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bixtya [17]
3 years ago
9

Which steps will verify that a parallelogram is a rectangle? Check all that apply.

Mathematics
2 answers:
avanturin [10]3 years ago
8 0

Answer;

-Calculate the lengths of the diagonals, and show that they are equal.

-Calculate the slopes of every side, and show that adjacent sides are perpendicular.

Explanation;

-A parallelogram is a quadrilateral with 2 pairs of opposite, equal and parallel sides. If the diagonals of a parallelogram are congruent, then it’s a rectangle and also if a parallelogram contains a right angle, then it’s a rectangle.

UkoKoshka [18]3 years ago
5 0

Answer:

1st and last

Step-by-step explanation:

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Solve the system of equations using elimination 2x+y=3 and 3x-y=12
Eddi Din [679]
2x + y = 3
3x - y = 12
Add both equations
5x = 15, x = 3
2(3) + y = 3
6 + y = 3, y = -3
Solution: x = 3, y = -3... or (3,-3)
3 0
2 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
Pls help me out it’s urgent <br> Pls don’t give out fake answers <br> WILL MARK BRAINLIEST !
pantera1 [17]
Hope I can help you!

6 0
2 years ago
I need help figuring this question out
Marysya12 [62]
ANSWER: (4,26)
Comment below if you have any questions or want an explanation:
7 0
2 years ago
40 x 125% step it out
AlladinOne [14]

Answer:

0.50 or 50%

Step-by-step explanation:

40 x 125 = X

1.25 as a decimal or 125%

40 x 1.25 = 0.50

0.50 --->50%

40 x 125% = 50%

3 0
1 year ago
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