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Virty [35]
3 years ago
14

FREEE POINTS ANSWER THIS Congruent figures have the same size and same shape.

Mathematics
2 answers:
kvasek [131]3 years ago
8 0

Yes, this means that they are congruent. A shape is congruent if the have the same size and shape! Hope this helps! Thanks for the free points :)

Shtirlitz [24]3 years ago
3 0

Answer:

true

Step-by-step explanation:

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Will mark brainlist for correct answer​
andrezito [222]

Answer:

I would say not here D.

Step-by-step explanation:

Because I found the area of the base first *the square* which is 4 x 11 = 44 and the rectangles are  4 x 14= 56 x 2 since there are 2 rectangles is 112. Then the triangles are 11x14=154/2=77 x 2 since there are 2 triangles is 154. Then we add up all those numbers 44+112+154= 310 not 308 *C*.

Hope I was able to help!

3 0
3 years ago
Help with this question, please! I'm so confused!! The answer with the red arrow is incorrect!
Andre45 [30]

Answer:

arc XY = 110°

Step-by-step explanation:

81 = 1/2(arc XY - arc WZ)

2 * 81 = arc XY - arc WZ

162 = arc XY - 52

arc XY = 162 - 52

arc XY = 110°

6 0
3 years ago
Brand A: $5 for 80 pounds of sand Brand B: $14 for 175 pound bag Which is the better buy?
nevsk [136]

Answer:

Brand A is the better buy

Step-by-step explanation:

bag Which is the better buy?

Step one

Brand A: $5 for 80 pounds of sand

Let us find the unit cost

Unit cost =5/80

=$0.0625 per pound

Brand B: $14 for 175 pound

The unit cost is

=14/175

=$0.08 per pound

4 0
3 years ago
Can you please help me find the surface area
arlik [135]

Answer:

392

Step-by-step explanation:

A = BH + 2LS + BL

H = 8

S= 7

L=16

B = 7

7(8) + 2(7)16 + 7(16) = 392

6 0
4 years ago
Find the area of the surface. the part of the cylinder y^2+z^2=9 that lies above the rectangle with vertices (0,0),(4,0),(0,2),
Mice21 [21]
We can parameterize this part of the cylinder using

\mathbf s(u,v)=\langleu,3\cos v,3\sin v\rangle

with 0\le u\le4 and \cos^{-1}\dfrac23\le v\le\dfrac\pi2. Then the surface element is

\mathrm d\mathbf S=\|\mathbf s_u\times\mathbf s_v\|\,\mathrm du\,\mathrm dv
\mathrm d\mathbf S=3\,\mathrm du\,\mathrm dv

so the area is given by the surface integral, where D denotes the part of the cylinder in question,

\displaystyle\iint_D\mathrm d\mathbf S=3\int_{v=\cos^{-1}(2/3)}^{v=\pi/2}\int_{u=0}^{u=4}\mathrm du\,\mathrm dv=6\pi

Probably less work would involve a simpler geometric approach, but it doesn't hurt to practice setting up a proper surface integral.
4 0
3 years ago
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