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pickupchik [31]
3 years ago
12

Pick the geometric term that matches the real-world object: a deck of cards.

Mathematics
2 answers:
malfutka [58]3 years ago
6 0

Answer: Rectangular Prism

==================================

Explanation:

A rectangle is a four sided object that has all four angles at 90 degrees (aka right angles). Imagine that the cards are designed as such they do not have their corners rounded off, but each corner is a 90 degree angle. Stacking 52 congruent rectangles leads to a 3D solid known as a rectangular prism. The base is a rectangle that is congruent to any of the 52 cards. This stack must be aligned and not allowed to have any of the cards offset (ie forming a sort of "leaning tower"). You can think of a "rectangular prism" as a "box" which is probably a more common term for it; however, "rectangular prism" is probably what your teacher will want since it likely involves geometric terms you've learned in your lessons.

Ray Of Light [21]3 years ago
6 0

Answer:

Rectangular Prism :)

Step-by-step explanation:

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Is 5 over 12 greater than 3 over 4
Leno4ka [110]


no, 3/4 and 5/12 can have a common denominator of 12, multiply the top and bottom by 3... 5/12 and 9/12


Hope this helped!!! :)


5 0
3 years ago
Solve 5x +3>19 what is it
tatyana61 [14]

Answer:

x > 3 1/5

Step-by-step explanation:

5x +3>19

Subtract 3 from each side

5x +3-3>19-3

5x >16

Divide each side by 5

5x/5 > 16/5

x > 16/5

x > 3 1/5

7 0
4 years ago
HELP PLEASE
Eva8 [605]

Answer:

Step-by-step explanation:

C 40x^6

8 0
3 years ago
Solve the equation.<br><br> 2(3z-2)+8=34<br><br> A. z=3<br> B. z=-3<br> C. z=-4<br> D. z=5
levacccp [35]
<span>2(3z-2)+8=34
6z - 4 + 8 = 34
6z + 4 = 34
6z = 34 -4
6z = 30
z =30/6
z = 5

answer </span><span>D. z=5</span>
6 0
3 years ago
Read 2 more answers
Sin(60-theta)sin(60+theta)​
ehidna [41]

Step-by-step explanation:

\sin(60 -  \theta)  \sin(60 +  \theta)  \\  =  \{ \sin(60)  \cos( \theta)  -  \sin( \theta)   \cos(60)  \} \{ \sin(60)  \cos( \theta)  +  \cos(60)  \sin( \theta)  \} \\  =  \{ \frac{ \sqrt{3} }{2}  \cos( \theta)  -  \frac{1}{2}  \sin( \theta)  \} \{ \frac{ \sqrt{3} }{2}  \cos( \theta)  +  \frac{1}{2}  \sin( \theta)  \} \\  \\

from difference of two squares:

{ \boxed{(a - b)(a + b) = ( {a}^{2} -  {b}^{2} ) }}

therefore:

=  \{ {( \frac{ \sqrt{3} }{2}) }^{2}  { \cos }^{2}  \theta \} -  \{ {( \frac{ \sqrt{3} }{2} )}^{2}  { \sin }^{2}  \theta \} \\  \\  =  \frac{3}{4}  { \cos }^{2}  \theta -  \frac{3}{4}  { \sin}^{2}  \theta

factorise out ¾ :

=  \frac{3}{4} ( { \cos }^{2}  \theta  -  { \sin}^{2}   \theta) \\  \\  = { \boxed{ \frac{3}{4}  \cos(2 \theta) }}

3 0
3 years ago
Read 2 more answers
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