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3 years ago

What is the smallest angle of rotational symmetry for a square

Mathematics

2 answers:

mars1129 [50]3 years ago

7 0

When we rotate a figure and there is no change in the shape of the figure then it has rotational symmetry.

We know that the order of rotation for a square is 4.

Hence, we have $\frac{360}{4} =90^{\circ}$

Thus, the angle of rotational symmetry of square are

$90^{\circ}, 180^{\circ}, 270^{\circ}$

Hence, the minimum angle of rotational symmetry is $90^{\circ}$

Therefore, the minimum angle of rotational symmetry for a square is 90 degrees.

Basile [38]3 years ago

4 0

Is it 90 degrees because all square have a 90 degree angle?

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Can someone give me the answers and step by step instructions please??

professor190 [17]

Answer:

$-1,4,-7,10,...$ neither

$192,24,3,\frac{3}{8},...$ geometric progression

$-25,-18,-11,-4,...$ arithmetic progression

Step-by-step explanation:

Given:

sequences: $-1,4,-7,10,...$

$192,24,3,\frac{3}{8},...$

$-25,-18,-11,-4,...$

To find: which of the given sequence forms arithmetic progression, geometric progression or neither of them

Solution:

A sequence forms an arithmetic progression if difference between terms remain same.

A sequence forms a geometric progression if ratio of the consecutive terms is same.

For $-1,4,-7,10,...$ :

$4-(-1)=5\\-7-4=-11\\10-(-7)=17\\So,\,\,4-(-1)\neq -7-4\neq 10-(-7)$

Hence,the given sequence does not form an arithmetic progression.

$\frac{4}{-1}=-4\\\frac{-7}{4}=\frac{-7}{4}\\\frac{10}{-7}=\frac{-10}{7}\\So,\,\,\frac{4}{-1}\neq \frac{-7}{4}\neq \frac{10}{-7}$

Hence,the given sequence does not form a geometric progression.

So, $-1,4,-7,10,...$ is neither an arithmetic progression nor a geometric progression.

For $192,24,3,\frac{3}{8},...$ :

$\frac{24}{192}=\frac{1}{8}\\\frac{3}{24}=\frac{1}{8}\\\frac{\frac{3}{8}}{3}=\frac{1}{8}\\So,\,\,\frac{24}{192}=\frac{3}{24}=\frac{\frac{3}{8}}{3}$

As ratio of the consecutive terms is same, the sequence forms a geometric progression.

For $-25,-18,-11,-4,...$ :

$-18-(-25)=-18+25=7\\-11-(-18)=-11+18=7\\-4-(-11)=-4+11=7\\So,\,\,-18-(-25)=-11-(-18)=-4-(-11)$

As the difference between the consecutive terms is the same, the sequence forms an arithmetic progression.

3 0

3 years ago

If DE=6x what is the perimeter of the triangle in terms of x? <br><br> A 6<br> B 18 <br> C 18x

tatyana61 [14]

May be another Lala La La La La La La La La La La La La La La La La La La la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la la

6 0

3 years ago

ANSWER BOTH QUESTIONS FOR BRAINLIEST TY :)

Alchen [17]

Answer:

<h2>Angle X; 39</h2><h2>Angle Y; 129</h2>

Step-by-step explanation:

180-90

This gives us the measurement of 51+x.

90-51

This is x as well as the angle that is across from it.

x+90=y

39+90=129

y=129.

to find the last angle, we add 90 to 51. This gives us 141.

So, now we add up all of the angles to double check everything. If they all add up to 360 then we are correct.

39+39+51+90+141=

360.

This means that we have solved everything correctly and that these are the correct answers.

I know that this is really confusing but the answers that you need are at the top so hopefully this helped!

4 0

3 years ago

Read 2 more answers

Among 9 electrical components exactly one is known not to function properly. if 4 components are randomly selected, find the pro

seraphim [82]

Answer:

0.4444

Step-by-step explanation:

Use the following property to ease the calculation:

P(At least one)=1-P(None)

Total number of electrical components: 9

Number that does not function well :1

Number that functions well : 8

We have $^8C_4=70$ ways to to choose 4 good components from 8.

We have $^9C_4=126$ ways to choose 4 components from a total of 9.

If all function properly then none is bad, we $^1C_0=1$ way to do this.

P(At least one)= $1-\frac{^8C_4*^1C_0}{^9C_4}$

P(At least one)= $1-\frac{70*1}{126}$

P(At least one)=0.4444

4 0

3 years ago

Read 2 more answers

Use mathematical induction to prove the statement is true for all positive integers n, or show why it is false:

kondaur [170]

$\text{Proof by induction:}$
$\text{Test that the statement holds or n = 1}$

$LHS = (3 - 2)^{2} = 1$
$RHS = \frac{6 - 4}{2} = \frac{2}{2} = 1 = LHS$
$\text{Thus, the statement holds for the base case.}$

$\text{Assume the statement holds for some arbitrary term, n= k}$
$1^{2} + 4^{2} + 7^{2} + ... + (3k - 2)^{2} = \frac{k(6k^{2} - 3k - 1)}{2}$

$\text{Prove it is true for n = k + 1}$
$RTP: 1^{2} + 4^{2} + 7^{2} + ... + [3(k + 1) - 2]^{2} = \frac{(k + 1)[6(k + 1)^{2} - 3(k + 1) - 1]}{2} = \frac{(k + 1)[6k^{2} + 9k + 2]}{2}$

$LHS = \underbrace{1^{2} + 4^{2} + 7^{2} + ... + (3k - 2)^{2}}_{\frac{k(6k^{2} - 3k - 1)}{2}} + [3(k + 1) - 2]^{2}$
$= \frac{k(6k^{2} - 3k - 1)}{2} + [3(k + 1) - 2]^{2}$
$= \frac{k(6k^{2} - 3k - 1) + 2[3(k + 1) - 2]^{2}}{2}$
$= \frac{k(6k^{2} - 3k - 1) + 2(3k + 1)^{2}}{2}$
$= \frac{k(6k^{2} - 3k - 1) + 18k^{2} + 12k + 2}{2}$
$= \frac{k(6k^{2} - 3k - 1 + 18k + 12) + 2}{2}$
$= \frac{k(6k^{2} + 15k + 11) + 2}{}$
$= \frac{(k + 1)[6k^{2} + 9k + 2]}{2}$
$= \frac{(k + 1)[6(k + 1)^{2} - 3(k + 1) - 1]}{2}$
$= RHS$

Since it is true for n = 1, n = k, and n = k + 1, by the principles of mathematical induction, it is true for all positive values of n.

3 0

3 years ago