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RSB [31]
3 years ago
8

I need help with surface area for a pyrimid please

Mathematics
4 answers:
IrinaVladis [17]3 years ago
6 0
Okay do find the area of a triangle and then add it to the area of the rectangles
exis [7]3 years ago
6 0
SA = Base Area • 2 + base perimeter • distance

SA = 1/2(12•8)+(12+10+10)•10
1/2(96)+(32•10)
48+320
SA= 368
Svetlanka [38]3 years ago
5 0
SA = Base Area • 2 + base perimeter • distance

SA = 1/2(12•8)+(12+10+10)•10
1/2(96)+(32•10)
48+320
SA= 368
kifflom [539]3 years ago
5 0
SA = Base Area • 2 + base perimeter • distance

SA = 1/2(12•8)+(12+10+10)•10
1/2(96)+(32•10)
48+320
SA= 368
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That's correct. Now you gotta plug in -5 in for x and get both angles.
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4 years ago
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NEED HELP FAST PLEASE
Bess [88]

The answer is 39 and I know because I took the test and got it right

3 0
3 years ago
Please help solve for x
Ludmilka [50]

Answer:

8.49

Step-by-step explanation:

there is a little formula related to the famous formula of Pythagoras.

it says that the height of a triangle is the square root of the product of both segments of the baseline (the segments the height splits the baseline into).

so, x is actuality the height of the triangle.

x = sqrt(3×24) = sqrt(72) = 8.49

7 0
3 years ago
What is the equation for a circle with a center at (-2,-4) that passes through the point (3,8)?
Margarita [4]
Check the picture below.

so, the center of the circle is the midpoint of that diametrical segment, and half that length is the radius.

\bf ~~~~~~~~~~~~\textit{middle point of 2 points }
\\\\
\begin{array}{ccccccccc}
&&x_1&&y_1&&x_2&&y_2\\
%  (a,b)
&&(~ -2 &,& -4~) 
%  (c,d)
&&(~ 3 &,& 8~)
\end{array}\qquad
%   coordinates of midpoint 
\left(\cfrac{ x_2 +  x_1}{2}\quad ,\quad \cfrac{ y_2 +  y_1}{2} \right)
\\\\\\
\left( \cfrac{3-2}{2}~~,~~\cfrac{8-4}{2} \right)\implies \left( \frac{1}{2}~,~2 \right)\impliedby center

\bf ~~~~~~~~~~~~\textit{distance between 2 points}
\\\\
\begin{array}{ccccccccc}
&&x_1&&y_1&&x_2&&y_2\\
%  (a,b)
&&(~ -2 &,& -4~) 
%  (c,d)
&&(~ 3 &,& 8~)
\end{array}~~~ 
%  distance value
d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2}
\\\\\\
d=\sqrt{[3-(-2)]^2+[8-(-4)]^2}\implies d=\sqrt{(3+2)^2+(8+4)^2}
\\\\\\
d=\sqrt{25+144}\implies d=\sqrt{169}\implies d=13\qquad\qquad \qquad  \stackrel{radius}{\frac{13}{2}}

\bf \textit{equation of a circle}\\\\ 
(x- h)^2+(y- k)^2= r^2
\qquad 
center~~(\stackrel{\frac{1}{2}}{ h},\stackrel{2}{ k})\qquad \qquad 
radius=\stackrel{\frac{13}{2}}{ r}
\\\\\\
\left( x-\frac{1}{2} \right)^2+(y-2)^2=\left( \frac{13}{2} \right)^2\implies \left( x-\frac{1}{2} \right)^2+(y-2)^2=\frac{169}{4}

3 0
3 years ago
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I really need some help so pls someone help me
Gemiola [76]

Answer:

its b ez

Step-by-step explanation:

5 0
3 years ago
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