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Stells [14]
3 years ago
13

Solve for x if z+4/5=11

Mathematics
1 answer:
Ede4ka [16]3 years ago
3 0

Solve for z. Isolate the z. Note the equal sign. What you do to one side, you do to the other side.

Subtract 4/5 from both sides

z + 4/5 (-4/5) = 11 (-4/5)

z = 10 5/5 - 4/5

Simplify

z = 10 1/5

10 1/5 is your answer for z

hope this helps

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6 0
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
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lara31 [8.8K]

Answer:

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Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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