+1) \\\\\\ y-3=-\cfrac{1}{3}x-\cfrac{1}{3}\implies y=-\cfrac{1}{3}x-\cfrac{1}{3}+3\implies y=-\cfrac{1}{3}x+\cfrac{8}{3}

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