All categories

3 years ago

(50000)^2×(0.00002)^3\(200)×(0.0000001)

Mathematics

1 answer:

andre [41]3 years ago

8 0

Answer:

Step-by-step explanation:

concepts used is law of indices

where

(a^b)^c = a^bc

a^m/a^n = a^(m-n)

a^m *a^n = a^(m+n)

a^0 = 1

_______________________________________

50000)^2 = (5*10^4)^2 = 25*10^8

(0.00002)^3 = (2*10^-5)^3 = 8*10^-15

(0.0000001) = 10^-7

now

(50000)^2×(0.00002)^3\(200)×(0.0000001) = 25*10^8 * 8*10^-15/200*10^-7

= 200*10^8-15-(-7)/200 = 10^-7 +7 = 10^0 = 1

Thus,1 is the answer

You might be interested in

Anyone please help me to do 4 ) f

slavikrds [6]

I think your missing one which is G

7 0

3 years ago

What is the corrct answer

Mademuasel [1]

D is the right answer

4 0

3 years ago

Show that 3 · 4^n + 51 is divisible by 3 and 9 for all positive integers n.

-BARSIC- [3]

Answer:

Step-by-step explanation:

Hello, please consider the following.

$3\cdot 4^n+51=3\cdot 4^n+3\cdot 17=3(4^n+17)$

So this is divisible by 3.

Now, to prove that this is divisible by 9 = 3*3 we need to prove that

$4^n+17$ is divisible by 3. We will prove it by induction.

Step 1 - for n = 1

4+17=21= 3*7 this is true

Step 2 - we assume this is true for k so $4^k+17$ is divisible by 3

and we check what happens for k+1

$4^{k+1}+17=4\cdot 4^k+17=3\cdot 4^k + 4^k+17$

$3\cdot 4^k$ is divisible by 3 and

$4^k+17$ is divisible by 3, by induction hypothesis

So, the sum is divisible by 3.

Step 3 - Conclusion

We just prove that $4^n+17$ is divisible by 3 for all positive integers n.

Thanks

4 0

3 years ago

How do I solve this?

Anni [7]

I hope this helps you

4^3/3.x^2/3.y^3/3

4.y.x^2/3

3 0

3 years ago

how to find the area of each face of an equilateral triangular prism that has a surface area of 500 square units

KATRIN_1 [288]

Answer:Break that up

Step-by-step explanation:See how you can break up 500 to make it easier for yourself

7 0

3 years ago

Read 2 more answers