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3 years ago

Which of the following polygons can form a regular tessellation?

Mathematics

1 answer:

anyanavicka [17]3 years ago

3 0

Answer:

option A.

Step-by-step explanation:

The regular polygons that can be used to form a regular tessellation are an equilateral triangle, a square, and a regular hexagon.

HOPE THIS HELPS.

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#7 please help!!!!!!!!!!!!!!!!!!!!!!!!!

Ira Lisetskai [31]

If they all painted at the same rate the equation is this

mothers 8
daughters 12 at 220 sq ft

mothers 6
daughters 8 at 152 sq ft

220 +152 =372 divided by 34 (because they all painted at the same rate)=10.94 sq ft they each painted

4 0

3 years ago

K is the midpoint of PQ, P has

Pachacha [2.7K]

Answer:

Coordinates of Q $(x_2,y_2) \:are\: \mathbf{(7,16)}$

Option D is correct option.

Step-by-step explanation:

We are given:

K is the midpoint of PQ

Coordinates of P = (-9,-4)

Coordinates of K = (-1,6)

We need to find coordinates of Q $(x_2,y_2)$

We will use the formula of midpoint: $Midpoint=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$

We are given midpoint K and $x_1,y_1$ the coordinates of P we need to find $x_2,y_2$ the coordinates of Q.

$Midpoint=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})\\(-1,6)=(\frac{-9+x_2}{2},\frac{-4+y_2}{2})\\$

Now, we can write

$-1=\frac{-9+x_2}{2}, 6=\frac{-4+y_2}{2}\\Simplifying:\\-2=-9+x_2\:,\: 12=-4+y_2\\-2+9=x_2\:,\: 12+4=+y_2\\x_2=7\:,\:y_2=16$

So, we get coordinates of Q $(x_2,y_2) \:are\: \mathbf{(7,16)}$

Option D is correct option.

6 0

3 years ago

You are comparing to wireless services. Verizon changes $35 a month and $5 per gigabyte of data.

ikadub [295]

A) Gigabytes = x
B) 35+5x = 25+10x
10 = 5x
2 = x
C) After 2 gigabytes, the wireless services will cost the same ($45)

7 0

3 years ago

Kierston needs to paint a wall of the youth center. She knows that 1 can of paint covers an area of 2.2 square meters. Kierston

Step2247 [10]

Answer:

See Explanation

Step-by-step explanation:

Given

$1\ can = 2.2m^2$

Required

Determine the number of cans for the wall

The dimension of the wall is not given. So, I will use the following assumed values:

$Length=20m$

$Width = 44m$

First, calculate the area of the wall

$Area = Length * Width$

$Area = 20m * 44m$

$Area = 880m^2$

If $1\ can = 2.2m^2$

Then $x = 880m^2$

Cross Multiply:

$x * 2.2m^2 = 1 * 880m^2$

$x * 2.2m^2 = 880m^2$

$x * 2.2 = 880$

Make x the subject

$x = \frac{880}{2.2}$

$x = 400$

400 cans using the assume dimensions.

So, all you need to to is, get the original values and follow the same steps

4 0

3 years ago

Help Me Please! Will reward Branliest!

alexgriva [62]

Answer:

Hey the person from the guy or gurl who told me to join the zoom. I wasn't rlly gonna join. I think its thrid option

8 0

3 years ago