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aev [14]
3 years ago
15

Can a rectangle be a square?

Mathematics
2 answers:
DENIUS [597]3 years ago
5 0
No, to be a square each side has to be equal. but all squares are rectangles.
Nastasia [14]3 years ago
5 0

Answer:

A rectangle can be tall and thin, short and fat or all the sides can have the same length. ... Thus every square is a rectangle because it is a quadrilateral with all four angles right angles. However not every rectangle is a square, to be a square its sides must have the same length.

Step-by-step explanation:

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Please answer these<br> i need this
Len [333]

Answer:

1. no

2. 8

Step-by-step explanation:

1. 20 does not equal 4/5

2. 6x=48 divide 6 on both sides

x=8

5 0
3 years ago
Read 2 more answers
Find the value of the variable and BC, if B is between A and C.<br> AB = 5x, BC = 3x-1, AC = 47
schepotkina [342]

Answer:

17

Step-by-step explanation:

AC = AB + BC

Since we know what each variable equals, we can write out the equation as shown:

47 = 5x + 3x - 1

Combining like terms:

47 = 8x - 1

Add 1 on both sides:

48 = 8x

Divide 8 on both sides:

X = 6

Now that we know what X equals, we can solve BC.

BC = 3(6) - 1

BC = 18 - 1

BC = 17

Therefore BC = 17.

4 0
3 years ago
What is 123 as a percent????
Anarel [89]

Answer:

12,300%

Step-by-step explanation:

7 0
3 years ago
Write your question here (Keep it simple and clear to get the best answer) write 6.4 to the nearest whole number
Alina [70]
It would be 6 because it it under .5 so you wouldn't round up   
8 0
3 years ago
Read 2 more answers
A) If the 5th term of an A.P. is double the 7th term, show that the sum of the 17 terms zero.​
gayaneshka [121]

Answer:

see explanation

Step-by-step explanation:

The nth term of an AP is

a_{n} = a₁ + (n - 1)d

where a₁ is the first term and d the common difference

Given a₅ is double a₇ , then

a₁ + 4d = 2(a₁ + 6d) , that is

a₁ + 4d = 2a₁ + 12d ( subtract a₁ from both sides )

4d = a₁ + 12d ( subtract 12d from both sides )

- 8d = a₁

The sum of n terms of an AP is

S_{n} = \frac{n}{2} [ 2a₁ + (n - 1)d ] , substitute values

S_{17} = \frac{17}{2} ( 2(- 8d) + 16d)

     = 8.5(- 16d + 16d)

     = 8.5 × 0

     = 0

5 0
2 years ago
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