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3 years ago

6

ken paid $12 for two magazines. The cost of each magazine was a multiple of $3. What are the possible prices of the magazines?

2 answers:

Lapatulllka [165]3 years ago

8 0

You can divide 12 by 2, and you'll get 6. 6 is also a multiple of 3. So $6 is the price of each magazine.

Hope it helped!

lesya [120]3 years ago

7 0

Well $\frac{12 dollars}{2 magazines}$ is $6 per magazine and 6 is a multiple of 3, so that's a possibility.<span />

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3 years ago

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olasank [31]

Answer:

Option A.

Step-by-step explanation:

According to triangle inequality theorem, a triangle exists if the sum of two sides is greater that third side.

Apply this theorem on each combination.

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$8x+4>18$

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$x>\dfrac{14}{8}$

$x>\dfrac{7}{4}$ ...(1)

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Case III: $6x+18>2x+4$

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$x>-\dfrac{14}{4}$

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On combining (1), (2) and (3), we get

$\dfrac{7}{4}$

Therefore, the correct option is A.

3 0

3 years ago

What is the unit ratio for a button of 16 grams and 2 milliliters?

Reika [66]

Answer:

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Step-by-step explanation:

Since, the unit ratio is a fraction with denominator 1,

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3 years ago

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Arte-miy333 [17]

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Step-by-step explanation:

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8 0

3 years ago

Read 2 more answers

The radius of a spherical balloon is measured as 20 inches, with a possible error of 0.03 inch. Use differentials to approximate

soldi70 [24.7K]

Answer:

a) $V = 33510.322\,in^{3}$ , b) $A_{s} = 5026.548\,in^{2}$ , c) $\% V = 0.450\,\%$ , $\%A_{s} = 0.300\,\%$ .

Step-by-step explanation:

The volume and the surface area of the sphere are, respectively:

$V = \frac{4}{3}\pi \cdot r^{3}$

$A_{s} = 4\pi \cdot r^{2}$

a) The volume of the sphere is:

$V = \frac{4}{3}\pi \cdot (20\,in)^{3}$

$V = 33510.322\,in^{3}$

b) The surface area of the sphere is:

$A_{s} = 4\pi \cdot (20\,in)^{2}$

$A_{s} = 5026.548\,in^{2}$

c) The total differentials for volume and surface area of the sphere are, respectively:

$\Delta V = 4\pi\cdot r^{2}\,\Delta r$

$\Delta V = 4\pi \cdot (20\,in)^{2}\cdot (0.03\,in)$

$\Delta V = 150.796\,in^{3}$

$\Delta A_{s} = 8\pi\cdot r \,\Delta r$

$\Delta A_{s} = 8\pi \cdot (20\,in)\cdot (0.03\,in)$

$\Delta A_{s} = 15.080\,in^{2}$

Relative errors are presented hereafter:

$\%V = \frac{\Delta V}{V}\times 100\%$

$\%V = \frac{150.796 \,in^{3}}{33510.322\,in^{3}}\times 100\,\%$

$\% V = 0.450\,\%$

$\% A_{s} = \frac{\Delta A_{s}}{A_{s}}\times 100\,\%$

$\% A_{s} = \frac{15.080\,in^{2}}{5026.548\,in^{2}}\times 100\,\%$

$\%A_{s} = 0.300\,\%$

4 0

3 years ago