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3 years ago

Which of the following figures has 90 rotation symmetry?

Mathematics

1 answer:

garik1379 [7]3 years ago

6 0

It would be a rectangle

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Change to a mixed number. 13/3

miss Akunina [59]

Answer:

4 and 1/3

Step-by-step explanation:

3 can go into 13 4 times with one left over

my rank is expert

hope this helps

5 0

2 years ago

Round 87,365,219 to the nearest million

OLga [1]

Answer:

87 000 000

Step-by-step explanation:

3 0

2 years ago

D=(107-115)^2+(-49+-69)^2

ladessa [460]

Answer:

d=13988 or d=1.3988 x 10^4

Step-by-step explanation:

d=(-8)^2+(-118)^2

d=64+13924

d=13988

6 0

2 years ago

PLEASE HELP WILL MARK BRAINLEIST

horsena [70]

Answer:

The picture shows the answer on the graph

Step-by-step explanation:

-2=-3/4×(x-6)

Distribute -3/4 through the parenthesis

-2= -3/4x+9/2

Multiply both sides of the equation by 4

-8=-3x+18

Move the variable to the left side and change its sign

3x-8=18

Move constant to the right side and change its sign

3x=18+8

Add the numbers

3x=26

Divide both sides of the equation by 3

x=26/3

Alternative Form- x=8 2/3 or x=8.6

7 0

3 years ago

Prove the following by induction. In each case, n is apositive integer.<br> 2^n ≤ 2^n+1 - 2^n-1 -1.

frutty [35]

<h2>Answer with explanation:</h2>

We are asked to prove by the method of mathematical induction that:

$2^n\leq 2^{n+1}-2^{n-1}-1$

where n is a positive integer.

Let us take n=1

then we have:

$2^1\leq 2^{1+1}-2^{1-1}-1\\\\i.e.\\\\2\leq 2^2-2^{0}-1\\\\i.e.\\2\leq 4-1-1\\\\i.e.\\\\2\leq 4-2\\\\i.e.\\\\2\leq 2$

Hence, the result is true for n=1.

Let us assume that the result is true for n=k

i.e.

$2^k\leq 2^{k+1}-2^{k-1}-1$

Now, we have to prove the result for n=k+1

i.e.

<u>To prove:</u> $2^{k+1}\leq 2^{(k+1)+1}-2^{(k+1)-1}-1$

Let us take n=k+1

Hence, we have:

$2^{k+1}=2^k\cdot 2\\\\i.e.\\\\2^{k+1}\leq 2\cdot (2^{k+1}-2^{k-1}-1)$

( Since, the result was true for n=k )

Hence, we have:

$2^{k+1}\leq 2^{k+1}\cdot 2-2^{k-1}\cdot 2-2\cdot 1\\\\i.e.\\\\2^{k+1}\leq 2^{(k+1)+1}-2^{k-1+1}-2\\\\i.e.\\\\2^{k+1}\leq 2^{(k+1)+1}-2^{(k+1)-1}-2$

Also, we know that:

$-2$

(

Since, for n=k+1 being a positive integer we have:

$2^{(k+1)+1}-2^{(k+1)-1}>0$ )

Hence, we have finally,

$2^{k+1}\leq 2^{(k+1)+1}-2^{(k+1)-1}-1$

Hence, the result holds true for n=k+1

Hence, we may infer that the result is true for all n belonging to positive integer.

i.e.

$2^n\leq 2^{n+1}-2^{n-1}-1$ where n is a positive integer.

6 0

3 years ago