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Mazyrski [523]
4 years ago
12

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
2 answers:
vagabundo [1.1K]4 years ago
8 0

Answer: Area (upper bound) = 4527.7056 cm²

               Perimeter (lower bound) = 278 cm

<u>Step-by-step explanation:</u>

The length and width of the rectangle have been ROUNDED to the nearest tenth. Let's calculate what their actual measurements could be:

LENGTH: rounded to 87.3,  actual is between 87.25 and 87.34

<em>87.25 is the lowest number it could be that would round it UP to 87.3</em>

<em>87.34 is the highest number it could be that would round DOWN to 87.3</em>

WIDTH: rounded to 51.8, actual is between 51.75 and 51.84

<em>51.75 is the lowest number it could be that would round it UP to 51.8</em>

<em>51.84 is the highest number it could be that would round DOWN to 51.8</em>

To find the Area of the upper bound, multiply the highest possible length and the highest possible width:

A = 87.34 × 51.84 = 4527.7056

To find the Perimeter of the lower bound, calculate the perimeter using the lowest possible length and the lowest possible width:

P = 2(87.25 + 51.75) = 278

viva [34]4 years ago
4 0

Answer:

Area=4529.0975

Perimeter=278

Step-by-step explanation:

same as the one above but with different area

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olchik [2.2K]

Answer: 4.2

Step-by-step explanation:

Here's the formula for how to find the hypotenuse: (a and b are the two legs for the triangle-- it doesn't matter which leg is a and which leg is b, but c MUST be the hypotenuse)

a^2+b^2=c^2

1.9^2+3.8^2=c^2

3.61+14.44=c^2

18.05=c^2

\sqrt{18.05} =c

c=4.2

6 0
3 years ago
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Please help i'm horrible at math ajdbvkdfjgvnfjglbdfg
SVETLANKA909090 [29]

Answer:

-29/3 i think

Step-by-step explanation:

7 0
3 years ago
Please help and thank you. ​
Marina CMI [18]

Answer:

4.

  • 400 ft elevation at 1 km and 3 km from the start
  • 300 ft/km average rate of change over the first 4 km

5.

  • length = 3.5 cm
  • width = 3.5 cm
  • height = 7.0 cm

6.

  • 6 mo: $1018.14
  • 5 yrs: $1196.89
  • avg incr: $3.28 per month

Step-by-step explanation:

<h3>4.</h3>

A graphing calculator is handy for solving problems involving cubic polynomials. You're interested in where the elevation is 400 ft. Since the value of f(x) is in hundreds of feet, you want to find x such that f(x) = 4.

Values of x where that is the case are x=1 and x=3, representing distances of 1 km and 3 km from the start of the road.

The average rate of change of elevation is the difference in elevation divided by the difference in distance from the start:

  average rate of change = (f(4) -f(0) hundred ft)/((4 - 0) km)

  = (19 -7)/4 hundred feet/km

  = 12/4  hundred feet/km = 300 ft/km

___

<h3>5.</h3>

You want to find x when ...

  A(x) = 122.5 cm²

  10x² = 122.5 cm² . . . . substitute the given expression for A(x)

  x² = 12.25 cm² . . . . . . divide by 10

  x = √(12.25 cm²) = 3.5 cm . . . . take the square root

The diagram tells you ...

  length = width = x = 3.5 cm

  height = 2x = 7.0 cm

___

<h3>6.</h3>

Evaluate the given expression for the different values of m:

  $1000·1.003^6 ≈ $1018.14 . . . . 6-month value

  $1000·1.003^60 ≈ $1196.89 . . . . 5-year value

The increase is $1196.89 -1000.00 = $196.89. That increase took place over 60 months, so the average increase per month is ...

  $196.89/(60 mo) ≈ $3.28 per mo . . . . average per month over 5 years

8 0
4 years ago
Help please! I am not very good at this!
diamong [38]
50/11 = 4.55
4 3/5 = 4.6
410% = 4.1

order from least to greatest
410%, 4.199, 50/11, 4 3/5,  4.61
7 0
3 years ago
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In the data set below, what is the lower quartile? <br><br> 2 2 3 5 6 6
MAXImum [283]
The median is 4
There are 3 numbers below 4. and the lower quartile is the middle number of these 3.
So its 2
6 0
3 years ago
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