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Artist 52 [7]
4 years ago
7

The lengths of a rectangle have been measured to the nearest tenth of a centimetre they are 87.3cm and 51.8cm what is the upper

bound for the area and the lower bound for the perimeter.
Mathematics
1 answer:
zmey [24]4 years ago
5 0

Answer:

87.3

Upper bound87.35

Lower bound87.25

51.8

Upper bound 51.85

Lower bound51.75

Step-by-step explanation:

Take the tenth of a centimeter and divide by 2. .05. This is the error margin. The upper bound is .05 more then the pro final length and the lower bound are .05 less than the original

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sammy [17]

Answer:

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the new one for d: 25/24 and 36/24

Step-by-step explanation:

4/7 x 3/3 equals 12/21

3/5 x 5/5 equals 15/25

11/9 x 3/3 equals 33/27

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3 years ago
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ladessa [460]

Answer:

The answer is 1/6

Step-by-step explanation:

So Think of a half and then divide it into 3 parts for a half it will be 1/3 but for a meter it will be 1/6 because there is 2 half's in 1 whole

Hope this helps!! :D

Please mark  brainliest if this helped you!

5 0
3 years ago
Marcus is wanting to prove that parallelogram ABCD is a square. He shows that opposite sides are congruent and opposite sides ar
faust18 [17]

Answer:

It cannot be concluded that ABCD is square.

Step-by-step explanation:

parallelogram is a quadrilateral whose opposite side are parallel

it may or may not have angles right angled.

For a quadrilateral to be square following condition should be satisfied

1: opposite side parallel

2: all side congruent

3: All angles right angle.

____________________________________________

coming back  to question

given

Marcus shows that opposite sides are congruent and opposite sides are parallel.

He says that opposite sides are congruent but no information is given about adjacent side. Adjacent side should also be congruent

Also , criteria of right angle is not also mentioned which is mandatory to prove shape of square.

Based on above information,

ABCD can also be

Rectangle and Rhombus, as both also  have  opposite sides  congruent and opposite sides  parallel.

Hence, it cannot be concluded that ABCD is square.

7 0
3 years ago
prove the following identity: sec x csc x(tan x + cot x) = 2+tan^2 x + cot^2 x please provide a proof in some shape form or fash
wolverine [178]

Answer:

Step-by-step explanation:

Hello,

<u><em>Is this equality true ?</em></u>

sec x csc x(tan x + cot x) = 2+tan^2 x + cot^2 x

<u>1. let 's estimate the left part of the equation</u>

sec(x)csc(x)(tan(x) + cot(x)) =\dfrac{1}{cos(x)sin(x)}*(\dfrac{sin(x)}{cos(x)}+\dfrac{cos(x)}{sin(x)})\\\\=\dfrac{1}{cos(x)sin(x)}*(\dfrac{sin^2(x)+cos^2(x)}{sin(x)cos(x)})\\\\=\dfrac{1}{cos(x)sin(x)}*(\dfrac{1}{sin(x)cos(x)})\\\\\\=\dfrac{1}{cos^2(x)sin^2(x)}

<u>1. let 's estimate the right part of the equation</u>

<u />2+tan^2(x) + cot^2(x)=2+\dfrac{sin^2(x)}{cos^2(x)}+\dfrac{cos^2(x)}{sin^2(x)}\\\\=\dfrac{2cos^2(x)sin^2(x)+cos^4(x)+sin^4(x)}{cos^2(x)sin^2(x)}\\\\=\dfrac{(cos^2(x)+sin^2(x))^2}{cos^2(x)sin^2(x)}\\\\=\dfrac{1^2}{cos^2(x)sin^2(x)}\\\\=\dfrac{1}{cos^2(x)sin^2(x)}<u />

This is the same expression

So

sec x csc x(tan x + cot x) = 2+tan^2 x + cot^2 x

hope this helps

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ANEK [815]

Answer:

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6 0
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