Question

The population of a city is given as 3.45×106 . A nearby city has a population of 1.3×105 . What is the sum of the populations of the two cities in scientific notation?

Answer 1

[AI-generated answer — Bot GPT 3.5]

The sum of the populations of the two cities is calculated by adding the two populations together.

3.45 × 10^6 + 1.3 × 10^5

To add these numbers, we need to have the same exponent for the powers of 10. We can do this by manipulating 1.3 × 10^5.

We can write 1.3 × 10^5 as 0.13 × 10^6, since dividing by 10 is equivalent to moving the decimal point one place to the left.

So, the sum of the populations is:

3.45 × 10^6 + 0.13 × 10^6 = 3.58 × 10^6

Thus, the sum of the populations of the two cities in scientific notation is 3.58 × 10^6.

Answer 2

An empty dump truck weighs about 2.8×105 pounds. How much does it weigh if 15,300 pounds of trash is added? The answer should be in scientific notation

Answer 3

[AI-generated answer — Bot GPT 3.5]

To find the weight of the dump truck after 15,300 pounds of trash is added, we need to add this weight to the weight of the empty dump truck.

Weight of the empty dump truck = 2.8 × 10^5 pounds
Weight of trash added = 15,300 pounds

Adding these weights together, we get:

2.8 × 10^5 + 15,300

To write this in scientific notation, we need to make the first term 10 raised to the same exponent as the second term. In this case, we need to write 15,300 in scientific notation.

15,300 = 1.53 × 10^4

Now, let's add the two terms:

2.8 × 10^5 + 1.53 × 10^4

To add these numbers, we can only add the coefficients if the exponents are the same. We need to convert 1.53 × 10^4 to have an exponent of 5.

1.53 × 10^4 = 0.153 × 10^5

Now we can add the coefficients:

2.8 × 10^5 + 0.153 × 10^5 = 2.953 × 10^5

Therefore, the weight of the dump truck after adding 15,300 pounds of trash is 2.953 × 10^5 pounds.

Answer 4

The half-life of uranium-235 is 700,000,000 years. The half-life of uranium-238 is 4.463×1011 years longer. What is the half-life of uranium-238? The answer should be in decimal form.

Answer 5

[AI-generated answer — Bot GPT 3.5]

To find the half-life of uranium-238, we can start by subtracting the additional years from the half-life of uranium-235:

Half-life of uranium-238 = Half-life of uranium-235 - Additional years

Half-life of uranium-235 = 700,000,000 years

Additional years = 4.463 × 10^11 years

Substituting these values into the equation:

Half-life of uranium-238 = 700,000,000 years - 4.463 × 10^11 years

To subtract these numbers, we need to rewrite 700,000,000 in scientific notation with the same exponent as 4.463 × 10^11.

700,000,000 = 7 × 10^8

Now we can subtract the numbers:

7 × 10^8 - 4.463 × 10^11

To subtract these numbers, the exponents need to be the same. We can convert 7 × 10^8 to have an exponent of 11.

7 × 10^8 = 0.7 × 10^11

Now we can subtract the coefficients:

0.7 × 10^11 - 4.463 × 10^11 = -3.763 × 10^11

Therefore, the half-life of uranium-238 is -3.763 × 10^11 years.

Answer 6

A mobile phone holds 1.28×1011 bytes of storage. You can also pay more money to get an additional 384,000,000,000 bytes. How much storage would your phone have if you buy the additional bytes? The answer should be in scientific notation.

Answer 7

[AI-generated answer — Bot GPT 3.5]

To find the total storage of the phone after buying the additional bytes, we need to add the additional bytes to the current storage:

Current storage = 1.28 × 10^11 bytes
Additional bytes = 384,000,000,000 bytes

Adding these together, we get:

1.28 × 10^11 + 384,000,000,000

To simplify this into scientific notation, we need to convert 384,000,000,000 into scientific notation.

384,000,000,000 = 3.84 × 10^11

Now we can add the two terms:

1.28 × 10^11 + 3.84 × 10^11

To add these numbers, we can only add the coefficients if the exponents are the same. We need to convert 1.28 × 10^11 to have an exponent of 11.

1.28 × 10^11 = 0.128 × 10^12

Now we can add the coefficients:

0.128 × 10^12 + 3.84 × 10^11 = 0.512 × 10^12

Therefore, the total storage of the phone after buying the additional bytes would be 0.512 × 10^12 bytes.