Mathematics, 05.05.2020 15:57 sterlingrobinson35

Shaun's tent (shown below) is a triangular prism. Find the surface area, including the floor, of his tent. \text{ m}^2 m 2 start text, space, m, end text, squared

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Answer from: sonnekaustin60

52.8 square m

Step-by-step explanation:

We are given that

Length of rectangular face of Shaun's tent=5 m

Breadth of rectangular face of Shaun's tent=3 m

Base of triangular face=1.5+1.5= 3m

Height of triangular face=2.6 m

Area of rectangle= $length\times breadth$

Using the formula

Area of 3 rectangular face= $3\times 5\times 3=45m^2$

Area of two triangular face = $2\times \frac{1}{2}\times base\times height$

Area of two triangular face= $3\times 2.6=7.8 m^2$

Area of tent=Area of 3 triangular faces+area of two triangular faces

Area of tent= 45+7.8=52.8m^2

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

Answer from: Quest

see below

step-by-step explanation:

7. there are 10 data points in the set. the middle point is between 5 and 6. the middle point in the top half is 8 and the middle point in the bottom half is 3. to find the iqr , take point 8 and subtract point 3

mid term point 8 = 100 point 3 = 88

iqr = 100-88 = 12

final point 8 = 93 point 3 = 78

iqr = 93 - 78 = 15

the final has the greatest iqr

6. b - there is a high data value that causes the data set to be asymmetrical for the males: the data for males are high and asymmetrical . there is an outlier for the males that skews the data

5. the iqr is a better measure of spread for the movies and basketball games. since the basketball game has an outlier, we do not want to use standard deviation for measurement.

3. the exam median is much higher than the class median. the median line is farther to the right on the exams

2. there is no way to tell what the means are from the box and whisker plots.

Answer from: Quest

x= -3 and $x=\frac{-3+/-3i\sqrt{3} }{2}$

step-by-step explanation:

to find the roots of a cubic, use division or factoring to find the factors then set equal to 0. as this is a difference of cubes, it factors into

$(x^3-a^3)=$ $(x-a)(x^2+ax+a^2)$

where a is the cube root of $a^3$ .

here this is 27 and the cube root of 27 is 3. so we write using the form:

$(x-3)(x^2+(3)x+3^2)\\(x-3)(x^2+3x+9)$ .

set each factor to 0 and solve for x.

x-3=0

x=3

the quadratic expression does not factor and must be solved using the quadratic formula.

$x=\frac{-b+/-\sqrt{b^2-4ac} }{2a}$

where a = 1, b=3, and c=9

$x=\frac{-3+/-\sqrt{3^2-4(1)(9)} }{2(1)} \\x=\frac{-3+/-\sqrt{9-36)} }{2} \\x=\frac{-3+/-\sqrt{-27)} }{2}\\x=\frac{-3+/-3i\sqrt{3} }{2}$

this cubic has two complex roots.

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