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3 years ago

Which of the following lists is in order from smallest to largest?

Mathematics

1 answer:

Stells [14]3 years ago

5 0

Answer:

The following list is in order from smallest to largest is $17\%,0.175,\frac{2}{11},\frac{1}{5}$

Step-by-step explanation:

Let the numbers be $\frac{2}{11},\frac{1}{5},17\%,0.175$

To find the list in order from smallest to largest:

$\frac{2}{11},\frac{1}{5},17\%,0.175$

Rewritting the above numbers as below

$0.18,0.5,\frac{17}{100},0.175$

$0.18,0.5,0.17,0.175$

Now arranging the numbers in order from smallest to largest

$0.17,0.175,0.18,0.5$

$\frac{17}{100},0.175,\frac{2}{11},\frac{1}{5}$

The arranged numbers can be written as

$17\%,0.175,\frac{2}{11},\frac{1}{5}$

The following list is the order from smallest to largest is $17\%,0.175,\frac{2}{11},\frac{1}{5}$

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Last year,raul earned $17,500 working as a freelance writer and 15,250 as a waiter the income tax he must pay is 18% what is the

Sveta_85 [38]

Answer: $5,895

Step-by-step explanation:

First you want to add the earnings in order to find the total value being taxed.

17,500 + 15,250 = 32,750

We can then find the percent as a decimal by dividing it by 100:

$\frac{18}{100} = .18$

We can then multiply the total earnings by this decimal to find the total income tax:

$32,750 * .18 = 5,895$

Hope this helps

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1 year ago

You need a 70% alcohol solution. On hand, you have a 420 mL of a 85% alcohol mixture. How much pure water will you need to add t

Nimfa-mama [501]

Answer:

Pure water =74.1 mL

Step-by-step explanation:

You need 420 mL of a 70% alcohol solution.

420 mL of solution and 70% of it should be alcohol.

So,

420 * 70/100 mL

= 420 * 0.7

= 294 mL of alcohol and the rest should be water.

On hand, you have a 85% alcohol mixture.

The alcohol can only come from the 85% mixture. So, how much of 85% alcohol do we need to get 294 mL of alcohol:

Let

y = alcohol solution needed

85% of y = 294

85/100 * y = 294

y = 294 ÷ 85/100

y = 294 × 100/85

y = 29,400/85

y = 345.88235294117

Approximately, 345.9mL

The rest of 420 mL solution has to be pure water

Pure water = 420 mL - 345.9mL

= 74.1 mL

Pure water =74.1 mL

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3 years ago

Maridel wants to find out how many of her classmates plan to come to the next football game. There are 800 students in her schoo

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In 2010, the world's largest pumpkin weighed 1,810 kilograms. An average-sized pumpkin weighs 5,000 grams.

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A trapezoidal prism undergoes a dilation. The surface area of the pre-image is 35 m². The surface area of the image is 315 m². T

Westkost [7]

The original object is called pre-image, and its dilated version is called image. The height of the image for the considered case is 17.4 m

<h3>How does dilation affect length, area, and volume of an object?</h3>

Suppose a figure (pre image) is dilated (dilated image) by scale factor of k.

So, if a side of the figure is of length L units, and that of its similar figure is of M units, then:

L = k × M

where 'k' will be called as scale factor.

The linear things grow linearly like length, height etc.

The quantities which are squares or multiple of linear things twice grow by square of scale factor. Thus, we need to multiply or divide by k² to get each other corresponding quantity from their similar figures' quantities.

So area of first figure = k² × area of second figure

Similarly,

Volume of first figure = k³ × volume of second figure.

It is because we will need to multiply 3 linear quantities to get volume, which results in k getting multiplied 3 times, thus, cubed.

For this case, we're given that;

Surface area of pre-image = 35 m²
Surface area of dilated image = 315 m²
Height of pre-image = 5.8 m
Height of dilated image.

Let the scale fator of dilation be k.

Then:

So area of pre-image = k² × area of dilated image

Thus, we get:

$35 = k^2 \times 315\\\\k = \sqrt{\dfrac{35}{315}} = 1/3$

(took positive root as scaling is done by non-negative factors as only magnitude matters mostly)

Thus, we get:

Height of pre-image = (1/3) × Height of dilated image (say h)

$5.8 = (1/3) \times h\\h = 3 \times 5.8 = 17.4 \rm \: m$

Thus, the original object is called pre-image, and its dilated version is called image. The height of the image for the considered case is 17.4 m

Learn more about dilation here:

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