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3 years ago

Which polygon will not tessellate a plane?

Mathematics

2 answers:

netineya [11]3 years ago

7 0

Answer:

Regular Octagon

Step-by-step explanation:

There are only three regular shapes that tessellate – the square, the equilateral triangle, and the regular hexagon. All other regular shapes, like the regular pentagon and regular octagon, do not tessellate on their own.

BabaBlast [244]3 years ago

3 0

I think regular octagon

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Simplify and solve for the unknown. Use order of operations as needed.<br>8×2+5^2-y=2(y+1)+6

Ket [755]

Answer:

y = 11

Step-by-step explanation:

8 × 2 + 5²- y = 2(y + 1) + 6

8 × 2 + 5²- y = 2y + 2 + 6

8 × 2 + 25 - y = 2y + 2 + 6

16 + 25 - y = 2y + 2 + 6

41 - y = 2y + 8

41 (- y + y) = (2y + y) + 8

41 = 3y + 8

41 - 8 = 3y (+ 8 - 8)

33 = 3y

33/3 = 3y/3

y = 11

3 0

2 years ago

The translation of ABCD to A'B'C'D'<br> is given by (×+[?].y-[ ]).

s2008m [1.1K]

Answer:

(x + 1 , y - 3)

Step-by-step explanation:

Compare x-coordinate of A(-6,1) and A'(-5,-2)

-6 + a = -5

a = -5 +6 = 1

x- coordinate of A' = x coordinate of A + 1

Compare y-coordinate of A(-6,1) and A'(-5,-2)

1 + a =-2

a = -2 - 1 = -3

y- coordinate of A' = y coordinate of A - 3

A(-6 , 1) = A'(-6+1 , 1-3) = A'(-5, -2)

D(( -1 , 1) = D'(-1+1, 1 - 3) = D'(0,-2)

C(-2,3) =C'(-2+1, 3-3) = C'(-1,0)

B(-4,3)= B'(-4+1 , 3-3) = B'(-3,0)

7 0

3 years ago

Can someone help me? <br> Please

Ivahew [28]

Answer:

on the same dot but on the left sid

Step-by-step explanation:

3 0

3 years ago

I’m a bit confused in learning about congruent triangles

Tems11 [23]

Okay and... what are you confused about

8 0

3 years ago

Solve this system. Write your answer as an ordered pair. -5x + y = -3 3x - 8y = 24

Ivanshal [37]

Hello there, this is how we will solve your problem.\

here is the equation you gave us.

${\huge{\boxed{\mathbb{EQUATION}}}$

$\begin{bmatrix}-5x+y=-3\\ 3x-8y=24\end{bmatrix}$

we can isolate $x$ and say $x=\frac{-3-y}{5}$

Now here is our equation, $\begin{bmatrix}3\left(-\frac{-3-y}{5}\right)-8y=24\end{bmatrix}$ After we simplify it we get $\begin{bmatrix}\frac{9-37y}{5}=24\end{bmatrix}$ , solve this and we get that $y=-3$ , Before we said $x$ equals $-\frac{-3-y}{5}$ , So now we can plug in our value that $y$ equaled, and now we have $\frac{-3-(-3)}{5}$ . Two negatives equal a positive and we get that $x=0$ .

_________________

$y=-3$

$x=0$

7 0

2 years ago