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2 years ago

A line has a slope of 11 and passes through the point (2,4). PLSS HELPP I NEED IT

Mathematics

1 answer:

Ivenika [448]2 years ago

7 0

Answer:

y = 11x - 18

Step-by-step explanation:

Use the equation y = mx + b

m (the slope) = 11

x (the x-coordinate) = 2

y (the y-coordinate) = 4

Plug it in for our original equation:

4 = 11*2 + b

4 = 22 + b

4 - 22 = b

b = -18

Therefore, the answer is y = 11x - 18. The tricks to these problems are usually in the same format, when you are given the slope and a point it passes through, just plug in the slope and the x and y, and solve for b. Hope this helps!

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

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

A line has a slope of 11 and passes through the point (2,4). ​ PLSS HELPP I NEED IT

A line has a slope of 11 and passes through the point (2,4). PLSS HELPP I NEED IT