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xeze [42]
3 years ago
7

A triangular prism. Rectangle A has a base of 14 meters and height of 8 meters. Rectangle B has a base of 14 meters and height o

f 6 meters. Rectangle C has a base of 14 meters and height of 10 meters. Triangles D and E have a base of 6 meters and height of 8 meters.
Which statements about the area of the faces of the triangular prism are true? Select all that apply.
Area of face A = 112 m2.
Area of face B = 60 m2.
Area of face C = 140 m2.
Area of face E = area of face D.
Area of face A = area of face C.
Mathematics
2 answers:
SCORPION-xisa [38]3 years ago
6 0

Answer:

a  c   d

Step-by-step explanation:

faltersainse [42]3 years ago
6 0

Answer:

ACD

Step-by-step explanation:

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4 years ago
Can someone plz help me
Illusion [34]
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V=15 cubic inches.

Second rectangular:
Doing the same like that to get:
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3 years ago
Find the slope of a line through (-15, -8) and (-17, -26)
ololo11 [35]

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65

Step-by-step explanation:

because this equals that and that proves it

8 0
3 years ago
1. Find the vertices and locate the foci for the hyperbola whose equation is given.
Irina18 [472]
\bf \cfrac{(x-{{ h}})^2}{{{ a}}^2}-\cfrac{(y-{{ k}})^2}{{{ b}}^2}=1
\qquad center\ ({{ h}},{{ k}})\qquad
 vertices\ ({{ h}}\pm a, {{ k}})\\\\
-----------------------------\\\\
\textit{now let's take a look at yours}
\\\\\\
49x2 - 16y2 = 784\implies \cfrac{49x^2}{784}-\cfrac{16y^2}{784}=1
\\\\\\
\cfrac{x^2}{16}-\cfrac{y^2}{49}=1\implies \cfrac{(x-0)^2}{4^2}-\cfrac{(y-0)^2}{7^2}=1
\\\\\\
recall\implies center\ ({{ h}},{{ k}})\qquad vertices\ ({{ h}}\pm a, {{ k}})
\\\\\\


\bf \textit{now, for the foci, the foci are "c" distance from the center point}\\\\\
whereas\qquad c=\sqrt{a^2+b^2}\qquad \textit{ that is }\qquad  h\pm \sqrt{a^2+b^2}

notice your "a" and "b" components, to get the distance "c" from the center to either foci and the vertices, of course, are h + a, k and h - a, k
5 0
3 years ago
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