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Alchen [17]
3 years ago
13

The tree frog population in your back yard doubles every two weeks. Currently, there are 10 frogs in your back yard. Assuming th

at there are 4 weeks in a month how many tree frogs will there be in 3 months?
Mathematics
2 answers:
Minchanka [31]3 years ago
6 0
There should be 80 but I'm not 100%.......
sergiy2304 [10]3 years ago
3 0

Answer:

640

Step-by-step explanation:

So we want to know how many frogs will there be in 3 months if the frog population doubles every two weeks and there are 10 frogs in the start. So lets see how many weeks are there in 3 months if there is 4 weeks in one month: 3 * 4 weeks = 12 weeks. So every two weeks the population doubles and that means the population will double 6 times in 12 weeks. 20 + 40 + 80 + 160 + 320 640 = 1260 frogs.

You might be interested in
What is the slope of the line?
Katyanochek1 [597]

Answer:

1

Step-by-step explanation:

slope (k) is the rise over run of the line

a way of finding the slope if you have two dots of the line is:

k \:  =  \frac{y2 - y2}{x2 - x2}

here the dots that belong to the line that you can most easily see are:

A(-6,0)

B(0,6)

by inserting the values into the euquation you get:

\frac{6 - 0}{0 - ( - 6)}  =  \frac{6}{6}  = 1

good luck with your maths solving

6 0
2 years ago
Dan recently bought a new mountain bike for $550. His friend works
Wittaler [7]
$990
550 x .80 = 440
$550 + $440 = $990
8 0
3 years ago
Cody hiked at an average speed of 1 mile per hour for 5 hours on Saturday. He hiked an average speed of 2 miles per hour for 3 h
ivanzaharov [21]

Answer:

Step 1: Multiply 1x5 Step 2: Multiply 2x3 Step 3: Add the two products

d=v*t

d=distance

v=velocity(speed)

t=time

distance is equal to the product of velocity and time, equation form;

d=vt

Any questions please feel free to ask. Thanks!

3 0
3 years ago
The floor of the restaurant is a square of sides 28m. The floor is covered with square tiles of sides 40m. Calculate the total n
lisov135 [29]

Answer:

196000 tiles.

Step-by-step explanation:

The 28 m is 2800 cm. So a 28 x 28 meter square would be 2800 x 2800 centimeter square.

The tiles are 40 cm.

If you divide 40 from (2800 x 2800). (2800 x 2800) ÷40 = 196000

So 196000 tiles.

I'm not entirely sure if that's correct, but if it helps that's awesome.

4 0
2 years ago
Prove that $5^{3^n} + 1$ is divisible by $3^{n + 1}$ for all nonnegative integers $n.$
Viktor [21]

When n=0, we have

5^{3^0} + 1 = 5^1 + 1 = 6

3^{0 + 1} = 3^1 = 3

and of course 3 | 6. ("3 divides 6", in case the notation is unfamiliar.)

Suppose this is true for n=k, that

3^{k + 1} \mid 5^{3^k} + 1

Now for n=k+1, we have

5^{3^{k+1}} + 1 = 5^{3^k \times 3} + 1 \\\\ ~~~~~~~~~~~~~ = \left(5^{3^k}\right)^3 + 1^3 \\\\ ~~~~~~~~~~~~~ = \left(5^{3^k} + 1\right) \left(\left(5^{3^k}\right)^2 - 5^{3^k} + 1\right)

so we know the left side is at least divisible by 3^{k+1} by our assumption.

It remains to show that

3 \mid \left(5^{3^k}\right)^2 - 5^{3^k} + 1

which is easily done with Fermat's little theorem. It says

a^p \equiv a \pmod p

where p is prime and a is any integer. Then for any positive integer x,

5^3 \equiv 5 \pmod 3 \implies (5^3)^x \equiv 5^x \pmod 3

Furthermore,

5^{3^k} \equiv 5^{3\times3^{k-1}} \equiv \left(5^{3^{k-1}}\right)^3 \equiv 5^{3^{k-1}} \pmod 3

which goes all the way down to

5^{3^k} \equiv 5 \pmod 3

So, we find that

\left(5^{3^k}\right)^2 - 5^{3^k} + 1 \equiv 5^2 - 5 + 1 \equiv 21 \equiv 0 \pmod3

QED

5 0
1 year ago
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