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2 years ago

The bases of a right prism are isosceles trapezoids with bases of 10 ft and 18 ft

Mathematics

1 answer:

likoan [24]2 years ago

4 0

Answer:

540 ft^2.

Step-by-step explanation:

The area of the trapezoid = h/2 (10 + 18) = 14h.

By Pythagoras the height h = √(5^2 - 4^2) = 3.

So the area of the 2 trapezoidal bases = 2 * 14*3

= 84 ft^2.

Now we calculate the area of the four lateral rectangular sides:

= 10*12 + 18*12 + 2*5*12

= 456 ft^2.

Total area = 456 + 54

= 540 ft^2.

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a slide on the playground is 12 and 1/2 FEET long..... its 3 feet and 7 inches longer then the small slide.... how long is the s

11111nata11111 [884]

You simply have to add the two numbers together, however, you have to convert everything into inches only. You know that there are 12 inches in one foot. To calculate the length of the longer slide, it is simply doing 12 feet TIMES 12 (since there are 12 inches in one foot). This makes 144 inches. Add 1/2 feet in there, this is adding 1/2 of 12 inches, which is 12/2=6 inches. The long slide is therefore 144+6= 150 inches. Now you want to convert 3 feet and 7 inches into inches only. 3 feet TIMES 12 = 3x12=36 inches. Add 7 inches to this, and you will get 36+7=43 inches. To know how long the small slide is, subtract 43 inches from 150 inches, and this will give you 150-43=107 inches. So, the small slide is 107 inches. :)

3 0

3 years ago

1. The figure shows the regular triangular pyramid SABC. The base of the pyramid has an edge AB = 6 cm and the side wall has an

Musya8 [376]

Given:

• AB = 6 cm

• SM = √15 cm

Let's solve for the following:

• 1) the base elevation AM.

Given that we have a regular triangular pyramid, the length of the three bases are equal.

AB = BC = AC

BM = BC/2 = 6/2 = 3 cm

To solve for AM, which is the height of the base, apply Pythagorean Theorem:

$\begin{gathered} AM=\sqrt{AB^2-BM^2} \\ \\ AM=\sqrt{6^2-3^2} \\ \\ AM=\sqrt{36-9} \\ \\ AM=\sqrt{27} \\ \\ AM=5.2\text{ cm} \end{gathered}$

The base elevation of the pyramid is 5.2 cm.

• (2)., The elevation SO.

To find the elevation of the pyramid, apply Pythagorean Theorem:

$SO=\sqrt{SM^2-MO^2}$

Where:

SM = √15 cm

MO = AM/2 = 5.2/2 = 2.6 cm

Thus, we have:

$\begin{gathered} SO=\sqrt{(\sqrt{15})^2-2.6^2} \\ \\ SO=\sqrt{15-6.76} \\ \\ SO=2.9\text{ cm} \end{gathered}$

Length of SO = 2.9 cm

• (3). Area of the base:

To find the area of the triangular base, apply the formula:

$A=\frac{1}{2}*BC*AM$

Thus, we have:

$\begin{gathered} A=\frac{1}{2}*6^*5.2 \\ \\ A=15.6\text{ cm}^2 \end{gathered}$

The area of the base is 15.6 square cm.

• (4). Area of the side surface.

Apply the formula:

$SA=\frac{1}{2}*p*h$

Where:

p is the perimeter

h is the slant height, SM = √15 cm

Thus, we have:

$\begin{gathered} A=\frac{1}{2}*(6*3)*\sqrt{15} \\ \\ A=34.86\text{ cm}^2 \end{gathered}$

• (5). Total surface area:

To find the total surface area, apply the formula:

$TSA=base\text{ area + area of side surface}$

Where:

Area of base = 15.6 cm²

Area of side surface = 34.86 cm²

TSA = 15.6 + 34.86 = 50.46 cm²

The total surface area is 50.46 cm²

• (6). Volume:

To find the volume, apply the formula:

$V=\frac{1}{3}*area\text{ of base *height}$

Where:

Area of base = 15.6 cm²

Height, SO = 2.9 cm

Thus, we have:

$\begin{gathered} V=\frac{1}{3}*15.6*2.9 \\ \\ V=15.08\text{ cm}^3 \end{gathered}$

The volume is 15.08 cm³.

ANSWER:

• 1.) 5.2 cm

• 2.) 2.9 cm

• 3.) 15.6 cm²

• 4.) 34.86 cm²

• (5). 50.46 cm²

• 6). 15.08 cm³.

7 0

8 months ago

Use the equation machine to help you solve the

ale4655 [162]

Answer: k=8

Step-by-step explanation:

6 0

2 years ago

Read 2 more answers

Two gubernatorial aspirants amp in two different states in Nigeria have probabilities of 2/9 and 4/11 respectively of winning an

allsm [11]

Probability of First = 2/9

Probability of Second = 4/11

Probability of both winning = 2/9 + 4/11

= 22/99 + 36/99

= 58/99

Answer is 58/99 probability of them both winning

Must click thanks and mark brainliest

8 0

2 years ago

The following table shows the percent increase of donations made on behalf of a non-profit organization for the period of 1984 t

pashok25 [27]

Giving the table below which shows the percent increase of donations made on behalf of a non-profit organization for the period of 1984 to 2003.

Year: 1984 1989 1993 1997 2001 2003

Percent: 7.8 16.3 26.2 38.9 49.2 62.1

The scatter plot of the data is attached with the x-axis representing the number of years after 1980 and the y-axis representing the percent increase of donations made on behalf of a non-profit organization.

To find the equation for the line of regression where the x-axis representing the number of years after 1980 and the y-axis representing the percent increase of donations made on behalf of a non-profit organization.
$\begin{center}
\begin{tabular}{ c| c| c| c| }
 x & y & x^2 & xy \\ [1ex] 
 4 & 7.8 & 16 & 31.2 \\ 
 9 & 16.3 & 81 & 146.7 \\ 
13 & 26.2 & 169 & 340.6 \\ 
17 & 38.9 & 289 & 661.3 \\ 
21 & 49.2 & 441 & 1,033.2 \\ 
23 & 62.1 & 529 & 1,428.3 \\ [1ex]
\Sigma x=87 & \Sigma y=200.5 & \Sigma x^2=1,525 & \Sigma xy=3,641.3 
\end{tabular}
\end{center}$

Recall that the equation of the regression line is given by
$y=a+bx$
where
$a= \frac{(\Sigma y)(\Sigma x^2)-(\Sigma x)(\Sigma xy)}{n(\Sigma x^2)-(\Sigma x)^2} = \frac{200.5(1,525)-87(3,641.3)}{6(1,525)-(87)^2} \\ \\ = \frac{305,762.5-316793.1}{9,150-7,569} = \frac{-11,030.6}{1,581} =-6.977$
and
$b= \frac{n(\Sigma xy)-(\Sigma x)(\Sigma y)}{n(\Sigma x^2)-(\Sigma x)^2} = \frac{6(3,641.3)-(87)(200.5)}{6(1,525)-(87)^2} \\ \\ = \frac{21,847.8-17,443.5}{9,150-7,569} = \frac{4,404.3}{1,581} =2.7858$

Thus, the equation of the regresson line is given by
$y=-6.977+2.7858x$

The graph of the regression line is attached.

Using the equation, we can predict the percent donated in the year 2015. Recall that 2015 is 35 years after 1980. Thus x = 35.

The percent donated in the year 2015 is given by
$-6.977+2.7858(35)=-6.977+97.503=90.526$

Therefore, the percent donated in the year 2015 is predicted to be 90.5

7 0

3 years ago