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3 years ago

What is 3/8 times 2/9? Answers: A 5/17 B 5/12 C 1/17 D 1/12

Mathematics

1 answer:

belka [17]3 years ago

3 0

$\frac{3}{8}\cdot\frac{2}{9}=\frac{1}{4}\cdot\frac{1}{3}=\frac{1}{12}\\\\Answer:D$

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user100 [1]

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3 years ago

Read 2 more answers

1.) Natalie is a lab technician and earns $10.65 per hour. What are her wages for 40 Hours?

Contact [7]

Answer:

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Step-by-step explanation:

10.65*40=426

426+50=476

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3 years ago

SOLVE FOR M <br><br> −4m+7=−5

Masteriza [31]

Answer:

m=3

Step-by-step explanation:

-4m+7= -5

Combine like terms.Put +7 on other side.

-4m= -5-7

-4m= -12

It's -4 times m = -12 right ?

So the answer is m=3 because m= -12/-4

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3 years ago

The length of the subway is 850 miles. If .15 inch represents 45 miles, what is the length of the subway on the map?

Scilla [17]

Answer:

The length of the subway on the map is $2.83\ inches$

Step-by-step explanation:

we know that

The scale on the map is equal to the length of the subway on the map divided by the length of the subway in the actual

Let

z------> the scale on the map

x------> the length of the subway on the map

y------> the length of the subway in the actual

$z=\frac{x}{y}$

In this problem we have

$z=\frac{0.15}{45}\frac{inches}{miles}$

$y=850\ miles$

substitute and solve for x

$\frac{0.15}{45}=\frac{x}{850}$

$x=850*\frac{0.15}{45}=2.83\ inches$

6 0

4 years ago

Read 2 more answers

A rectangular box 4 meters long, 3 meters wide and 2 meters tall. On the box is a cat, and the box is floating in water such tha

Vikentia [17]

Answer:

Mass of the cat = 16800kg

Step-by-step explanation:

The volume of the box = its length * its breath * its height- 4m × 3m × 2m = 24 $m^{3}$

The standard density of water is 1000 Kg/ $m^{3}$

By Archimedes principle, the mass of a body floating body is equivalent to the mass of the volume liquid displaced

in this case we have

Mass of water displaced = Density of the water × Volume of the water displaced = 1000 Kg/ $m^{3}$ × 24 $m^{3}$ = 24000kg

The mass of the box = Box density × Box volume = 24 $m^{3}$ × 300kg/ $m^{3}$ = 7200 kg Hence the mass of water displaced = mass of the foating box + mass of th cat

24000kg = mass of cat +7200kg

mass of cat = 24000kg - 7200kg = 16800kg

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4 years ago