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drek231 [11]
3 years ago
9

Which of the following polygons cannot be used to form a regular tessellation?A. Equilateral triangleB. Regular octagonC. Square

D. Regular hexagon
Mathematics
1 answer:
Nataly_w [17]3 years ago
5 0
A- Equilateral triangle.
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When 605,000 is written in scientific notation, what will be the value of the exponent?
son4ous [18]
6.05 x 10^5

I think this is the answer
3 0
3 years ago
Given: MPNA pyramid All edges congruent AO = 36, AO ⊥ (MNP) Find: m∠AMO, AΔANM
Iteru [2.4K]

Answer:

m∠AMO ≈ 54.7°

AΔANM = 486√3

Step-by-step explanation:

The edges are congruent, so all four faces are congruent equilateral triangles.  We'll say the length of each edge is 2r.

The height of the pyramid h is given to be 36.

The perpendicular distance from O to line MP is called the apothem (a).  Using 30-60-90 triangles, b = 2a and r = a√3.

Use cosine to find m∠AMO.

cos(∠AMO) = b / (2r)

cos(∠AMO) = (2a) / (2a√3)

cos(∠AMO) = 1 / √3

m∠AMO ≈ 54.7°

Use Pythagorean theorem to find the apothem.

(2r)² = b² + h²

(2a√3)² = (2a)² + 36²

12a² = 4a² + 1296

8a² = 1296

a² = 162

a = 9√2

So the edge length is:

2r = 2√3 (9√2)

2r = 18√6

The area of the equilateral triangle ΔANM is half the apothem times the perimeter:

A = ½aP

A = ½ (9√2) (3 × 18√6)

A = 243√12

A = 486√3

6 0
3 years ago
On a recent day, 8 euros were worth $9 and 24 euros were worth $27. Write an equation of the form y=kx to show the between the n
g100num [7]
$9= k•8 euros so k=9/8=1.125
So for every 1 euro we have $1.125
y=1.125•x

$27=k•24 euros so k=28/24=1.166666667
y=1.67•x

5 0
3 years ago
Check all solutions to the equation. If there are no solutions, check "None."<br> x=-81
Elodia [21]

Answer:

none

Step-by-step explanation:

If a squared variable is equal to a negative number, all of the solutions are imaginary

There are no real solutions

8 0
3 years ago
Read 2 more answers
a line passes through the point -2 and five and has a slope of 2/3 points A(x,3) and B(-2,y)lie on the line
olasank [31]
A and B lie on the line, yes, but what specifically are you supposed to do?  Looks like your problem statement was cut off before you'd finished typing it in.

You say your line passes thru (-2,5) and has a slope of 2/3?  Then, using the point-slope formula,

y-5 = (2/3)(x+2)  This is the general equation for your line.

Now let's play around with B(-2,y).  Suppose we subst. the x-coordinate of B, which is -2, into the equation   y-5 = (2/3)(x+2); we get y-5 = (2/3)(-2+2) = 0.  This tells us that y must be 5.  But we already knew that!!

So, please review the original problems with its instructions and this discussion and tell me what you need to know from this point on.



7 0
3 years ago
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