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2 years ago

How long is 110 minutes

Mathematics

2 answers:

sveticcg [70]2 years ago

6 0

Answer:

<h2>ANSWER</h2>

1.833 hours

Step-by-step explanation:

Divide the time value by 60

$10 \div 60 = 1.833$

alina1380 [7]2 years ago

6 0

Answer:
1 hour 50 minutes
Step-by-step explanation:
Hope this helped :))

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What is the length of leg s of the triangle below?

Katyanochek1 [597]

8, as it is a 45-45-90 triangle, and if it has a side x not on the hypotenuse, the hypotenuse is x√2, and the other side is also x

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3 years ago

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What is the relationship between the sine and cosine of complementary angles? a They are the same. b They are reciprocals. c The

yanalaym [24]

Answer:

A. They are the same.

Step-by-step explanation:

The sine of any acute angle is equal to its cosine.

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3 years ago

John read 1/3 of his book on the first day and 1/4 of the book on the second day. What fraction of the book does John have left

eduard

Answer:

5/12

Step-by-step explanation:

The whole book is 1.

So This will be your equation.

1/3 + 1/4 = 1

4/12 +3/12 = 1

7/12 = 1

Now you minus 7/12 from 1

1-7/12 is how much is left

12/12-7/12= 5/12

8 0

3 years ago

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In 1859, 24 rabbits were released into the wild in Australia, where they had no natural predators. Their population grew exponen

Mashcka [7]

Answer:

a) P' = P

$P(t) = 24e^{0.693t}$ where t is step of 6 months

b) 7.7 years

c)1064.67 rabbits/year

Step-by-step explanation:

The differential equation describing the population growth is

$\frac{dP}{dt} = P$

Where t is the range of 6 months, or half of a year.

P(t) would have the form of

$P(t) = P_0e^{kt}$

where $P_0 = 24$ is the initial population

After 6 month (t = 1), the population is doubled to 48

$P(1) = 24e^k = 48$

$e^k = 2$

$k = ln(2) = 0.693$

Therefore $P(t) = 24e^{0.693t}$

where t is step of 6 months

b. We can solve for t to get how long it takes to get to a population of 1,000,000:

$24e^{0.693t} = 1000000$

$e^{0.693t} = 1000000 / 24 = 41667$

$0.693t = ln(41667) = 10.64$

$t = 10.64 / 0.693 = 15.35$

So it would take 15.35 * 0.5 = 7.7 years to reach 1000000

c. $P' = P_0ke^{kt}$

We need to resolve for k if t is in the range of 1 year. In half of a year (t = 0.5), the population is 48

$24e^{0.5k) = 48$

$0.5k = ln2 = 0.693$

$k = 1.386$

Therefore, $P' = 1.386*24e^{1.386t}$

At the mid of the 3rd year, where t = 2.5, we can calculate P'

$P' = 1.386*24e^{1.386*2.5} = 1064.67$ rabbits/year

4 0

3 years ago

What is the sum of 56+64 writen as the product of its gcf and another sum

jenyasd209 [6]

<span>Look for the sum of 56 and 64, written as the product of its GCF and another sum.
First, let’s find the greatest common factor of both given numbers:
=> 56 = 1, 2, 4, 7, 8, 13, 28 and 56
=> 64 = 1, 2, 4, 8, 16, 32, and 64
Now, we need to find the greatest common factor between the two numbers. The GCF of the 2 numbers is 8.
=> 56 / 8 = 7
=> 64 / 8 = 8
=> 56 + 64 = 120
=> (8 x 7) + (8 x 8)
=> 56 + 64
=> 120.</span><span>

</span>

3 0

3 years ago