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densk [106]
3 years ago
6

Jada made a scale drawing of a house. The scale of the drawing was 1 centimeter = 5 meters. What is the scale factor of the draw

ing?
Mathematics
1 answer:
zlopas [31]3 years ago
4 0

Answer:

1 : 500

Step-by-step explanation:

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At the start of the month, the value of an investment was $73.42. By the end of the month, the value of the investment changed b
Aleonysh [2.5K]

Answer:

0.8%.

Step-by-step explanation:

Given:

At the start of the month, the value of an investment was $73.42.

By the end of the month, the value of the investment changed by a loss of $13.53.

Question asked:

What was the value, in dollars, of the investment at the end of the month?

What was the percent loss?

Solution:

At the start of the month, the value of an investment = $73.42.

Loss = $13.53.

<u><em>The value, of the investment at the end of the month = The value of an investment at the start of the month - loss</em></u>

The value, of the investment at the end of the month = $73.42 - $13.53

                                                                                         = $59.89

Thus, the value, of the investment at the end of the month is $59.89.

Now, we will find percent loss:-

Percent\ loss = \frac{Loss}{value\ of \ investment\ at \ start}

                    =\frac{59.89}{73.42} \\=0.81\%

Thus, percent loss on investment by the end of the month is 0.8%.

7 0
3 years ago
Select all the pairs that could be reasonable approximations for the diameter and circumference of a cirlce 5 meters and 22 mete
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Answer:

Option a) circle 5 meters and 22 meters

Step-by-step explanation:

We are given the following information in the question:

A pair of diameter and the circumference is given. We have to find a correct approximations for the diameter and circumference.

a) circle 5 meters and 22 meters

\text{Diameter} = 5\text{ meters}\\\text{Circumference} = \pi d = 3.14\times 5 = 15.7\text{ meters}

b) 19 inches and 50 inches

\text{Diameter} = 19\text{ inches}\\\text{Circumference} = \pi d = 3.14\times 19 = 59.66\text{ inches}

c) 33 centimeters and 80 centimeters

\text{Diameter} = 33\text{ centimeters}\\\text{Circumference} = \pi d = 3.14\times 33 =103.62\text{ centimeters}

Thus, no pair gives a reasonable approximation. Only the circle with diameter 5 and circumference 22 meters have closest approximation.

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What is there to solve 
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