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suter [353]
3 years ago
12

A paperweight is shaped like a triangular pyramid. The base is an equilateral triangle. Find the surface area of the paperweight

. Thanks soooo much!

Mathematics
1 answer:
melomori [17]3 years ago
5 0

Answer:

<h2>8.3 in²</h2>

Step-by-step explanation:

We have

three triangles with the base <em>b = 2 in.</em> and the height <em>h₁ = 2.2 in.</em>

one triangle with the base <em>b = 2 in. </em>and the height <em>h₂ = 1.7 in.</em>

The formula of an area of a triangle:

A=\dfrac{b\cdot h}{2}

<em>b</em><em> - base</em>

<em>h</em><em> - height</em>

Substitute:

A_1=\dfrac{(2)(2.2)}{2}=\dfrac{4.4}{2}=2.2\ in^2

A_2=\dfrac{(2)(1.7)}{2}=1.7\ in^2

S.A.=3A_1+A_2

S.A.=3(2.2)+1.7=6.6+1.7=8.3\ in^2

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3.<br> a. Domain<br> b. Range<br> c. Function?
77julia77 [94]

Answer:

hiii

Step-by-step explanation:

5 0
3 years ago
The batteries from a certain manufacturer have a mean lifetime of 850 hours, with a standard deviation of 70 hours, assuming tha
matrenka [14]

To answer (a), we will need to find the Z value for 710 hours and 990 hours.

\begin{gathered} For\text{ 710} \\ Z\text{ = }\frac{x-\mu}{\sigma} \\ \text{    = }\frac{710-850}{70} \\ \text{    = -2} \\ p\text{ =0.0228 } \\ For\text{ 990:} \\ Z\text{ = }\frac{990-850}{70} \\ \text{ =2} \\ p\text{ = 0.9772} \end{gathered}\begin{gathered} To\text{ find P \lparen–2 \le Z \le 2\rparen} \\ =\text{ 0.9772 -0.0228} \\ =\text{ 0.9544} \\ =95.44\% \end{gathered}

B) 68% of data lies between one standard deviation of the mean.

850 +70 = 920

850 - 70 = 780

5 0
1 year ago
Answer to this question?
Svetradugi [14.3K]
$88.50 you just add btw.just letting u know
6 0
3 years ago
Read 2 more answers
Compare 7⁄8 with 14⁄16 using ( &lt;, &gt;, =). A. none of the above B. 7⁄8 &gt; 14⁄16 C. 7⁄8 = 14⁄16 D. 7⁄8 &lt; 14⁄16
Kryger [21]
7/8 = 0.875
14/16 = 0.875

Therefore, 7/8 = 14/16
7 0
3 years ago
Cyruer.
Maslowich

Answer:

Hello,

Step-by-step explanation:

Area of the sphere:

S=4*\pi *r^2

Volume of the sphere:

V=\dfrac{4*\pi *r^3}{3}

\dfrac{S}{3} =\dfrac{V}{7} \\\\\dfrac{4*\pi *r^2}{3} =\dfrac{4*\pi *r^3}{3*7} \\\\1=\dfrac{r}{7} \\\\\\r=7\\\\great \ circle: 2*\pi* r=14*\pi\\

Volume=\dfrac{4*\pi *7^3}{3} =\dfrac{1372*\pi }{3} \approx{1436,7550...(cm^3)}

4 0
2 years ago
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