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2 years ago

What is the perimeter of the triangle below? Round to the nearest hundredth.

Mathematics

1 answer:

hoa [83]2 years ago

3 0

I think - 14 or 14 but I am not sure.

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Solve by completing the square:

laiz [17]

The answer is 4 trust me bro

8 0

2 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

2 years ago

What is the value of y

Ilya [14]

Answer: C

Step-by-step explanation: A straight angle is 180 degrees:

180-130 = 50

Triangle angles equal up to 180.

50 + 2y =180

y =65

5 0

2 years ago

Read 2 more answers

Ben is taking three AP classes and two regular classes. His AP classes count twice as much as his regular classes in his GPA. Ea

Naddika [18.5K]

Is it a trick question. What was his grades?

4 0

2 years ago

Whats the best answer

Volgvan

Answer:

$\displaystyle\sqrt[3]{2x}-6\sqrt[3]{x}$

Step-by-step explanation:

$7\sqrt[3]{2x}-3\sqrt[3]{16x}-3\sqrt[3]{8x}=7\sqrt[3]{2x}-3\sqrt[3]{8}\sqrt[3]{2x}-3\sqrt[3]{8}\sqrt[3]{x}\\\\=7\sqrt[3]{2x}-3\cdot 2\sqrt[3]{2x}-3\cdot 2\sqrt[3]{x}=(7-6)\sqrt[3]{2x}-6\sqrt[3]{x}\\\\=\sqrt[3]{2x}-6\sqrt[3]{x}$

8 0

2 years ago