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3 years ago

I need to know how to do this

Mathematics

2 answers:

Colt1911 [192]3 years ago

8 0

So 3+1=4 then you do the half you know 2 hlafs=whole so thats 1 and 4+1=5
5 is the answer

Lena [83]3 years ago

4 0

It is 5.
first, you add 3 and 1 and you get 4. next, you add 1/2 and 1/2 and you get 1.
add 4 and 1 and you get 5

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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

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

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

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3 years ago

9a- 24 >48 what is the solution

Artyom0805 [142]

Answer:

The solution to the given inequality is a > 8.

Step-by-step explanation:

For this problem, we have to solve for the inequality.

9a - 24 > 48

add 24 on both sides of the inequality.

9a > 72

Divide 9 on both sides of the inequality.

a > 8

So, your solution would be a > 8.

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