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3 years ago

What is 19 divided by 20 using long division & decimals?

Mathematics

1 answer:

andrew-mc [135]3 years ago

3 0

Answer:

0.95

(Don't have paper, because I accidentally threw away paper )

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Find the missing length

krek1111 [17]

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3 years ago

Expand the given power using the Binomial Theorem. (10k – m)5

agasfer [191]

Answer:

$(10k - m)^{5}=100000k-50000k^{4}m+10000k^{3}m^{2}-1000k^{2}m^{3}+50km^{4}-m^{5}$

Step-by-step explanation:

* Lets explain how to solve the problem

- The rule of expand the binomial is:

$(a+b)^{n}=(a)^{n}+nC1(a)^{n-1}(b)+nC2(a)^{n-2}(b)^{2}+nC3(a)^{n-3}(b)^{3}+...............+(b)^{5}$

∵ The binomial is $(10k-m)^{5}$

∴ a = 10k , b = -m and n = 5

∴ $(10k-m)^{5}=(10k)^{5}+5C1(10k)^{4}(-m)+5C2(10k)^{3}(-m)^{2}+5C3(10k)^{2}(-m)^{3}+5C4(10k)^{1} (-m)^{4}+5C5(10k)^{0}(-m)^{5}$

∵ 5C1 = 5

∵ 5C2 = 10

∵ 5C3 = 10

∵ 5C4 = 5

∵ 5C5 = 1

∴ $(10k-m)^{5}=100000k^{5}+(5)(10000)k^{4}(-m)+(10)(1000)k^{3}(m^{2})+(10)(100)k^{2}(-m^{3})+5(10k)^{1} (m^{4})+(10k)^{0}(-m^{5})$

∴ $(10k-m)^{5}=100000k^{5}-50000)k^{4}m+10000k^{3}m^{2}-1000k^{2}m^{3}+50km^{4}-m^{5}$

5 0

3 years ago

3 of the 30 computers are out of service. What percentage of the computers are working?

agasfer [191]

Answer:

90%

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

the absolute value of the sum of two different integers with the same sign is 8. Pat says there are three pairs of integers that

Irina18 [472]

Disagree
They say there are three pairs of integers that match, well..
1+7=8
2+6=8
3+5=8

-1+-7=-8
-2+-6=-8
-3+-5=-8
The same applies to negative integers because the absolute value of -3 and 3 is still 3. No matter the negative or not. So that gives us 6 pairs not 3.

4 0

3 years ago

-4x=16 what is x in the equation

adell [148]

Answer: x=-4

Step-by-step explanation:

-4x=16

Divide by -4 on each side to cancel it out

x=-4

4 0

3 years ago