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vekshin1
3 years ago
15

17/23 as a decimal rounded to the nearest 10th

Mathematics
2 answers:
zmey [24]3 years ago
4 0

Answer:

 17 23 as decimal is approximately: 0.73913 (Rounded to fifth decimal place)

Step-by-step explanation:

bazaltina [42]3 years ago
4 0

Answer:

0.7

Step-by-step explanation:

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Mike and talia each make a scale drawing of the right side of their house. The scale of the side of the actual house to Mikes dr
Nutka1998 [239]

The width and height of Talia's Scale drawing are :

  • Width = 4.35 in
  • Height = 5.7 in

We can obtain the actual length of the side of the house from Mike's drawing :

  • Mike's scale = 3 ft to 1 in

Actual measurements:

  • Scale ratio × model drawing
  • Height, h = 9.5 × 3 = 28.5 feets
  • Width, w = 7.25 × 3 = 21.75 feets

Measurement of Talia's Scale drawing :

  • Talia's scale = 5 ft to 1 in
  • Actual measurement ÷ scale
  • Height = 28.5 ÷ 5 = 5.7 in
  • Width = 21.75 ÷ 5 = 4.35 in

The width and height of Talia's scale drawing are 5.7 and 4.35 respectively.

Learn more : brainly.com/question/17388747?referrer=searchResults

5 0
3 years ago
Evaluate. 9.75 + 4.625 + (-9.75) + 2.6=?
marusya05 [52]
The answer would be 7.225.
4 0
3 years ago
Read 2 more answers
What operation do you use to convert 50 cups to gallons?
zzz [600]
Division can be used. For example

50/16=3.125gallons

or you could use multiplication

50*1/16=3.125gallons


5 0
3 years ago
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23−11 Enter your answer as a fraction in simplest form by filling in the boxes.<br> 30 30
mamaluj [8]

23 - 11 = 12

12 = 12/1

6 0
3 years ago
Please help this question.
sergejj [24]

Answer:

a). 59.049°C

b). 2.1179 seconds

Step-by-step explanation:

Expression representing the final temperature after decrease in temperature of the metal from 100°C to T°C is,

T = 100(0.9)^{x}

where x = duration of cooling

a). Temperature when x = 5 seconds

T = 100(0.9)⁵

  = 59.049

  ≈ 59.049°C

b). If the temperature of the metal decreases from 100°C to 80°C

Which means for T = 80°C we have to calculate the duration of cooling 'x' seconds

80 = 100(0.9)^{x}

0.8 = (0.9)^{x}

By taking log on both the sides

log(0.8) =log[(0.9)^{x}]

-0.09691 = x[log(0.9)]

-0.09691 = -0.045757x

x = \frac{0.09691}{0.045757}

x = 2.1179

x ≈ 2.1179 seconds

3 0
3 years ago
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