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Gemiola [76]
3 years ago
10

To which set does -1/2 belong? A whole numbers only B rational numbers only

Mathematics
2 answers:
Jlenok [28]3 years ago
8 0

Answer:

B

Step-by-step explanation:

Because -1/2 is not a whole number so b is the only answer left

Alex3 years ago
6 0
Answer is B as whole numbers are not written as fractional form
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Round 299 to the nearest hundred. Enter your answer in the box below.
Troyanec [42]

Answer:

300

Step-by-step explanation:

hundreds  tens  ones

2                   9      9

We look at the tens

It is 5 or higher so we round up the hundreds place

2 becomes 3

300

7 0
3 years ago
Read 2 more answers
A snail can crawl 2/5 of a meter in a minute.
gavmur [86]
For 2/5 meter, it takes = 1 min.
For 1 meter, it is = 5/2 min.

Now, for 6 meters, it would be: 5/2 * 6 = 30/2 = 15

In short, Your Answer would be 15 minutes

Hope this helps!
6 0
3 years ago
Read 2 more answers
Approximate the change in the volume of a sphere when its radius changes from r​ = 40 ft to r equals 40.05 ft (Upper V (r )equal
alexgriva [62]

Answer:

The change in the volume of a sphere whose radius changes from 40 feet to 40.05 feet is approximately 1005.310 cubic feet.

Step-by-step explanation:

The volume of the sphere (V), measured in cubic feet, is represented by the following formula:

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

Where r is the radius of the sphere, measured in feet.

The change in volume is obtained by means of definition of total difference:

\Delta V = \frac{\partial V}{\partial r}\Delta r

The derivative of the volume as a function of radius is:

\frac{\partial V}{\partial r} = 4\pi \cdot r^{2}

Then, the change in volume is expanded:

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

If r = 40\,ft and \Delta r = 40\,ft-40.05\,ft = 0.05\,ft, the change in the volume of the sphere is approximately:

\Delta V \approx 4\pi\cdot (40\,ft)^{2}\cdot (0.05\,ft)

\Delta V \approx 1005.310\,ft^{3}

The change in the volume of a sphere whose radius changes from 40 feet to 40.05 feet is approximately 1005.310 cubic feet.

7 0
3 years ago
The expedition team plan a final practice run to test the range of their communication equipment. One member travels a distance
podryga [215]

Answer:

We need to sketch the problem first.

Find the size of angle R.

One member travels a distance of 12km due north. Another team member heads 50degree east of north and travels a distance of 10km.

If se substute 50° of 180 we have

180-50=130°

The distance between the two team members is the missing side.

We know two sides and included angle, so we use the cosine rule.

A2+b2+c2-2bcCosA

= 102+122-(2x10x12xcos130°)

=100+144-(-147.08)

=100+144+147.08

=391.08

A==SQRT391.08

=19.775

19.7km

Step-by-step explanation:

Download docx
4 0
3 years ago
A rectangle has long sides measuring 4 cm and short sides measuring 3 cm. What is the area of this rectangle?
bearhunter [10]
Area of this rectangle
4×3=12cm^2

formula
long side×short side=area of this rectangle
8 0
3 years ago
Read 2 more answers
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